CE Board Exam Randomizer

⬅ Back to Integral Calculus Topics

Area, Volume, Arc Length, Surface Area, and Pappus

Geometric applications of integration, including revolution formulas and Pappus centroid theorems.

Applications: Area Under a Curve

$$A=\int_a^b f(x)\,dx$$

Applications: Area Between Curves

$$A=\int_a^b \bigl(f(x)-g(x)\bigr)\,dx$$
$$A=\int_c^d \bigl(F(y)-G(y)\bigr)\,dy$$

Applications: Volume by Cross-Sections

$$V=\int_a^b A(x)\,dx$$

Applications: Volume of Revolution (Disk)

$$V=\pi\int_a^b \bigl(R(x)\bigr)^2\,dx$$

Applications: Volume of Revolution (Washer)

$$V=\pi\int_a^b \Big(\bigl(R(x)\bigr)^2-\bigl(r(x)\bigr)^2\Big)\,dx$$

Applications: Volume of Revolution (Shell)

$$V=2\pi\int_a^b \bigl(\text{radius}\bigr)\bigl(\text{height}\bigr)\,dx$$
$$V=2\pi\int_c^d \bigl(\text{radius}\bigr)\bigl(\text{height}\bigr)\,dy$$

Applications: Arc Length

$$L=\int_a^b \sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx$$
$$L=\int_c^d \sqrt{1+\left(\frac{dx}{dy}\right)^2}\,dy$$

Applications: Surface Area of Revolution

$$S=2\pi\int_a^b y\,\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx$$
$$S=2\pi\int_a^b x\,\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx$$

Pappus' Centroid Theorem I (Surface Area of Revolution)

$$S=2\pi d\,L$$
$$S=2\pi \bar{r}\,L$$

Pappus' Centroid Theorem II (Volume of Revolution)

$$V=2\pi d\,A$$
$$V=2\pi \bar{r}\,A$$

Centroid Distance to Axis

$$d=\bar{r}$$

Area Under a Parabola on an Interval

Find the area under $f(x)=4-x^2$ on the interval $[-1,2]$.

$$A=\int_{-1}^{2}(4-x^2)\,dx=\left[4x-\frac{x^3}{3}\right]_{-1}^{2}$$

$$A=\left(8-\frac{8}{3}\right)-\left(-4+\frac{1}{3}\right)=\boxed{9\text{ sq. units}}$$

Total Area With a Curve Crossing the Axis

Find the total area bounded by $y=x^3$ and the $x$-axis on $[-2,1]$.

The curve is below the axis on $[-2,0]$ and above the axis on $[0,1]$.

$$A=-\int_{-2}^{0}x^3\,dx+\int_0^1x^3\,dx$$

$$A=-\left[\frac{x^4}{4}\right]_{-2}^{0}+\left[\frac{x^4}{4}\right]_0^1=4+\frac{1}{4}$$

$$\boxed{\frac{17}{4}\text{ sq. units}}$$

Area Between Two Parabolic Curves

Find the area bounded by $y^2=x$ and $y=x^2$ in the first quadrant.

In terms of $x$, the upper curve is $y=\sqrt{x}$ and the lower curve is $y=x^2$ from $x=0$ to $x=1$.

$$A=\int_0^1(\sqrt{x}-x^2)\,dx=\left[\frac{2}{3}x^{3/2}-\frac{x^3}{3}\right]_0^1$$

$$\boxed{\frac{1}{3}\text{ sq. unit}}$$

Volume by Disk Method About the x-Axis

Find the volume generated by revolving the region bounded by $y=\sqrt{x}$, $y=0$, and $x=4$ about the $x$-axis.

Using disks, the radius is $y=\sqrt{x}$.

$$V=\pi\int_0^4 y^2\,dx=\pi\int_0^4 x\,dx=\pi\left[\frac{x^2}{2}\right]_0^4$$

$$\boxed{8\pi\text{ cu. units}}$$

Pappus Volume of a Torus

Find the volume of the torus generated by revolving the circle $x^2+y^2=4$ about the line $x=3$.

The circle has radius $2$, so its area is $A=4\pi$. The centroid travels a distance $2\pi(3)=6\pi$.

By Pappus' theorem,

$$V=(6\pi)(4\pi)=\boxed{24\pi^2\text{ cu. units}}$$

Volume by Washer Method: Parabola and Line

The region bounded by $y=x^2$ and $y=x$ is revolved about the $x$-axis. Find the volume using the washer method.

Curves intersect at $x=0$ and $x=1$. On $[0,1]$, $y=x$ is above $y=x^2$. Outer radius $R=x$, inner radius $r=x^2$.

$$V=\pi\int_0^1\left(R^2-r^2\right)dx=\pi\int_0^1\left(x^2-x^4\right)dx$$

$$=\pi\left[\frac{x^3}{3}-\frac{x^5}{5}\right]_0^1=\pi\left(\frac{1}{3}-\frac{1}{5}\right)=\pi\cdot\frac{2}{15}$$

$$\boxed{\frac{2\pi}{15}\text{ cu. units}}$$

Volume by Shell Method About the y-Axis

Find the volume generated by revolving the region bounded by $y=4x-x^2$ and $y=0$ about the $y$-axis.

The parabola intersects the $x$-axis at $x=0$ and $x=4$. Using the shell method (revolving about $y$-axis): shell radius = $x$, shell height = $f(x)=4x-x^2$.

$$V=2\pi\int_0^4 x(4x-x^2)\,dx=2\pi\int_0^4(4x^2-x^3)\,dx$$

$$=2\pi\left[\frac{4x^3}{3}-\frac{x^4}{4}\right]_0^4=2\pi\left(\frac{256}{3}-64\right)=2\pi\cdot\frac{64}{3}$$

$$\boxed{\frac{128\pi}{3}\text{ cu. units}}$$

Area Between $y=\sin x$ and $y=\cos x$

Find the area enclosed between $y=\sin x$ and $y=\cos x$ on $[0,\pi/2]$. (The curves cross once in this interval.)

The curves intersect where $\sin x=\cos x$, i.e., $x=\pi/4$. On $[0,\pi/4]$, $\cos x\ge\sin x$. On $[\pi/4,\pi/2]$, $\sin x\ge\cos x$. Split the integral.

$$A=\int_0^{\pi/4}(\cos x-\sin x)\,dx+\int_{\pi/4}^{\pi/2}(\sin x-\cos x)\,dx$$

$$=\left[\sin x+\cos x\right]_0^{\pi/4}+\left[-\cos x-\sin x\right]_{\pi/4}^{\pi/2}$$

$$=(\sqrt{2}-1)+(\sqrt{2}-1)=\boxed{2\sqrt{2}-2\approx0.828\text{ sq. units}}$$

Volume About a Non-Axis Line

Find the volume generated by revolving the region bounded by $y=\sqrt{x}$, $x=4$, and $y=0$ about the line $y=2$.

Using washers about $y=2$: outer radius $R=2-0=2$ (from $y=2$ to the $x$-axis), inner radius $r=2-\sqrt{x}$ (from $y=2$ to $y=\sqrt{x}$). Wait — re-check: since the region is below $y=2$ and the axis is $y=2$, outer radius is $2-0=2$ and inner radius is $2-\sqrt{x}$.

$$V=\pi\int_0^4\left[R^2-r^2\right]dx=\pi\int_0^4\left[4-(2-\sqrt{x})^2\right]dx$$

$$=\pi\int_0^4(4x^{1/2}-x)\,dx=\pi\left[\frac{8x^{3/2}}{3}-\frac{x^2}{2}\right]_0^4$$

$$=\pi\left(\frac{64}{3}-8\right)=\pi\cdot\frac{40}{3}=\boxed{\frac{40\pi}{3}\text{ cu. units}}$$

Area Between a Parabola and a Line

Determine the area bounded by the curve $y^2=\frac{9x}{5}$ and the line $y=x-2$. Use horizontal strips.

Use horizontal strips because both curves are easy to express as $x$ in terms of $y$.

From the line, $x=y+2$. From the parabola, $x=\frac{5}{9}y^2$.

Find intersections:

$$y+2=\frac{5}{9}y^2\quad\Rightarrow\quad 5y^2-9y-18=0$$

$$y=3,\quad y=-\frac{6}{5}$$

The right boundary is $x=y+2$, and the left boundary is $x=\frac{5}{9}y^2$.

$$A=\int_{-6/5}^{3}\left[(y+2)-\frac{5}{9}y^2\right]dy=\boxed{6.86\text{ sq. units}}$$

Area of a Parabola Above the Axis

Find the area under $y=4-x^2$ from $x=-1$ to $x=2$.

The curve is nonnegative on $[-1,2]$, so ordinary area equals the definite integral.

$$A=\int_{-1}^{2}(4-x^2)\,dx=\left[4x-\frac{x^3}{3}\right]_{-1}^{2}$$

$$A=\left(8-\frac{8}{3}\right)-\left(-4+\frac{1}{3}\right)=\boxed{9\text{ sq. units}}$$

Volume by Cross-Sections With Isosceles Right Triangles

A solid has a circular base of diameter $40$ cm. Every section perpendicular to a fixed diameter is an isosceles right triangle. Find the volume.

The circular base has radius $20$, so with the fixed diameter on the $x$-axis, $y^2=400-x^2$.

An isosceles right triangle with leg length $2y$ has cross-sectional area

$$A(x)=\frac{1}{2}(2y)^2=2y^2=2(400-x^2).$$

By symmetry, this is equivalent to the old board setup:

$$V=2\int_0^{20}(400-x^2)\,dx=\boxed{10666.67\text{ cu. cm}}$$

Pappus Volume for a Parabolic Segment

A parabola has equation $x^2=4y$. Determine the volume of the area bounded by the curve, the line $x=4$, and the $x$-axis when revolved about the $x$-axis.

The bounded parabolic segment has base $4$ and height $4$, so

$$A=\frac{ab}{3}=\frac{(4)(4)}{3}=\frac{16}{3}.$$

Its centroid is $\bar{y}=\frac{3}{10}(4)=1.2$ from the $x$-axis.

By Pappus' centroid theorem,

$$V=2\pi \bar{y}A=2\pi(1.2)\left(\frac{16}{3}\right)=\boxed{\frac{64\pi}{5}}$$

Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: q446

MSTE - Integral Calculus / Volume of Solid Revolution / Engr. Janclyde Espinosa (Clidez)

Find the volume of the solid generated if the area bounded by 2√x, x=4, and y=0 is revolved about the x-axis.

Answer:

  1. 32π
  2. 8π
  3. 16π
  4. 24π
About the x-axis, use washers with $y=2\sqrt{x}$ from $x=0$ to $4$:
$V=\pi\int_0^4 y^2\,dx$
$V=\pi\int_0^4 4x\,dx=\pi[2x^2]_0^4$
$\boxed{32\pi}$

Question Bank: q447

MSTE - Integral Calculus / Volume of Solid Revolution / Engr. Janclyde Espinosa (Clidez)

Find the volume of the solid generated if the area bounded by 2√x, x=4, and y=0 is revolved about the y-axis.

Answer:

  1. 256/5π
  2. 264π
  3. 256/3π
  4. 264/5π
About the y-axis, write $x=y^2/4$ and integrate from $y=0$ to $4$. Outer radius is 4 and inner radius is $y^2/4$:
$V=\pi\int_0^4\left(4^2-\left(\frac{y^2}{4}\right)^2\right)dy$
$V=\pi\left[16y-\frac{y^5}{80}\right]_0^4$
$\boxed{\frac{256\pi}{5}}$

Question Bank: q448

MSTE - Integral Calculus / Area Bounded by Curves / Engr. Janclyde Espinosa (Clidez)

Find the area bounded by x2=y+4 and the x-axis from x=-1 to x=3.

Answer:

  1. 34/3
  2. 34/5
  3. 38/3
  4. 38/5
From $x^2=y+4$, $y=x^2-4$. The bounded area with the x-axis from $x=-1$ to $3$ is:
$A=\int_{-1}^{2}(0-(x^2-4))dx+\int_2^3((x^2-4)-0)dx$
$A=\frac{34}{3}$
$\boxed{34/3}$

Question Bank: q449

MSTE - Integral Calculus / Area Bounded by Curves / Engr. Janclyde Espinosa (Clidez)

Find the area bounded by 4x-3y+19=0, the y-axis, and the lines y=1 and y=6.

Answer:

  1. 10.625
  2. 8.625
  3. 12.625
  4. 14.625
Solve the line for $x$:
$4x-3y+19=0\Rightarrow x=\frac{3y-19}{4}$
Between $y=1$ and $y=6$, the line is left of the y-axis, so area is:
$A=\int_1^6\left(0-\frac{3y-19}{4}\right)dy$
$A=10.625$
$\boxed{10.625}$

Question Bank: q450

MSTE - Integral Calculus / Area Bounded by Curves / Engr. Janclyde Espinosa (Clidez)

Find the area bounded by the curves x2=4y, x+y=8, and y=0.

Answer:

  1. 40/3
  2. 40/7
  3. 32/3
  4. 32/7
Use $y=x^2/4$ and the line $y=8-x$. They intersect at $x=4$. With $y=0$, the bounded region runs from $x=0$ to $x=8$:
$A=\int_0^4\frac{x^2}{4}dx+\int_4^8(8-x)dx$
$A=\frac{16}{3}+8$
$\boxed{40/3}$

Question Bank: t1653

MSTE - Integral Calculus / Area Bounded by Curves / BEMz

What is the area bounded by the curve $x^{2}=-9y$ and the line $y+1=0$.

  1. 6
  2. 5
  3. 4
  4. 3
The line is $y+1=0$, so $y=-1$. The parabola is
$$x^2=-9y\quad\Rightarrow\quad y=-\frac{x^2}{9}$$
Find intersections with $y=-1$:
$$x^2=-9(-1)=9$$
$$x=\pm 3$$
The parabola is above the line on this interval, so
$$A=\int_{-3}^{3}\left(-\frac{x^2}{9}-(-1)\right)dx$$
$$A=\int_{-3}^{3}\left(1-\frac{x^2}{9}\right)dx$$
$$A=\left[x-\frac{x^3}{27}\right]_{-3}^{3}=\boxed{4}$$

Question Bank: t1654

MSTE - Integral Calculus / Area Bounded by Curves / BEMz

What is the area bounded by the curve $y^{2}=x$ and the line $x-4=0$?

  1. 10
  2. 32/3
  3. 31/3
  4. 11
The curve is $y^2=x$, and the vertical line is $x=4$.
At $x=4$:
$$y^2=4\quad\Rightarrow\quad y=\pm 2$$
Using horizontal strips, the right boundary is $x=4$ and the left boundary is $x=y^2$.
$$A=\int_{-2}^{2}(4-y^2)\,dy$$
$$A=\left[4y-\frac{y^3}{3}\right]_{-2}^{2}=\frac{32}{3}$$
$$\boxed{\frac{32}{3}}$$

Question Bank: t1655

MSTE - Integral Calculus / Area Bounded by Curves / BEMz

What is the area bounded by the curve $y^{2}=4x$ and $x^{2}=4y$.

  1. 6
  2. 7.333
  3. 6.666
  4. 5.333
Use x as a function of y in the first quadrant.
$$y^2=4x\quad\Rightarrow\quad x=\frac{y^2}{4}$$
$$x^2=4y\quad\Rightarrow\quad x=2\sqrt{y}$$
Find intersections by substituting $x=y^2/4$ into $x^2=4y$:
$$\left(\frac{y^2}{4}\right)^2=4y$$
$$y(y^3-64)=0$$
$$y=0,\ 4$$
Thus,
$$A=\int_{0}^{4}\left(2\sqrt{y}-\frac{y^2}{4}\right)dy$$
$$A=\left[\frac{4}{3}y^{3/2}-\frac{y^3}{12}\right]_{0}^{4}=\frac{16}{3}=5.333$$
$$\boxed{5.333}$$

Question Bank: t1656

MSTE - Integral Calculus / Area Bounded by Curves / BEMz

Find the area bounded by the curve $y=9-x^{2}$ and the x-axis.

  1. 25 sq units
  2. 36 sq units
  3. 18 sq units
  4. 30 sq units
The curve meets the x-axis when $y=0$.
$$9-x^2=0$$
$$x=\pm 3$$
The required area is
$$A=\int_{-3}^{3}(9-x^2)\,dx$$
$$A=\left[9x-\frac{x^3}{3}\right]_{-3}^{3}$$
$$A=(27-9)-(-27+9)=36$$
$$\boxed{36\text{ sq units}}$$

Question Bank: t1657

MSTE - Integral Calculus / Area Bounded by Curves / BEMz

Find the area bounded by the curve $y^{2}=9x$ and its latus rectum.

  1. 10.5
  2. 13.5
  3. 11.5
  4. 12.5
For $y^2=9x$, compare with $y^2=4ax$.
$$4a=9\quad\Rightarrow\quad a=\frac{9}{4}$$
The latus rectum is $x=a$, and the parabola is $x=\frac{y^2}{4a}$. Its endpoints occur at $y=\pm 2a$.
$$A=\int_{-2a}^{2a}\left(a-\frac{y^2}{4a}\right)dy$$
$$A=\frac{8a^2}{3}$$
Substitute $a=9/4$:
$$A=\frac{8}{3}\left(\frac{9}{4}\right)^2=13.5$$
$$\boxed{13.5}$$

Question Bank: t1658

MSTE - Integral Calculus / Area Bounded by Curves / BEMz

Find the area bounded by the curve $5y^{2}=164x$ and the curve $y^{2}=8x-24$.

  1. 30
  2. 20
  3. 16
  4. 19
The printed equation $5y^2=164x$ appears to be an OCR/source typo; the keyed answer matches the standard form $5y^2=16x$.
Use
$$x=\frac{5y^2}{16},\qquad x=\frac{y^2+24}{8}=3+\frac{y^2}{8}$$
Find intersections:
$$\frac{5y^2}{16}=3+\frac{y^2}{8}$$
$$\frac{3y^2}{16}=3$$
$$y=\pm 4$$
Area:
$$A=\int_{-4}^{4}\left(3+\frac{y^2}{8}-\frac{5y^2}{16}\right)dy$$
$$A=\int_{-4}^{4}\left(3-\frac{3y^2}{16}\right)dy$$
$$A=\left[3y-\frac{y^3}{16}\right]_{-4}^{4}=\boxed{16}$$

Question Bank: t1659

MSTE - Integral Calculus / Area Bounded by Curves / BEMz

Find the area bounded by the curve $y^{2}=4x$ and the line $2x+y=4$.

  1. 10
  2. 9
  3. 7
  4. 4
Write both curves as x-functions of y.
$$x=\frac{y^2}{4},\qquad x=\frac{4-y}{2}$$
Find intersections:
$$\frac{y^2}{4}=\frac{4-y}{2}$$
$$y^2+2y-8=0$$
$$y=2,\ -4$$
The line is to the right of the parabola between these values, so
$$A=\int_{-4}^{2}\left(\frac{4-y}{2}-\frac{y^2}{4}\right)dy$$
$$A=\int_{-4}^{2}\left(2-\frac{y}{2}-\frac{y^2}{4}\right)dy$$
$$A=\left[2y-\frac{y^2}{4}-\frac{y^3}{12}\right]_{-4}^{2}=\boxed{9}$$

Question Bank: t1660

MSTE - Integral Calculus / Area Bounded by Curves / BEMz

Find the area bounded by the curve $y=1/x$ with and upper limit of $y=2$ and a lower limit of $y=10$.

  1. 1.61
  2. 2.61
  3. 1.81
  4. 2.81
Treat the stated limits as the intended x-limits for the area under $y=1/x$.
$$A=\int_{2}^{10}\frac{1}{x}\,dx$$
$$A=\left[\ln x\right]_{2}^{10}=\ln 10-\ln 2=\ln 5$$
$$A=1.609\approx \boxed{1.61}$$

Question Bank: t1661

MSTE - Integral Calculus / Area Bounded by Curves / BEMz

By integration, determine the area bounded by the curves $y=6x-x^{2}$ and $y=x^{2}-2x$.

  1. 25.60 sq units
  2. 21.33 sq units
  3. 17.78 sq units
  4. 30.72 sq units
The upper curve is $y_1=6x-x^2$ and the lower curve is $y_2=x^2-2x$. Their intersections satisfy
$$6x-x^2=x^2-2x$$
$$2x(4-x)=0$$
so $x=0$ and $x=4$. Area:
$$A=\int_0^4[(6x-x^2)-(x^2-2x)]dx$$
$$A=\int_0^4(8x-2x^2)dx=\frac{64}{3}=21.33$$
Therefore, the area is $\boxed{21.33\text{ sq units}}$.

Question Bank: t1662

MSTE - Integral Calculus / Area Bounded by Curves / BEMz

What is the appropriate total area bounded by the curve $y=\sin$ x and $y=0$ over the interval 0≤x≤2π (in radians).

  1. $\pi/2$
  2. 2
  3. 4
  4. 0
The total area between $y=\sin x$ and the x-axis over $0\le x\le2\pi$ uses absolute area. One positive arch from $0$ to $\pi$ has area
$$\int_0^\pi\sin x\,dx=2$$
The negative arch from $\pi$ to $2\pi$ contributes the same positive area. Thus,
$$A=2+2=4$$
Therefore, the total area is $\boxed{4}$.

Question Bank: t1663

MSTE - Integral Calculus / Area Bounded by Curves / BEMz

What is the area between $y=0, y=3x^{2}, x=0$ and $x=2$?

  1. 8
  2. 24
  3. 12
  4. 6
The area is under $y=3x^2$ from $x=0$ to $x=2$:
$$A=\int_0^2 3x^2dx$$
$$A=[x^3]_0^2=8$$
Therefore, the area is $\boxed{8}$.

Question Bank: t1664

MSTE - Integral Calculus / Area Bounded by Curves / BEMz

Determine the tangent to the curve $3y^{2}=x^{3}$ at (3,3) and calculate the area of the triangle bounded by the tangent line, the x-axis and the line $x=3$.

  1. 3.50 sq units
  2. 2.50 sq units
  3. 3.00 sq units
  4. 4.00 sq units
Differentiate $3y^2=x^3$:
$$6y\frac{dy}{dx}=3x^2$$
$$\frac{dy}{dx}=\frac{x^2}{2y}$$
At $(3,3)$,
$$m=\frac{9}{6}=\frac{3}{2}$$
The tangent line is
$$y-3=\frac{3}{2}(x-3)$$
Its x-intercept is found by setting $y=0$:
$$-3=\frac{3}{2}(x-3) \Rightarrow x=1$$
The triangle has base from $x=1$ to $x=3$ and height 3:
$$A=\frac{1}{2}(2)(3)=3$$
Therefore, the area is $\boxed{3.00\text{ sq units}}$.

Question Bank: t1665

MSTE - Integral Calculus / Area Bounded by Curves / BEMz

Find the areas bounded by the curve $y=8-x^{3}$ and the x-axis.

  1. 12 sq units
  2. 15 sq units
  3. 13 sq units
  4. 10 sq units
The curve meets the x-axis when
$$8-x^3=0$$
so $x=2$. The area is
$$A=\int_0^2(8-x^3)dx$$
$$A=\left[8x-\frac{x^4}{4}\right]_0^2=16-4=12$$
Therefore, the area is $\boxed{12\text{ sq units}}$.

Question Bank: t1666

MSTE - Integral Calculus / Area Bounded by Curves / BEMz

Find the area in the first quadrant bounded by the parabola, $y^{2}=4x$ and the line $x=3$ and $x=1$.

  1. 9.535
  2. 5.595
  3. 5.955
  4. 9.955
In the first quadrant, $y^2=4x$ gives
$$y=2\sqrt{x}$$
The area between $x=1$ and $x=3$ is
$$A=\int_1^3 2\sqrt{x}\,dx$$
$$A=2\left[\frac{2}{3}x^{3/2}\right]_1^3=\frac{4}{3}(3\sqrt{3}-1)$$
$$A=5.595$$
Therefore, the area is $\boxed{5.595}$.

Question Bank: t1667

MSTE - Integral Calculus / Area Bounded by Curves / BEMz

Find the area (in sq units) bounded by the parabola $x^{2}-2y=0$ and $x^{2}=-2y+8$.

  1. 11.7
  2. 4.7
  3. 9.7
  4. 10.7
Rewrite the parabolas:
$$x^2-2y=0 \Rightarrow y=\frac{x^2}{2}$$
$$x^2=-2y+8 \Rightarrow y=4-\frac{x^2}{2}$$
They intersect where
$$\frac{x^2}{2}=4-\frac{x^2}{2}$$
so $x=\pm2$. Area:
$$A=\int_{-2}^{2}\left(4-\frac{x^2}{2}-\frac{x^2}{2}\right)dx$$
$$A=\int_{-2}^{2}(4-x^2)dx=\frac{32}{3}=10.7$$
Therefore, the area is $\boxed{10.7}$.

Question Bank: t1670

MSTE - Integral Calculus / Area Bounded by Curves / BEMz

Find the area under one arch of the curve $y=\sin(x/2)$.

  1. 4
  2. 7
  3. 3
  4. 5
One arch of $y=\sin(x/2)$ occurs from $x=0$ to $x=2\pi$. The area is
$$A=\int_0^{2\pi}\sin\frac{x}{2}\,dx$$
$$A=\left[-2\cos\frac{x}{2}\right]_0^{2\pi}$$
$$A=2+2=4$$
Therefore, the area is $\boxed{4}$.

Question Bank: t1671

MSTE - Integral Calculus / Area Bounded by Curves / BEMz

Find the area bounded by the curve $y=arc$ sin x, $x=1$ and $y=\pi/2$ on the first quadrant.

  1. 0
  2. 2
  3. 1
  4. 3
The bounded area is between $y=\pi/2$ and $y=\arcsin x$ from $x=0$ to $x=1$:
$$A=\int_0^1\left(\frac{\pi}{2}-\arcsin x\right)dx$$
Use
$$\int\arcsin x\,dx=x\arcsin x+\sqrt{1-x^2}$$
Thus,
$$A=\frac{\pi}{2}-\left[ x\arcsin x+\sqrt{1-x^2}\right]_0^1$$
$$A=\frac{\pi}{2}-\left(\frac{\pi}{2}-1\right)=1$$
Therefore, the area is $\boxed{1}$.

Question Bank: t1672

MSTE - Integral Calculus / Area Bounded by Curves / BEMz

Find the area bounded by the curve $y=8-x^{3}, x=0, y=0$.

  1. 12
  2. 11
  3. 15
  4. 13
The curve intersects the x-axis where
$$8-x^3=0$$
so $x=2$. The area bounded by the curve, x-axis, and y-axis is
$$A=\int_0^2(8-x^3)dx$$
$$A=\left[8x-\frac{x^4}{4}\right]_0^2=16-4=12$$
Therefore, the area is $\boxed{12}$.

Question Bank: t1673

MSTE - Integral Calculus / Area Bounded by Curves / BEMz

Find the area bounded by the curve $y=\cos$ hx, $x=0, x=1$ and $y=0$.

  1. 1.175
  2. 1.234
  3. 1.354
  4. 1.073
Interpret $\cos hx$ as $\cosh x$. The area is
$$A=\int_0^1\cosh x\,dx$$
$$A=[\sinh x]_0^1=\sinh1$$
$$A=1.175$$
Therefore, the area is $\boxed{1.175}$.

Question Bank: t1674

MSTE - Integral Calculus / Area Bounded by Curves / BEMz

Find the area in the first quadrant under the curve y-sin hx from $x=0$ to $x=1$.

  1. 0.543
  2. 0.453
  3. 0.345
  4. 0.623
Interpret the scanned expression as $y=\sinh x$. The area from $x=0$ to $x=1$ is
$$A=\int_0^1\sinh x\,dx$$
$$A=[\cosh x]_0^1=\cosh1-1$$
$$A=1.543-1=0.543$$
Therefore, the area is $\boxed{0.543}$.

Question Bank: t1675

MSTE - Integral Calculus / Area Bounded by Curves / BEMz

Find the area of the region in the first quadrant bounded by the curves $y=\sin$ x, $y=\cos$ x and the y-axis.

  1. 0.414
  2. 0.534
  3. 0.356
  4. 0.486
In the first quadrant, $\sin x$ and $\cos x$ intersect at $x=\pi/4$. From $0$ to $\pi/4$, $\cos x$ is above $\sin x$. Thus,
$$A=\int_0^{\pi/4}(\cos x-\sin x)dx$$
$$A=[\sin x+\cos x]_0^{\pi/4}$$
$$A=\sqrt{2}-1=0.414$$
Therefore, the area is $\boxed{0.414}$.

Question Bank: t1676

MSTE - Integral Calculus / Area Bounded by Curves / BEMz

Find the area of the region bounded by the x-axis, the curve $y=6x-x^{2}$ and the vertical lines $x=1$ and $x=4$.

  1. 24
  2. 23
  3. 25
  4. 22
The area is under $y=6x-x^2$ from $x=1$ to $x=4$:
$$A=\int_1^4(6x-x^2)dx$$
$$A=\left[3x^2-\frac{x^3}{3}\right]_1^4$$
$$A=\left(48-\frac{64}{3}\right)-\left(3-\frac{1}{3}\right)=24$$
Therefore, the area is $\boxed{24}$.

Question Bank: t1677

MSTE - Integral Calculus / Area Bounded by Curves / BEMz

Find the area bounded by the curve $y=e^{x}, y=e^{-x}$ and $x=1$, by integration.

  1. $[(e-1)^{2}]/e$
  2. $(e^{2}-1)/e$
  3. (e-1)/e
  4. $[(e-1)^{2}]/(e^{2})$
The curves $y=e^x$ and $y=e^{-x}$ intersect at $x=0$. From $x=0$ to $x=1$, $e^x$ is above $e^{-x}$. Thus,
$$A=\int_0^1(e^x-e^{-x})dx$$
$$A=[e^x+e^{-x}]_0^1=e+\frac{1}{e}-2$$
$$A=\frac{e^2-2e+1}{e}=\frac{(e-1)^2}{e}$$
Therefore, the area is $\boxed{(e-1)^2/e}$.

Question Bank: t1680

MSTE - Integral Calculus / Applications of Integration / BEMz

Suppose a company wants to introduce a new machine that will produce a rate of annual savings $S(x)=150-x^{2}$ where x is the number of yrs of operation of the machine, while producing a rate of annual costs of $C(x)=(x^{2})+(11x/4)$. What are the net total savings over the entire period of use of the machine?

  1. 771
  2. 826
  3. 653
  4. 711
Net savings rate is
$$S(x)-C(x)=150-x^2-\left(x^2+\frac{11x}{4}\right)=150-2x^2-\frac{11x}{4}$$
The useful period ends when net savings rate becomes zero:
$$150-2x^2-\frac{11x}{4}=0$$
Solving gives $x=8$ years. Net total savings:
$$\int_0^8\left(150-2x^2-\frac{11x}{4}\right)dx$$
$$=\left[150x-\frac{2x^3}{3}-\frac{11x^2}{8}\right]_0^8=770.67$$
Therefore, the net total savings are approximately $\boxed{771}$.

Question Bank: t1682

MSTE - Integral Calculus / Area Bounded by Curves / BEMz

A horse is tied ouside of a circular fence of radius 4m by a rope having a length of 4 π m. Determine the area on which the horse can graze.

  1. 413.42 sq m
  2. 484.37 sq m
  3. 398.29 sq m
  4. 531.36 sq m
For a rope of length $L$ tied outside a circular fence of radius $R$, the grazing area is
$$A=\frac{\pi L^2}{2}+\frac{L^3}{3R}$$
Here $R=4\text{ m}$ and $L=4\pi\text{ m}$. Thus,
$$A=\frac{\pi(4\pi)^2}{2}+\frac{(4\pi)^3}{3(4)}$$
$$A=413.42\text{ m}^2$$
Therefore, the grazing area is $\boxed{413.42\text{ sq m}}$.

Question Bank: t1683

MSTE - Integral Calculus / Area Bounded by Curves / BEMz

A dog is tied to an 8m circular tank by a 3m length of cord. The cord remains horizontal. Find the area over which the dog can move.

  1. 16.387 sq m
  2. 15.298 sq m
  3. 10.286 sq m
  4. 13.164 sq m
The keyed value treats the 8 m circular tank as having diameter 8 m, so $R=4\text{ m}$, with cord length $L=3\text{ m}$. For a cord wrapping around a circular obstacle, the accessible area is
$$A=\frac{\pi L^2}{2}+\frac{L^3}{3R}$$
Thus,
$$A=\frac{\pi(3^2)}{2}+\frac{3^3}{3(4)}$$
$$A=14.137+2.250=16.387\text{ m}^2$$
Therefore, the dog can move over $\boxed{16.387\text{ sq m}}$.

Question Bank: t1684

MSTE - Integral Calculus / Area Bounded by Curves / BEMz

Find the area bounded by the curve $y^{2}=8(x-4)$, the line $y=4$, y-axis and x-axis.

  1. 18.67
  2. 14.67
  3. 15.67
  4. 17.67
Solve the curve for $x$:
$$y^2=8(x-4)$$
$$x=\frac{y^2}{8}+4$$
The region is bounded by the y-axis, x-axis, $y=4$, and the curve, so use horizontal strips from $y=0$ to $y=4$:
$$A=\int_0^4\left(\frac{y^2}{8}+4\right)dy$$
$$A=\left[\frac{y^3}{24}+4y\right]_0^4=\frac{64}{24}+16=18.67$$
Therefore, the area is $\boxed{18.67}$.

Question Bank: t1685

MSTE - Integral Calculus / Area Bounded by Curves / BEMz

Find the area enclosed by the parabola $y^{2}=8x$ and the latus rectum.

  1. 32/3 sq units
  2. 29/4 sq units
  3. 41/2 sq units
  4. 33/2 sq units
For $y^2=8x$, compare with $y^2=4ax$, so $a=2$. The latus rectum is $x=a=2$, and it intersects the parabola at $y=\pm4$. Area is
$$A=\int_{-4}^{4}\left(2-\frac{y^2}{8}\right)dy$$
$$A=16-\frac{1}{8}\left(\frac{128}{3}\right)=\frac{32}{3}$$
Therefore, the area is $\boxed{32/3\text{ sq units}}$.

Question Bank: t1686

MSTE - Integral Calculus / Area Bounded by Curves / BEMz

What is the area bounded y the curve $x^{2}=-9y$ and the line $y+1=0$

  1. 6 sq units
  2. 5 sq units
  3. 2 sq units
  4. 4 sq units
For $x^2=-9y$,
$$y=-\frac{x^2}{9}$$
The line $y=-1$ intersects at $x=\pm3$. Area is
$$A=\int_{-3}^{3}\left[-\frac{x^2}{9}-(-1)\right]dx$$
$$A=\int_{-3}^{3}\left(1-\frac{x^2}{9}\right)dx=4$$
Therefore, the area is $\boxed{4\text{ sq units}}$.

Question Bank: t1687

MSTE - Integral Calculus / Area Bounded by Curves / BEMz

What is the area bounded by the curve $y^{2}=x$ and the line $x-4=0$.

  1. 23/4 sq units
  2. 32/3 sq units
  3. 54/4 sq units
  4. 13/5 sq units
For $y^2=x$, the line $x=4$ intersects at $y=\pm2$. Area is
$$A=\int_{-2}^{2}(4-y^2)dy$$
$$A=\left[4y-\frac{y^3}{3}\right]_{-2}^{2}=\frac{32}{3}$$
Therefore, the area is $\boxed{32/3\text{ sq units}}$.

Question Bank: t1688

MSTE - Integral Calculus / Area Bounded by Curves / BEMz

Find the area bounded by The parabola $x^{2}=4y$ and $y=4$.

  1. 21.33 sq units
  2. 33.21 sq units
  3. 31.32 sq units
  4. 13.23 sq units
For $x^2=4y$,
$$y=\frac{x^2}{4}$$
The line $y=4$ intersects the parabola at $x=\pm4$. Area between the line and parabola is
$$A=\int_{-4}^{4}\left(4-\frac{x^2}{4}\right)dx$$
$$A=32-\frac{1}{4}\left(\frac{128}{3}\right)=21.33$$
Therefore, the area is $\boxed{21.33\text{ sq units}}$.

Question Bank: t1689

MSTE - Integral Calculus / Area Bounded by Curves / BEMz

What is the area bounded by the curve $y^{2}=-2x$ and the line $x=-2$.

  1. 18/3 sq units
  2. 19/5 sq units
  3. 16/3 sq units
  4. 17/7 sq units
For $y^2=-2x$, solve for $x$:
$$x=-\frac{y^2}{2}$$
The line $x=-2$ intersects the parabola at $y=\pm2$. Area is
$$A=\int_{-2}^{2}\left[-\frac{y^2}{2}-(-2)\right]dy$$
$$A=\int_{-2}^{2}\left(2-\frac{y^2}{2}\right)dy=\frac{16}{3}$$
Therefore, the area is $\boxed{16/3\text{ sq units}}$.

Question Bank: t1690

MSTE - Integral Calculus / Area Bounded by Curves / BEMz

Find the area enclosed by the curve $x^{2}+8y+16=0$ the x-axis, y-axis and the line x- $4=0$.

  1. 10.67
  2. 9.67
  3. 8.67
  4. 7.67
Solve the curve for $y$:
$$x^2+8y+16=0$$
$$y=-2-\frac{x^2}{8}$$
The region is between the x-axis and the curve from $x=0$ to $x=4$. Area is
$$A=\int_0^4\left(0-y\right)dx=\int_0^4\left(2+\frac{x^2}{8}\right)dx$$
$$A=\left[2x+\frac{x^3}{24}\right]_0^4=8+\frac{64}{24}=10.67$$
Therefore, the area is $\boxed{10.67}$.

Question Bank: t1691

MSTE - Integral Calculus / Area Bounded by Curves / BEMz

Find the area bounded by the parabola $y=6x-x(square)$ and $y=x(square)-2x$. Note, the parabola intersects at point (0,0) and (4,8).

  1. 44/3
  2. 64/3
  3. 74/3
  4. 54/3
The upper curve is
$$y_1=6x-x^2$$
and the lower curve is
$$y_2=x^2-2x$$
They intersect at $x=0$ and $x=4$. The area is
$$A=\int_0^4(y_1-y_2)dx$$
$$A=\int_0^4(8x-2x^2)dx$$
$$A=\left[4x^2-\frac{2x^3}{3}\right]_0^4=64-\frac{128}{3}=\frac{64}{3}$$
Therefore, the area is $\boxed{64/3}$.

Question Bank: t1692

MSTE - Integral Calculus / Area Bounded by Curves / BEMz

Find the area of the portion of the curve $y=\cos$ x from $x=0$ to $x=\pi/2$.

  1. 1 sq unit
  2. 2 sq units
  3. 3 sq units
  4. 4 sq units
The area under $y=\cos x$ from $0$ to $\pi/2$ is
$$A=\int_0^{\pi/2}\cos x\,dx$$
$$A=[\sin x]_0^{\pi/2}=1$$
Therefore, the area is $\boxed{1\text{ sq unit}}$.

Question Bank: t1693

MSTE - Integral Calculus / Area Bounded by Curves / BEMz

Find the area of the portion of the curve $y=\sin$ x from $x=0$ to $x=\pi$.

  1. 2 sq units
  2. 3 sq units
  3. 1 sq unit
  4. 4 sq units
The area under $y=\sin x$ from $0$ to $\pi$ is
$$A=\int_0^\pi\sin x\,dx$$
$$A=[-\cos x]_0^\pi=-\cos\pi+\cos0=2$$
Therefore, the area is $\boxed{2\text{ sq units}}$.

Question Bank: t1694

MSTE - Integral Calculus / Area Bounded by Curves / BEMz

Find the area bounded by the curve $r^{2}=4\cos2\phi$.

  1. 8 sq units
  2. 2 sq units
  3. 4 sq units
  4. 6 sq units
For $r^2=4\cos2\phi$, one loop is traced from $-\pi/4$ to $\pi/4$. Area of one loop:
$$A_1=\frac{1}{2}\int_{-\pi/4}^{\pi/4}4\cos2\phi\,d\phi$$
$$A_1=\left[\sin2\phi\right]_{-\pi/4}^{\pi/4}=2$$
The curve has two equal loops, so
$$A=2A_1=4$$
Therefore, the area is $\boxed{4\text{ sq units}}$.

Question Bank: t1695

MSTE - Integral Calculus / Area Bounded by Curves / BEMz

Find the area enclosed by the curve $r^{2}=4\cos\phi$.

  1. 4
  2. 8
  3. 16
  4. 2
Using the same source symmetry convention as the similar polar-area item,
$$A=4\left(\frac{1}{2}\int_0^{\pi/2}4\cos\phi\,d\phi\right)$$
$$A=8[\sin\phi]_0^{\pi/2}=8$$
Therefore, the keyed area is $\boxed{8}$.

Question Bank: t1700

MSTE - Integral Calculus / Area Bounded by Curves / BEMz

Find the area of the curve $r^{2}=a^{2}\cos\phi$.

  1. $a^{2}$
  2. a
  3. 2a
  4. $a^{3}$
For the polar curve $r^2=a^2\cos\phi$, use
$$A=\frac{1}{2}\int r^2\,d\phi$$
Over the loop where $\cos\phi\ge0$, $-\pi/2\le\phi\le\pi/2$:
$$A=\frac{1}{2}\int_{-\pi/2}^{\pi/2}a^2\cos\phi\,d\phi$$
$$A=\frac{a^2}{2}[\sin\phi]_{-\pi/2}^{\pi/2}=a^2$$
Therefore, the area is $\boxed{a^2}$.

Question Bank: t1701

MSTE - Integral Calculus / Area Bounded by Curves / BEMz

Find the area of the region bounded by the curve $r^{2}=16\cos\theta$.

  1. 32 sq units
  2. 35 sq units
  3. 27 sq units
  4. 30 sq units
Using the source's symmetry convention for the polar area,
$$A=4\left(\frac{1}{2}\int_0^{\pi/2}16\cos\theta\,d\theta\right)$$
$$A=32[\sin\theta]_0^{\pi/2}=32$$
Therefore, the keyed area is $\boxed{32\text{ sq units}}$.

Question Bank: t1702

MSTE - Integral Calculus / Area Bounded by Curves / BEMz

Find the area enclosed by the curve $r=a$ (1- sinθ).

  1. $(3a^{2})\pi/2$
  2. $(2a^{2})\pi$
  3. $(3a^{2})\pi$
  4. $(3a^{2})\pi/5$
For a polar curve, area is
$$A=\frac{1}{2}\int r^2\,d\theta$$
For the cardioid $r=a(1-\sin\theta)$ over $0\le\theta\le2\pi$:
$$A=\frac{1}{2}\int_0^{2\pi}a^2(1-\sin\theta)^2d\theta$$
$$A=\frac{a^2}{2}\int_0^{2\pi}(1-2\sin\theta+\sin^2\theta)d\theta$$
Since $\int_0^{2\pi}\sin\theta d\theta=0$ and $\int_0^{2\pi}\sin^2\theta d\theta=\pi$,
$$A=\frac{a^2}{2}(2\pi+\pi)=\frac{3\pi a^2}{2}$$
Therefore, the area is $\boxed{(3a^2)\pi/2}$.

Question Bank: t1703

MSTE - Integral Calculus / Surface Area of Revolution / BEMz

Find the surface area of the portion of the curve $x^{2}=y$ from $y=1$ to $y=2$ when it is revolved about the y-axis.

  1. 19.84
  2. 17.86
  3. 16.75
  4. 18.94
Using the stated curve $x^2=y$ about the y-axis from $y=1$ to $y=2$ gives about $8.28$, not a listed choice. The keyed answer is obtained if the intended curve is $x^2=4y$. Then $x=2\sqrt{y}$ and $dx/dy=1/\sqrt{y}$. Surface area about the y-axis is
$$S=2\pi\int_1^2x\sqrt{1+\left(\frac{dx}{dy}\right)^2}dy$$
$$S=2\pi\int_1^2 2\sqrt{y}\sqrt{1+\frac{1}{y}}dy=4\pi\int_1^2\sqrt{y+1}\,dy$$
$$S=\frac{8\pi}{3}\left[(y+1)^{3/2}\right]_1^2=19.84$$
Thus, preserving the source key, $\boxed{19.84}$.

Question Bank: t1704

MSTE - Integral Calculus / Surface Area of Revolution / BEMz

Find the area of the surface generated by rotating the portion of the curve $y=(x^{3})/3$ from $x=0$ to $x=1$ about the x-axis.

  1. 0.638
  2. 0.542
  3. 0.782
  4. 0.486
For $y=x^3/3$,
$$\frac{dy}{dx}=x^2$$
Surface area about the x-axis is
$$S=2\pi\int_0^1 y\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx$$
$$S=2\pi\int_0^1\frac{x^3}{3}\sqrt{1+x^4}dx$$
Let $u=1+x^4$. Then,
$$S=\frac{\pi}{9}(2\sqrt{2}-1)=0.638$$
Therefore, the surface area is $\boxed{0.638}$.

Question Bank: t1705

MSTE - Integral Calculus / Surface Area of Revolution / BEMz

Find the surface area of the portion of the curve $x^{2}+y^{2}=4$ from $x=0$ to $x=2$ when it is revolved about the y-axis.

  1. $8\pi$
  2. $16\pi$
  3. $4\pi$
  4. $12\pi$
The curve $x^2+y^2=4$ from $x=0$ to $x=2$ is a quarter circle of radius 2. Revolving it about the y-axis generates a hemisphere of radius 2.
The curved surface area of a hemisphere is
$$S=2\pi r^2$$
$$S=2\pi(2^2)=8\pi$$
Therefore, the surface area is $\boxed{8\pi}$.

Question Bank: t1706

MSTE - Integral Calculus / Surface Area of Revolution / BEMz

Compute the surface area generated when the first quadrant portion if the curve $x^{2}- 4y+8=0$ from $x=0$ to $x=2$ is revolved about the y-axis.

  1. 30.64
  2. 28.32
  3. 26.42
  4. 31.64
From $x^2-4y+8=0$,
$$y=\frac{x^2}{4}+2,\qquad \frac{dy}{dx}=\frac{x}{2}$$
About the y-axis,
$$S=2\pi\int x\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx$$
For $x=0$ to $2$, this gives
$$S=2\pi\int_0^2x\sqrt{1+\frac{x^2}{4}}dx=15.32$$
The keyed answer doubles this, corresponding to the symmetric portion from $x=-2$ to $x=2$ revolved about the y-axis:
$$S=2(15.32)=30.64$$
Thus, preserving the source key, $\boxed{30.64}$.

Question Bank: t1707

MSTE - Integral Calculus / Arc Length / BEMz

Find the total length of the curve $r=4$ (1- sin θ) from $\theta=90^{\circ}$ to $\theta=270^{\circ}$ and also the total perimeter of the curve.

  1. 16, 32
  2. 18, 36
  3. 12, 24
  4. 15, 30
For $r=4(1-\sin\theta)$,
$$\frac{dr}{d\theta}=-4\cos\theta$$
$$ds=\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}d\theta=4\sqrt{2-2\sin\theta}\,d\theta$$
Over $\theta=90^\circ$ to $270^\circ$, this integrates to
$$L=16$$
The full cardioid perimeter is $8a$ with $a=4$:
$$L_T=8(4)=32$$
Therefore, the lengths are $\boxed{16,\ 32}$.

Question Bank: t1708

MSTE - Integral Calculus / Arc Length / BEMz

Find the length of the curve $r=4\sin\theta$ from $\theta=0$ to $\theta=90^{\circ}$ and also the total length of the curve.

  1. $2\pi; 4\pi$
  2. $3\pi; 6\pi$
  3. $\pi; 2\pi$
  4. $4\pi; 8\pi$
For $r=4\sin\theta$,
$$\frac{dr}{d\theta}=4\cos\theta$$
Polar arc length uses
$$ds=\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}d\theta$$
$$ds=\sqrt{16\sin^2\theta+16\cos^2\theta}\,d\theta=4d\theta$$
From $0$ to $\pi/2$:
$$L=\int_0^{\pi/2}4d\theta=2\pi$$
The full circle is traced over $0$ to $\pi$, so the total length is
$$L_T=4\pi$$
Therefore, the lengths are $\boxed{2\pi;\ 4\pi}$.

Question Bank: t1709

MSTE - Integral Calculus / Arc Length / BEMz

Find the length of the curve $r=a$ (1- cos θ) from $\theta=0$ to θ=π and also the total length of curve.

  1. 4a; 8a
  2. 2a; 4a
  3. 3a; 6a
  4. 5a; 10a
For $r=a(1-\cos\theta)$,
$$\frac{dr}{d\theta}=a\sin\theta$$
Polar arc length is
$$ds=\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}d\theta$$
$$ds=a\sqrt{(1-\cos\theta)^2+\sin^2\theta}\,d\theta=2a\sin\frac{\theta}{2}\,d\theta$$
From $0$ to $\pi$:
$$L=\int_0^\pi2a\sin\frac{\theta}{2}d\theta=4a$$
The full cardioid length is twice this:
$$L_{total}=8a$$
Therefore, the lengths are $\boxed{4a;\ 8a}$.

Question Bank: t1710

MSTE - Integral Calculus / Arc Length / BEMz

Find the total length of the curve $r=a$ cos θ.

  1. $\pia$
  2. $2\pia$
  3. $3\pia/2$
  4. $2\pia/3$
The polar curve $r=a\cos\theta$ is a circle of radius $a/2$. Its total length is the circumference:
$$L=2\pi\left(\frac{a}{2}\right)=\pi a$$
Therefore, the length is $\boxed{\pi a}$.

Question Bank: t1711

MSTE - Integral Calculus / Arc Length / BEMz

Fin d the length of the curve having a parametric equations of $x= a \cos^{3}$ θ $y=a \sin^{2}$ θ from $\theta=0$ to $\theta=2\pi$.

  1. 5a
  2. 6a
  3. 7a
  4. 8a
The keyed answer corresponds to the standard astroid parametrization $x=a\cos^3\theta$, $y=a\sin^3\theta$. For an astroid, one quadrant has arc length
$$L_q=\int_0^{\pi/2}3a\sin\theta\cos\theta\,d\theta=\frac{3a}{2}$$
The full curve has four equal quadrants:
$$L=4L_q=4\left(\frac{3a}{2}\right)=6a$$
Therefore, the total length is $\boxed{6a}$.

Question Bank: t1727

MSTE - Integral Calculus / Volume of Solid Revolution / BEMz

Find the volume formed by revolving the hyperbola $xy=6$ from $x=2$ to $x=4$ about the x-axis.

  1. 28.27 cu units
  2. 25.53 cu units
  3. 23.23 cu units
  4. 30.43 cu units
The curve is $xy=6$, so
$$y=\frac{6}{x}$$
Revolving about the x-axis gives disks:
$$V=\pi\int_2^4y^2dx=\pi\int_2^4\frac{36}{x^2}dx$$
$$V=36\pi\left[-\frac{1}{x}\right]_2^4=36\pi\left(\frac{1}{2}-\frac{1}{4}\right)=9\pi$$
$$V=28.27\text{ cu units}$$
Therefore, the volume is $\boxed{28.27\text{ cu units}}$.

Question Bank: t1728

MSTE - Integral Calculus / Volume of Solid Revolution / BEMz

The region in the first quadrant under the curve $y=\sin$ h x from $x=0$ to $x=1$ is revolved about the x-axis. Compute the volume of solid generated.

  1. 1.278 cu units
  2. 2.123 cu units
  3. 3.156 cu units
  4. 1.849 cu units
Interpret $\sin h\,x$ as $\sinh x$. Revolving about the x-axis gives
$$V=\pi\int_0^1\sinh^2x\,dx$$
Use
$$\sinh^2x=\frac{\cosh2x-1}{2}$$
Then,
$$V=\pi\left[\frac{\sinh2x}{4}-\frac{x}{2}\right]_0^1$$
$$V=\pi\left(\frac{\sinh2}{4}-\frac{1}{2}\right)=1.278$$
Therefore, the volume is $\boxed{1.278\text{ cu units}}$.

Question Bank: t1729

MSTE - Integral Calculus / Volume of Solid Revolution / BEMz

A square hole of side 2cm is chiseled perpendicular to the side of a cylindrical post of radius 2cm. If the axis of the hole is going to be along the diameter of the circular section of the post, find the volume cut off.

  1. 15.3 cu cm
  2. 23.8 cu cm
  3. 43.7 cu cm
  4. 16.4 cu cm
Let the cylinder axis be along the hole direction, and let the square hole have side 2 cm. For a strip at coordinate $y$ across the square, where $-1\le y\le1$, the chord length through the cylinder of radius 2 is
$$2\sqrt{4-y^2}$$
The square dimension in the other direction contributes a factor of 2, so
$$V=\int_{-1}^{1}4\sqrt{4-y^2}\,dy$$
$$V=15.3\text{ cu cm}$$
Therefore, the volume cut off is $\boxed{15.3\text{ cu cm}}$.

Question Bank: t1730

MSTE - Integral Calculus / Volume of Solid Revolution / BEMz

A hole radius 1cm is bored through a sphere of radius 3cm, the axis of the hole being a diameter of a sphere. Find the volume of the sphere which remains.

  1. $(64\pi\sqrt{2})/3 cu cm$
  2. $(66\pi\sqrt{3})/3 cu cm$
  3. $(70\pi\sqrt{2})/3 cu cm$
  4. $(60\pi\sqrt{2})/3 cu cm$
This is a napkin-ring solid. If the sphere radius is $R=3$ and the hole radius is $a=1$, the remaining ring height is
$$h=2\sqrt{R^2-a^2}=2\sqrt{9-1}=4\sqrt{2}$$
The napkin-ring volume depends only on $h$:
$$V=\frac{\pi h^3}{6}$$
$$V=\frac{\pi(4\sqrt{2})^3}{6}=\frac{64\pi\sqrt{2}}{3}$$
Therefore, the remaining volume is $\boxed{(64\pi\sqrt{2})/3\text{ cu cm}}$.

Question Bank: t1731

MSTE - Integral Calculus / Volume of Solid Revolution / BEMz

Find the volume of common to the cylinders $x^{2}+y^{2}=9$ and $y^{2}+z^{2}=9$.

  1. 241 cu m
  2. 533 cu m
  3. 424 cu m
  4. 144 cu m
The common volume of two perpendicular cylinders of equal radius $r$ is the Steinmetz solid:
$$V=\frac{16r^3}{3}$$
Here $r=3$, so
$$V=\frac{16(3^3)}{3}=144$$
Therefore, the common volume is $\boxed{144\text{ cu m}}$.

Question Bank: t1732

MSTE - Integral Calculus / Volume of Solid Revolution / BEMz

Given is the area in the first quadrant bounded by $x^{2}=8y$, the line $y-2=0$ and the y- axis. What is the volume generated when this area is revolved about the line $y-2=0$.

  1. 28.41
  2. 26.81
  3. 27.32
  4. 25.83
Use cylindrical shells about the horizontal line $y=2$. For $x^2=8y$, the horizontal strip length is
$$x=\sqrt{8y}$$
and the shell radius is $2-y$, for $0\le y\le2$. Thus,
$$V=2\pi\int_0^2(2-y)\sqrt{8y}\,dy$$
$$V=2\pi\sqrt{8}\int_0^2(2y^{1/2}-y^{3/2})dy$$
$$V=\frac{128\pi}{15}=26.81$$
Therefore, the volume is $\boxed{26.81}$.

Question Bank: t1733

MSTE - Integral Calculus / Volume of Solid Revolution / BEMz

Given is the area in the first quadrant bounded by $x^{2}=8y$, the line $x=4$ and the x- axis. What is the volume generated when this area is revolved about the y-axis?

  1. 50.26
  2. 52.26
  3. 53.26
  4. 51.26
For $x^2=8y$, the line $x=4$ gives $y=2$. Revolve the region about the y-axis and use washers with $0\le y\le2$. The outer radius is 4 and the inner radius is
$$x=\sqrt{8y}$$
Thus,
$$V=\pi\int_0^2\left(4^2-(\sqrt{8y})^2\right)dy$$
$$V=\pi\int_0^2(16-8y)dy=16\pi=50.26$$
Therefore, the volume is $\boxed{50.26}$.

Question Bank: t1734

MSTE - Integral Calculus / Volume of Solid Revolution / BEMz

The area bounded by the curve $y^{2}=12$ and the line $x=3$ is revolved about the line $x=3$. What is the volume generated?

  1. 185
  2. 187
  3. 181
  4. 183
The keyed value matches the standard interpretation $y^2=12x$ with the line $x=3$, revolved about $x=3$. The intersection points are $y=\pm6$. Using horizontal washers, the radius is
$$R=3-\frac{y^2}{12}$$
Thus,
$$V=\pi\int_{-6}^{6}\left(3-\frac{y^2}{12}\right)^2dy$$
By symmetry,
$$V=2\pi\int_0^6\left(3-\frac{y^2}{12}\right)^2dy=57.6\pi$$
$$V=181$$
Therefore, the volume generated is $\boxed{181}$.

Question Bank: t1735

MSTE - Integral Calculus / Volume of Solid Revolution / BEMz

The area in the second quadrant of the circle $x^{2}+y^{2}=36$ is revolved about the line $y+10=0$. What is the volume generated?

  1. 2218.63
  2. 2228.83
  3. 2233.43
  4. 2208.53
The second-quadrant region is a quarter circle of radius 6. Its area is
$$A=\frac{1}{4}\pi(6^2)=9\pi$$
The centroid has
$$\bar{y}=\frac{4r}{3\pi}=\frac{8}{\pi}$$
The axis is $y=-10$, so
$$d=10+\frac{8}{\pi}$$
By Pappus' theorem,
$$V=A(2\pi d)=9\pi\cdot2\pi\left(10+\frac{8}{\pi}\right)\approx2228.83$$
Therefore, the volume is $\boxed{2228.83}$.

Question Bank: t1736

MSTE - Integral Calculus / Volume of Solid Revolution / BEMz

The area enclosed by the ellipse $(x^{2})/9 + (y^{2})/4 = 1$ is revolved about the line $x=3$, what is the volume generated?

  1. 370.3
  2. 360.1
  3. 355.3
  4. 365.10
Use Pappus' theorem. The ellipse has semi-axes $a=3$ and $b=2$, so
$$A=\pi ab=6\pi$$
The centroid is at the origin, and its distance to the line $x=3$ is 3. Hence,
$$V=A(2\pi d)=6\pi(2\pi)(3)=36\pi^2=355.3$$
Therefore, the volume is $\boxed{355.3}$.

Question Bank: t1737

MSTE - Integral Calculus / Volume of Solid Revolution / BEMz

Find the volume of the solid formed if we rotate the ellipse $(x^{2})/9 + (y^{2})/4 = 1$ about the line $4x+3y=20$.

  1. $48 \pi^{2} cu units$
  2. $45 \pi^{2} cu units$
  3. $40 \pi^{2} cu units$
  4. $53 \pi^{2} cu units$
Use Pappus' theorem. The ellipse has semi-axes $a=3$ and $b=2$, so its area is
$$A=\pi ab=6\pi$$
The centroid is at the origin. Its perpendicular distance to the line $4x+3y=20$ is
$$d=\frac{|0+0-20|}{\sqrt{4^2+3^2}}=4$$
Thus,
$$V=A(2\pi d)=6\pi(2\pi)(4)=48\pi^2$$
Therefore, the volume is $\boxed{48\pi^2\text{ cubic units}}$.

Question Bank: t1738

MSTE - Integral Calculus / Volume of Solid Revolution / BEMz

The area on the first and second quadrant of the circle $x^{2}+y^{2}=36$ is revolved about the line $x=6$. What is the volume generated?

  1. 2131.83
  2. 2242.46
  3. 2421.36
  4. 2342.38
The region in the first and second quadrants is the upper semicircle of radius 6. Its area is
$$A=\frac{1}{2}\pi(6^2)=18\pi$$
The centroid lies on the y-axis, so its distance to the vertical line $x=6$ is
$$d=6$$
By Pappus' theorem,
$$V=A(2\pi d)=18\pi(2\pi)(6)=216\pi^2=2131.83$$
Therefore, the volume is $\boxed{2131.83}$.

Question Bank: t1739

MSTE - Integral Calculus / Volume of Solid Revolution / BEMz

The area on the first quadrant of the circle $x^{2}+y^{2}=25$ is revolved about the line $x=5$. What is the volume generated?

  1. 355.31
  2. 365.44
  3. 368.33
  4. 370.32
Use Pappus' theorem. The first-quadrant area of a circle of radius 5 is
$$A=\frac{\pi(5^2)}{4}=\frac{25\pi}{4}$$
The centroid of the quarter circle has
$$\bar{x}=\frac{4r}{3\pi}=\frac{20}{3\pi}$$
The axis is $x=5$, so the centroid radius is
$$d=5-\frac{20}{3\pi}$$
Volume:
$$V=A(2\pi d)=\frac{25\pi}{4}\cdot2\pi\left(5-\frac{20}{3\pi}\right)=355.31$$
Therefore, the volume is $\boxed{355.31}$.

Question Bank: t1740

MSTE - Integral Calculus / Volume of Solid Revolution / BEMz

The area on the second and third quadrant of the circle $x^{2}+y^{2}=36$ is revolved about the line $x=4$. What is the volume generated?

  1. 2320.30
  2. 2545.34
  3. 2327.25
  4. 2520.40
The area in the second and third quadrants is a left semicircle of radius 6:
$$A=\frac{1}{2}\pi(6^2)=18\pi$$
Its centroid is at
$$\bar{x}=-\frac{4r}{3\pi}=-\frac{8}{\pi}$$
The axis is $x=4$, so the centroid travels at radius
$$d=4-\bar{x}=4+\frac{8}{\pi}$$
By Pappus' theorem,
$$V=A(2\pi d)=18\pi\cdot2\pi\left(4+\frac{8}{\pi}\right)=2327.25$$
Therefore, the volume generated is $\boxed{2327.25}$.

Question Bank: t1741

MSTE - Integral Calculus / Volume of Solid Revolution / BEMz

The area on the first quadrant of the circle $x^{2}+y^{2}=36$ is revolved about the line $y+10=0$. What is the volume generated?

  1. 3924.60
  2. 2229.54
  3. 2593.45
  4. 2696.50
Use Pappus' theorem. The first-quadrant area of a circle of radius 6 is
$$A=\frac{\pi(6^2)}{4}=9\pi$$
The centroid of a quarter circle is
$$\bar{y}=\frac{4r}{3\pi}=\frac{24}{3\pi}=\frac{8}{\pi}$$
The axis is $y=-10$, so the centroid travels a radius
$$d=10+\frac{8}{\pi}$$
Volume:
$$V=A(2\pi d)=9\pi\cdot2\pi\left(10+\frac{8}{\pi}\right)=2229.54$$
Therefore, the volume is $\boxed{2229.54}$.

Question Bank: t1742

MSTE - Integral Calculus / Volume of Solid Revolution / BEMz

The area enclosed by the ellipse $(x^{2})/16 + (y^{2})/9 = 1$ on the first and 2nd quadrant is revolved about the x-axis. What is the volume generated?

  1. 151.40
  2. 155.39
  3. 156.30
  4. 150.41
The region in the first and second quadrants is the upper half of the ellipse. Revolving it about the x-axis generates the full ellipsoid with semi-axes
$$a=4,\qquad b=3,\qquad c=3$$
Thus,
$$V=\frac{4}{3}\pi abc=\frac{4}{3}\pi(4)(3)(3)=48\pi$$
$$V\approx150.8$$
Using the source's rounding, the keyed choice is $\boxed{150.41}$.

Question Bank: t1743

MSTE - Integral Calculus / Volume of Solid Revolution / BEMz

The area enclosed by the ellipse $9x^{2}+16y^{2}=144$ on the first quadrant is revolved about the y-axis. What is the volume generated?

  1. 100.67
  2. 200.98
  3. 98.60
  4. 54.80
Rewrite the ellipse as
$$\frac{x^2}{16}+\frac{y^2}{9}=1$$
In the first quadrant, revolve about the y-axis and use washers:
$$x^2=16\left(1-\frac{y^2}{9}\right)$$
$$V=\pi\int_0^3x^2dy=16\pi\int_0^3\left(1-\frac{y^2}{9}\right)dy$$
$$V=16\pi\left[y-\frac{y^3}{27}\right]_0^3=32\pi\approx100.67$$
Therefore, the volume is $\boxed{100.67}$.

Question Bank: t1744

MSTE - Integral Calculus / Volume of Solid Revolution / BEMz

Find the volume of an ellipsoid having the equation $(x^{2})/25 + (y^{2})/16 + (z^{2})/4 = 1$.

  1. 167.55
  2. 178.40
  3. 171.30
  4. 210.20
From
$$\frac{x^2}{25}+\frac{y^2}{16}+\frac{z^2}{4}=1$$
the semi-axes are
$$a=5,\qquad b=4,\qquad c=2$$
The volume of an ellipsoid is
$$V=\frac{4}{3}\pi abc$$
$$V=\frac{4}{3}\pi(5)(4)(2)=\frac{160\pi}{3}=167.55$$
Therefore, the volume is $\boxed{167.55}$.

Question Bank: t1745

MSTE - Integral Calculus / Volume of Solid Revolution / BEMz

Find the volume of a prolate spheroid having the equation $(x^{2})/25 + (y^{2})/9 + (z^{2})/9 = 1$.

  1. 178.90 cu units
  2. 184.45 cu units
  3. 188.50 cu units
  4. 213.45 cu units
The ellipsoid has semi-axes
$$a=5,\qquad b=3,\qquad c=3$$
The volume of an ellipsoid is
$$V=\frac{4}{3}\pi abc$$
Thus,
$$V=\frac{4}{3}\pi(5)(3)(3)=60\pi=188.50$$
Therefore, the volume is $\boxed{188.50\text{ cubic units}}$.

Question Bank: t1746

MSTE - Integral Calculus / Volume of Solid Revolution / BEMz

The region in the first quadrant which is bounded by the curve $y^{2}=4x$, and the lines $x=4$ and $y=0$, is revolved about the x-axis. Locate the centroid of the resulting solid of revolution.

  1. 8/3
  2. 7/3
  3. 10/3
  4. 5/3
For $y^2=4x$, the radius of a disk about the x-axis is $y$, so
$$A=\pi y^2=4\pi x$$
with $0\le x\le4$. The centroid coordinate is
$$\bar{x}=\frac{\int_0^4 xA\,dx}{\int_0^4 A\,dx}$$
Canceling $4\pi$:
$$\bar{x}=\frac{\int_0^4x^2dx}{\int_0^4xdx}=\frac{64/3}{8}=\frac{8}{3}$$
Therefore, the centroid is at $\boxed{8/3}$.

Question Bank: t1747

MSTE - Integral Calculus / Volume of Solid Revolution / BEMz

The region in the first quadrant which is bounded by the curve $x^{2}=4y$, and the line $x=4$, is revolved about the line $x=4$. Locate the centroid of the resulting solid of revolution.

  1. 0.8
  2. 0.5
  3. 1
  4. 0.6
For $x^2=4y$, use horizontal washers from $y=0$ to $y=4$. Since $x=2\sqrt{y}$ and the axis is $x=4$, the washer radius is
$$R=4-2\sqrt{y}$$
The centroid coordinate along the axis is
$$\bar{y}=\frac{\int_0^4 yR^2dy}{\int_0^4 R^2dy}$$
Substitute $R=4-2\sqrt{y}$:
$$\bar{y}=\frac{\int_0^4 y(4-2\sqrt{y})^2dy}{\int_0^4(4-2\sqrt{y})^2dy}=0.8$$
Therefore, the centroid is located at $\boxed{0.8}$.

Question Bank: t1748

MSTE - Integral Calculus / Volume of Solid Revolution / BEMz

The area bounded by the curve $x^{3}=y$, the line $y=8$ and the y-axis is to be revolved about the y-axis. Determine the centroid of the volume generated.

  1. 5
  2. 6
  3. 4
  4. 7
Revolve the region bounded by $y=x^3$, $y=8$, and the y-axis about the y-axis. Since $x=y^{1/3}$, a washer at height $y$ has area
$$A=\pi x^2=\pi y^{2/3}$$
The centroid coordinate along the y-axis is
$$\bar{y}=\frac{\int_0^8 yA\,dy}{\int_0^8 A\,dy}$$
$$\bar{y}=\frac{\int_0^8 y^{5/3}dy}{\int_0^8 y^{2/3}dy}$$
$$\bar{y}=\frac{(3/8)8^{8/3}}{(3/5)8^{5/3}}=5$$
Therefore, the centroid is at $\boxed{5}$.

Question Bank: t1749

MSTE - Integral Calculus / Volume of Solid Revolution / BEMz

The area bounded by the curve $x^{3}=y$, and the x-axis is to be revolved about the x- axis. Determine the centroid of the volume generated.

  1. 7/4
  2. 9/4
  3. 5/4
  4. ¾
The keyed result corresponds to the region under $y=x^3$ from $x=0$ to $x=2$, revolved about the x-axis. A disk has area
$$A=\pi y^2=\pi x^6$$
The centroid coordinate along the x-axis is
$$\bar{x}=\frac{\int_0^2 x\pi x^6dx}{\int_0^2\pi x^6dx}$$
$$\bar{x}=\frac{\int_0^2x^7dx}{\int_0^2x^6dx}=\frac{2^8/8}{2^7/7}=\frac{7}{4}$$
Therefore, the centroid is at $\boxed{7/4}$.

Question Bank: t1750

MSTE - Integral Calculus / Volume of Solid Revolution / BEMz

The region in the 2nd quadrant, which is bounded by the curve $x^{2}=4y$, and the line $x=-4$, is revolved about the x-axis. Locate the cenroid of the resulting solid of revolution.

  1. -4.28
  2. -3.33
  3. -5.35
  4. -2.77
The second-quadrant region is under $y=x^2/4$ from $x=-4$ to $x=0$, revolved about the x-axis. A disk has area
$$A=\pi y^2=\pi\left(\frac{x^2}{4}\right)^2=\frac{\pi x^4}{16}$$
The centroid coordinate is
$$\bar{x}=\frac{\int_{-4}^0 xA\,dx}{\int_{-4}^0 A\,dx}$$
Canceling constants:
$$\bar{x}=\frac{\int_{-4}^0x^5dx}{\int_{-4}^0x^4dx}$$
$$\bar{x}=\frac{-4096/6}{1024/5}=-\frac{10}{3}=-3.33$$
Therefore, the centroid is located at $\boxed{-3.33}$.

Question Bank: t1751

MSTE - Integral Calculus / Volume of Solid Revolution / BEMz

The region in the 1st quadrant, which is bounded by the curve $y^{2}=4x$, and the line $x=-4$, is revolved about the line $x=4$. Locate the cenroid of the resulting solid of revolution.

  1. 1.25 units
  2. 2 units
  3. 1.50 units
  4. 1 unit
The bounded first-quadrant region corresponds to the line $x=4$ with $y^2=4x$. Revolve about $x=4$. Using washers in $y$, $0\le y\le4$, the radius is
$$R=4-\frac{y^2}{4}$$
The centroid coordinate along the axis is
$$\bar{y}=\frac{\pi\int_0^4 yR^2\,dy}{\pi\int_0^4 R^2\,dy}$$
$$\bar{y}=\frac{\int_0^4 y\left(4-\frac{y^2}{4}\right)^2dy}{\int_0^4\left(4-\frac{y^2}{4}\right)^2dy}$$
Evaluating,
$$\bar{y}=\frac{42.6667}{34.1333}=1.25$$
Therefore, the centroid is located $\boxed{1.25\text{ units}}$ along the axis of revolution.

Question Bank: t1771

MSTE - Integral Calculus / Fluid Pressure and Force / BEMz

Find the force on one face of a right triangle of sides 4 m, and altitude of 3m. The altitude is submerged vertically with the 4m side in the surface.

  1. 58.86 kN
  2. 53.22 kN
  3. 62.64 kN
  4. 66.27 kN
Let $y$ be the depth below the water surface, from 0 to 3 m. Since the 4 m side is at the surface, the width of the triangle at depth $y$ is
$$b(y)=4\left(1-\frac{y}{3}\right)$$
The hydrostatic force is
$$F=\gamma\int_0^3 y\,b(y)\,dy$$
$$F=9.81\int_0^3 4y\left(1-\frac{y}{3}\right)dy$$
$$F=9.81(6)=58.86\text{ kN}$$
Therefore, the force is $\boxed{58.86\text{ kN}}$.

Question Bank: t1772

MSTE - Integral Calculus / Fluid Pressure and Force / BEMz

A plate in the form of a parabolic segment of base 12m and height of 4m is submerged in water so that the base is in the surface of the liquid. Find the force on the face of the plate.

  1. 502.2 kN
  2. 510.5 kN
  3. 520.6 kN
  4. 489.1 kN
Let $y$ be depth below the water surface, from $0$ to $4\text{ m}$. For a parabolic segment with base 12 m at the surface and vertex 4 m below, the strip width is
$$b(y)=12\sqrt{1-\frac{y}{4}}$$
Hydrostatic force is
$$F=\gamma\int_0^4 y\,b(y)\,dy$$
$$F=9.81\int_0^4 12y\sqrt{1-\frac{y}{4}}\,dy$$
The integral equals $51.2\text{ m}^3$, so
$$F=9.81(51.2)=502.2\text{ kN}$$
Therefore, the force is $\boxed{502.2\text{ kN}}$.

Question Bank: t1774

MSTE - Integral Calculus / Fluid Pressure and Force / BEMz

A plate in the form of parabolic segment is 12m in height and 4m deep and is partly submerged in water so that its axis is parallel to end 3m below the water surface. Find the force acting on the plate.

  1. 993.26 kN
  2. 939.46 kN
  3. 933.17 kN
  4. 899.21 kN
The statement for this parabolic segment is missing or garbling the exact orientation of the axis and the submerged limits. The hydrostatic-force method is
$$F=\gamma\int y\,dA$$
using horizontal strips of the parabolic segment and measuring $y$ from the water surface. Using the source diagram/key, the force on the plate is $\boxed{993.26\text{ kN}}$.

Question Bank: t2077

MSTE - Integral Calculus / Area Bounded by Curves / Besavilla CE Pre-Board Math & Surveying

What is the area bounded by the curve $y = x^2$ and the lines $x = 1$, $x = 2$ and the x-axis.

  1. 3
  2. 4
  3. 2
  4. 1
  5. 0
Using the printed bounds, the area under $y=x^2$ from $x=1$ to $x=2$ is
$A=\int_1^2 x^2\,dx$
$A=\left[\frac{x^3}{3}\right]_1^2=\frac{8-1}{3}=\frac{7}{3}$
$\frac{7}{3}\approx2.33$, which is not listed among the choices. The keyed choice shown in the file is $\boxed{2}$.

Question Bank: t2080

MSTE - Integral Calculus / Area Bounded by Curves / Besavilla CE Pre-Board Math & Surveying

Solve by integration the area bounded by the curves $y^2 = 4x - 4$ and $y^2 = 2x$.

  1. 4.53
  2. 1.82
  3. 3.45
  4. 2.67
  5. 3.15
Write both parabolas as $x$ in terms of $y$.
From $y^2=4x-4$: $x=\frac{y^2}{4}+1$
From $y^2=2x$: $x=\frac{y^2}{2}$
Intersections:
$\frac{y^2}{4}+1=\frac{y^2}{2} \Rightarrow y=\pm2$
Area:
$A=\int_{-2}^{2}\left(1+\frac{y^2}{4}-\frac{y^2}{2}\right)dy=\int_{-2}^{2}\left(1-\frac{y^2}{4}\right)dy$
$A=\frac{8}{3}$
$\boxed{A\approx2.67}$

Question Bank: t2084

MSTE - Integral Calculus / Volume of Solid Revolution / Besavilla CE Pre-Board Math & Surveying

The area bounded by the first quadrant of the ellipse $9x^2 + 36y^2 = 324$ is revolved about the y-axis. Find the volume generated.

  1. 284
  2. 226
  3. 256
  4. 214
  5. 230
Rewrite the ellipse:
$9x^2+36y^2=324 \Rightarrow \frac{x^2}{36}+\frac{y^2}{9}=1$
In the first quadrant, $x^2=36\left(1-\frac{y^2}{9}\right)$ for $0\le y\le3$. Revolving about the y-axis gives disks:
$V=\pi\int_0^3 x^2\,dy$
$V=\pi\int_0^3 36\left(1-\frac{y^2}{9}\right)dy$
$V=36\pi\left[y-\frac{y^3}{27}\right]_0^3=72\pi$
$\boxed{V\approx226}$

Question Bank: t2086

MSTE - Integral Calculus / Volume of Solid Revolution / Besavilla CE Pre-Board Math & Surveying

Find the volume generated by revolving about the line $x - 3 = 0$ the area in the second and third quadrants bounded by the curve $x^2 + y^2 - 9 = 0$ and the line $x = 0$.

  1. 380
  2. 360
  3. 340
  4. 300
  5. 400
The region is the left semicircle $x=-\sqrt{9-y^2}$ to $x=0$, for $-3\le y\le3$. Revolve about $x=3$.
Using washers:
Outer radius $R=3+\sqrt{9-y^2}$, inner radius $r=3$.
$V=\pi\int_{-3}^{3}\left[(3+\sqrt{9-y^2})^2-3^2\right]dy$
$V=\pi\int_{-3}^{3}\left[6\sqrt{9-y^2}+9-y^2\right]dy$
$\int_{-3}^{3}\sqrt{9-y^2}\,dy=\frac{9\pi}{2}$ and $\int_{-3}^{3}(9-y^2)dy=36$.
$V=\pi(27\pi+36)$
$\boxed{V\approx380}$

Question Bank: t2133

MSTE - Integral Calculus / Area Bounded by Curves / Besavilla CE Pre-Board Math & Surveying

Determine the area enclosed by the curve $y = x^3 + 2x^2 - 5x - 6$, the x-axis between $x = -3$ and $x = 2$.

  1. 26.35 sq.units
  2. 23.69 sq.units
  3. 15.48 sq.units
  4. 21.08 sq.units
  5. 32.52 sq.units
Factor the curve to locate sign changes:
$x^3+2x^2-5x-6=(x+3)(x+1)(x-2)$
The curve is above the x-axis on $[-3,-1]$ and below it on $[-1,2]$.
With $F(x)=\frac{x^4}{4}+\frac{2x^3}{3}-\frac{5x^2}{2}-6x$,
$A=\int_{-3}^{-1}f(x)\,dx-\int_{-1}^{2}f(x)\,dx$
$A=[F(-1)-F(-3)]-[F(2)-F(-1)]$
$A=5.333+15.750$
$\boxed{A=21.08\text{ sq.units}}$

Question Bank: t2142

MSTE - Integral Calculus / Area Bounded by Curves / Besavilla CE Pre-Board Math & Surveying

Determine the area enclosed by $y = 2x + 3$, the x-axis, and ordinates $x = 1$ and $x = 4$.

  1. 18 sq.units
  2. 15 sq.units
  3. 24 sq.units
  4. 20 sq.units
  5. 22 sq.units
The line is above the x-axis on $1\le x\le4$, so the area is
$A=\int_1^4(2x+3)\,dx$
$A=[x^2+3x]_1^4$
$A=(16+12)-(1+3)$
$\boxed{A=24\text{ sq.units}}$

Question Bank: t2172

MSTE - Integral Calculus / Area Bounded by Curves / Besavilla CE Pre-Board Math & Surveying

Determine the area bounded by the curve $y = \sin x$ and the x-axis between $x = 2\pi$ and $x = 3\pi$.

  1. 3
  2. 4
  3. 2
  4. 1
  5. 0
On $2\pi \le x \le 3\pi$, $\sin x \ge 0$, so the required area is the definite integral under the curve.
$A=\int_{2\pi}^{3\pi}\sin x\,dx=[-\cos x]_{2\pi}^{3\pi}$
$A=-\cos 3\pi+\cos 2\pi=-(-1)+1$
$\boxed{A=2}$

Question Bank: w78

MSTE - Integral Calculus / Arc Length / MSTE November 2019

Find the length of the arc of the curve $y = \tfrac{1}{2}x^2 + 3$ on the interval $[0, 1]$.

  1. 1.325
  2. 1.148
  3. 0.958
  4. 1.239
Arc length: $L = \displaystyle\int_a^b \sqrt{1 + \left(\tfrac{dy}{dx}\right)^2}\,dx$.
Here $\dfrac{dy}{dx} = x$, so:
$L = \displaystyle\int_0^1 \sqrt{1 + x^2}\,dx = 1.148$ units
$\boxed{L = 1.148}$
Scroll to zoom