Volumes by slicing
We used integrals to find areas by cutting a region into thin rectangles. Now we find volumes the same way: we cut a solid into thin slices and add their volumes.
The slicing method
A slice of bread is almost a thin cylinder. Its volume is the area of its face times its thickness.
Suppose a solid lies between $x=a$ and $x=b$. Cut it at $x$, straight across. The flat face you see is called a cross-section. Let $A(x)$ be its area.
A slice of thickness $\Delta x$ has volume about
$A(x)\,\Delta x$.
Adding the slices and taking the limit gives
$\displaystyle V=\int_a^b A(x)\,dx$.
Example 1: Show that a sphere of radius $r$ has volume $\dfrac{4}{3}\pi r^3$.
Solution:
Put the center of the sphere at the origin. Cut it at $x$, where $-r\le x\le r$. The cross-section is a circle. By the Pythagorean theorem, its radius is $\sqrt{r^2-x^2}$. So its area is
$A(x)=\pi(r^2-x^2)$.
Now integrate:
$\begin{align*}&V\\&=\int_{-r}^{r}\pi(r^2-x^2)\,dx\\&=\pi\left[r^2x-\dfrac{x^3}{3}\right]_{-r}^{r}\\&=\pi\left(\dfrac{2r^3}{3}+\dfrac{2r^3}{3}\right)\\&=\dfrac{4}{3}\pi r^3\end{align*}$
This is where the volume formula for a sphere comes from.
Solids of revolution: the disk method
Take the region under $y=f(x)$, from $a$ to $b$. Spin it around the $x$-axis. It sweeps out a solid, called a solid of revolution.
A cross-section at $x$ is a circle (a disk) with radius $f(x)$. Its area is $\pi[f(x)]^2$. So
$\displaystyle V=\int_a^b\pi[f(x)]^2\,dx$.
Example 2: The region under $y=\sqrt{x}$, from $x=0$ to $x=1$, is rotated about the $x$-axis. Find the volume of the solid.
Solution:
The radius at $x$ is $\sqrt{x}$, so the area of a disk is $\pi(\sqrt{x})^2=\pi x$.
$\displaystyle V=\int_0^1\pi x\,dx=\pi\left[\dfrac{x^2}{2}\right]_0^1=\dfrac{\pi}{2}$
See figure (a) above.
Example 3: Show that a cone with base radius $r$ and height $h$ has volume $\dfrac{1}{3}\pi r^2h$.
Solution:
Rotate the line
$y=\dfrac{r}{h}x$, for $0\le x\le h$,
about the $x$-axis. It sweeps out a cone with its point at the origin. At $x=h$, the radius is $r$.
$\begin{align*}&V\\&=\int_0^h\pi\left(\dfrac{r}{h}x\right)^2dx\\&=\dfrac{\pi r^2}{h^2}\int_0^h x^2\,dx\\&=\dfrac{\pi r^2}{h^2}\cdot\dfrac{h^3}{3}\\&=\dfrac{1}{3}\pi r^2h\end{align*}$
The washer method
Now take the region between two curves, with $f(x)\ge g(x)\ge 0$, and spin it around the $x$-axis. The solid has a hole in the middle. A cross-section is a washer: a disk with a hole, like a flat ring.
- The outer radius is $R=f(x)$.
- The inner radius is $r=g(x)$.
The area of a washer is the big disk minus the hole: $\pi R^2-\pi r^2$. So
$\displaystyle V=\int_a^b\pi\left([f(x)]^2-[g(x)]^2\right)dx$.
Be careful: square each radius first, then subtract. $\pi(R-r)^2$ is wrong.
Example 4: The region between $y=x$ and $y=x^2$ is rotated about the $x$-axis. Find the volume.
Solution:
The curves meet at $x=0$ and $x=1$. On $[0,1]$, $x\ge x^2$. So the outer radius is $x$, and the inner radius is $x^2$. See figure (b) above.
$\begin{align*}&V\\&=\int_0^1\pi\left(x^2-x^4\right)dx\\&=\pi\left[\dfrac{x^3}{3}-\dfrac{x^5}{5}\right]_0^1\\&=\pi\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\\&=\dfrac{2\pi}{15}\end{align*}$
Rotating about the $y$-axis
If the region is spun around the $y$-axis, the slices are horizontal. Then write the radius as a function of $y$, and integrate with respect to $y$:
$\displaystyle V=\int_c^d\pi[g(y)]^2\,dy$,
where the radius at height $y$ is $x=g(y)$.
Example 5: The region between $y=x^3$, the line $y=8$, and the $y$-axis is rotated about the $y$-axis. Find the volume.
Solution:
Solve for $x$: $x=y^{1/3}$. At height $y$, the slice is a disk with radius $y^{1/3}$. The heights go from $y=0$ to $y=8$.
$\begin{align*}&V\\&=\int_0^8\pi\left(y^{1/3}\right)^2dy\\&=\pi\int_0^8 y^{2/3}\,dy\\&=\pi\left[\dfrac{3}{5}y^{5/3}\right]_0^8\\&=\pi\cdot\dfrac{3}{5}\cdot 32\\&=\dfrac{96\pi}{5}\end{align*}$
Here $8^{5/3}=\left(\sqrt[3]{8}\right)^5=2^5=32$.
Summary
- Volume by slicing: $\displaystyle V=\int_a^b A(x)\,dx$, where $A(x)$ is the area of a cross-section.
- Disk method: $A(x)=\pi[\text{radius}]^2$.
- Washer method: $A(x)=\pi\left(R^2-r^2\right)$, outer radius squared minus inner radius squared.
- For rotation about the $y$-axis, use horizontal slices and integrate with respect to $y$.