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Integral Calculus lecture notes: Solids of revolution - volume by slicing - Disks

Eugene Kritchevski, Vanier College, last updated on March 9, 2016

The disk method

Suppose that the function $f$ is continuous and nonnegative on $[a,b]$ and let ${\cal R}$ be the region in the $(x,y)$ plane bounded by the graph of $y=f(x)$, the $x$-axis and the vertical lines $x=a$ and $x=b$, namely $${\cal R}=\set{(x,y): a\leq x\leq b \,\textrm{ and }\, 0\leq y\leq f(x) }.$$ Considering the $(x,y)$ plane as part of the three dimensional space, we rotate the region ${\cal R}$ around the $x$-axis to obtain a solid of revolution. Let us call the resulting $\cal S$ and our goal is to compute the volume of $\cal S$.

Move the rotate slider and try to imagine what the resulting solid of revolution will look like. Then check the box show solid to see the solid.
The volume of $\cal S$ can be computed by a procedure analogous to the computation of the area of the region $\cal {R}$. Recall that $\cal{R}$ is aproximated by rectangles. Analogously, $\cal{S}$ is approximated by disks. Recall the details of computing the area of the region $\cal R$ using right-endpoint Riemann sums. The interval $[a,b]$ is subdived into $n$ suntintervals of equal length $$\Delta x=\frac{b-a}{n}.$$ The $k^{\rm th}$ subinterval is $[x_{k-1},x_k]$ where $$x_k=a+k\Delta x.$$ The base of the $k^{\rm th}$ aproximating rectangle is $\Delta x$ and the height is given by $f(x_k)$. Thus the area of the $k^{\rm th}$ aproximating rectangle is $$A_k= f(x_k)\cdot \Delta x.$$
An approximation to the area of $\cal R$ is then given by the sum of the areas of the $n$ approximating rectangles: $$R_n=\sum_{k=1}^{n}A_k=\sum_{k=1}^{n}f(x_k) \Delta x.$$ The exact area of the region $\cal{R}$ is then given by $$area({\cal R})=\lim_{n\to\infty} R_n=\int_a^b f(x)\,dx.$$

The volume of the solid of revolution is found in a very similar way. We approximate the solid $\cal S$ by the solid $\cal S_n$, built from $n$ disks, such that $volume({\cal S_n})\to volume ({\cal S})$ as $n\to \infty$. The $k^{\rm th}$ disk has volume $$V_k=\pi\cdot radius^2\cdot width=\pi\cdot[f(x_k)]^2\cdot \Delta x.$$ Summing the volumes of the disks, we get $$volume({\cal S_n})=\sum_{k=1}^n V_k=\sum_{k=1}^n \pi\cdot[f(x_k)]^2\cdot \Delta x.$$ The key observation here is that $\sum_{k=1}^n \pi\cdot[f(x_k)]^2\cdot \Delta x$ is the right-endpoint Riemann sum for the definite integral $\int_a^b \pi [f(x)]^2\,dx$. It follows that $$volume({\cal S})=\lim_{n\to\infty} volume({\cal S_n})=\lim_{n\to\infty}\sum_{k=1}^n \pi\cdot[f(x_k)]^2\cdot \Delta x=\int_a^b \pi [f(x)]^2\,dx.$$ This gives us a very useful general formula:
The volume of the solid obtained by rotating about the $x$-axis the region under $y=f(x)$ from $a$ to $b$, is $$\int_a^b \pi [f(x)]^2\,dx$$


Example 1: Consider a regular cone with height $h$ and whose base is a circle or radius $r$. The cone can be obtained by rotating about the $x$-axis the (triangular) region under the linear graph $$y=f(x)=(r/h)x$$ from $a=0$ to $b=h$.



The volume of the cone is then $$volume=\int_0^h \pi[(r/h)x]^2\,dx=\frac{\pi r^2}{h^2}\int_0^h x^2\,dx=\frac{\pi r^2}{h^2}\frac{x^3}{3}\Big|_{x=0}^{x=h}=\frac{\pi r^2}{h^2}\frac{h^3}{3}=\frac{\pi r^2 h}{3}.$$

Example 2: Consider a sphere of radius $r$. The sphere can be obtained by rotating about the $x$-axis the (semicicular) region under the linear graph $$y=f(x)=\sqrt{r^2-x^2}$$ from $a=-r$ to $b=r$.



The volume of the sphere is then $$volume=\int_{-r}^r \pi[\sqrt{r^2-x^2}]^2\,dx=\int_{-r}^r \pi\of{r^2-x^2}\,dx=\pi\of{r^2 x -\frac{x^3}{3}}\Big|_{x=-r}^{x=r}=\pi\of{r^3 -\frac{r^3}{3}}-\pi\of{-r^3 -\frac{(-r))^3}{3}}=\frac{4}{3}\pi r^3.$$