Area, Volume, Arc Length, Surface Area, and Pappus
Geometric applications of integration, including revolution formulas and Pappus centroid theorems.
Geometric applications of integration, including revolution formulas and Pappus centroid theorems.
Find the area under $f(x)=4-x^2$ on the interval $[-1,2]$.
$$A=\int_{-1}^{2}(4-x^2)\,dx=\left[4x-\frac{x^3}{3}\right]_{-1}^{2}$$
$$A=\left(8-\frac{8}{3}\right)-\left(-4+\frac{1}{3}\right)=\boxed{9\text{ sq. units}}$$
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 $[0,1]$.
$$A=-\int_{-2}^{0}x^3\,dx+\int_0^1x^3\,dx$$
$$A=-\left[\frac{x^4}{4}\right]_{-2}^{0}+\left[\frac{x^4}{4}\right]_0^1=4+\frac{1}{4}$$
$$\boxed{\frac{17}{4}\text{ sq. units}}$$
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 $x=1$.
$$A=\int_0^1(\sqrt{x}-x^2)\,dx=\left[\frac{2}{3}x^{3/2}-\frac{x^3}{3}\right]_0^1$$
$$\boxed{\frac{1}{3}\text{ sq. unit}}$$
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 $y=\sqrt{x}$.
$$V=\pi\int_0^4 y^2\,dx=\pi\int_0^4 x\,dx=\pi\left[\frac{x^2}{2}\right]_0^4$$
$$\boxed{8\pi\text{ cu. units}}$$
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,
$$V=(6\pi)(4\pi)=\boxed{24\pi^2\text{ cu. units}}$$
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 $r=x^2$.
$$V=\pi\int_0^1\left(R^2-r^2\right)dx=\pi\int_0^1\left(x^2-x^4\right)dx$$
$$=\pi\left[\frac{x^3}{3}-\frac{x^5}{5}\right]_0^1=\pi\left(\frac{1}{3}-\frac{1}{5}\right)=\pi\cdot\frac{2}{15}$$
$$\boxed{\frac{2\pi}{15}\text{ cu. units}}$$
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 = $f(x)=4x-x^2$.
$$V=2\pi\int_0^4 x(4x-x^2)\,dx=2\pi\int_0^4(4x^2-x^3)\,dx$$
$$=2\pi\left[\frac{4x^3}{3}-\frac{x^4}{4}\right]_0^4=2\pi\left(\frac{256}{3}-64\right)=2\pi\cdot\frac{64}{3}$$
$$\boxed{\frac{128\pi}{3}\text{ cu. units}}$$
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.
$$A=\int_0^{\pi/4}(\cos x-\sin x)\,dx+\int_{\pi/4}^{\pi/2}(\sin x-\cos x)\,dx$$
$$=\left[\sin x+\cos x\right]_0^{\pi/4}+\left[-\cos x-\sin x\right]_{\pi/4}^{\pi/2}$$
$$=(\sqrt{2}-1)+(\sqrt{2}-1)=\boxed{2\sqrt{2}-2\approx0.828\text{ sq. units}}$$
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 $2-\sqrt{x}$.
$$V=\pi\int_0^4\left[R^2-r^2\right]dx=\pi\int_0^4\left[4-(2-\sqrt{x})^2\right]dx$$
$$=\pi\int_0^4(4x^{1/2}-x)\,dx=\pi\left[\frac{8x^{3/2}}{3}-\frac{x^2}{2}\right]_0^4$$
$$=\pi\left(\frac{64}{3}-8\right)=\pi\cdot\frac{40}{3}=\boxed{\frac{40\pi}{3}\text{ cu. units}}$$
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 $x=\frac{5}{9}y^2$.
$$A=\int_{-6/5}^{3}\left[(y+2)-\frac{5}{9}y^2\right]dy=\boxed{6.86\text{ sq. units}}$$
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.
$$A=\int_{-1}^{2}(4-x^2)\,dx=\left[4x-\frac{x^3}{3}\right]_{-1}^{2}$$
$$A=\left(8-\frac{8}{3}\right)-\left(-4+\frac{1}{3}\right)=\boxed{9\text{ sq. units}}$$
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:
$$V=2\int_0^{20}(400-x^2)\,dx=\boxed{10666.67\text{ cu. cm}}$$
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,
$$V=2\pi \bar{y}A=2\pi(1.2)\left(\frac{16}{3}\right)=\boxed{\frac{64\pi}{5}}$$