Let $$R$$ be the region under the curve $$y=f(x)$$ between $$x=a$$ and $$x=b$$ ($$0\leq a<b$$) (Figure 1(a)). In Section 9.2, we computed the volume of the solid obtained by revolving $$R$$ about the $$x$$-axis. Another way of generating a totally different solid is to revolve the region $$R$$ about the $$y$$-axis as shown in Figure 1(b). (a) (b)

Figure 1

To compute the volume of this solid, consider a vertical thin rectangle of height $$f(x)$$ and width $$dx$$. When this rectangle is revolved around the $$y$$-axis, it generates a hollow, thin-walled shell of radius $$x$$, height $$f(x)$$ and thickness $$dx$$ (Figure 2(a)).

If the cylindrical shell has been rolled out flat like a thin sheet of tin, it becomes a thin slab of height $f(x)$, thickness $dx$, and  length $$2\pi x$$, which is the circumference of the shell (Figure 2(b)). Therefore, the element of volume is $dV=\underbrace{2\pi x}_{\small \rm circumference}\underbrace{f(x)}_{\small \rm height}\underbrace{dx}_{\small \rm thickness}.$ The total volume is then obtained by adding up the columns of the infinitesimal shells. $V=\int_{a}^{b}dV=\int_{a}^{b}2\pi xf(x)dx.$

 (a) (b)

Figure 2

• In principle, the volume of this solid can also be obtained by considering thin disks generated by revolving infinitesimally thin horizontal rectangles; however, it often turns out to be more difficult because (1) the equation $$y=f(x)$$ has to be solved for $$x$$ in terms of $$y$$, and (2) the formula for the length of the horizontal rectangle may vary in the region. In such cases, we have to compute more than one integral.
• If the region is between two curves $$y=f(x)$$ and $$y=g(x)$$ (with $$f(x)\geq g(x)$$), then the height of the vertical rectangle, which is the same as the height of the cylindrical shell, is $$f(x)-g(x)$$ (Figure 3). Therefore, in this case $dV=2\pi x[f(x)-g(x)]dx$ and $V=2\pi\int_{a}^{b}x[f(x)-g(x)]dx.$

Figure 3

In general, we can write $\bbox[#F2F2F2,5px,border:2px solid black]{V=\int_{a}^{b}2\pi(\text{shell radius})(\text{area of thin rectangle})=\int_{a}^{b}2\pi\rho dA}$

Example 1
The region in the first quadrant between $$y=4-x^{2}$$ and $$y=x^{2}-4$$ is revolved about the $$y$$-axis. Find the volume of the resulting solid by the shell method.
Solution 1

Figure 4

Let $$y_{1}=4-x^{2}$$ and $$y_{2}=x^{2}-4$$. Then the height of the typical shell $$MN$$ is \begin{aligned} MN & =y_{1}-y_{2}\\ & =\left(4-x^{2}\right)-\left(x^{2}-4\right)\\ & =8-2x^{2}\end{aligned} So the volume of the shell shown in the figure is $dV=2\pi\rho\ dA,$ where $dA=MN\ dx=\left(8-2x^{2}\right)dx$ and $\rho=\text{the distance of the thin rectangle from the axis of rotation}=x.$ Therefore, $dV=2\pi x\left(8-2x^{2}\right)dx$ and the total volume is \begin{aligned} V & =\int_{0}^{2}dV=\int_{0}^{2}2\pi x\left(8-2x^{2}\right)dx\\ & =4\pi\int_{0}^{2}\left(4x-x^{3}\right)dx\\ & =4\pi\left[2x^{2}-\frac{1}{4}x^{4}\right]_{0}^{2}\\ & =4\pi\left(8-\frac{16}{4}\right)=16\pi.\end{aligned}

Example 2
The region inside the circle $$x^{2}+y^{2}=a^{2}$$ is revolved about the $$y$$-axis. Find the volume of the resulting solid (which is a sphere) by the shell method.
Solution 2

Figure 5

We write the equation of the circle as $y_{1}=\sqrt{a^{2}-x^{2}}\quad\text{and}\quad y_{2}=-\sqrt{a^{2}-x^{2}}.$ Then \begin{aligned} dA & =\text{area of the thin rectangle}\\ & =(y_{1}-y_{2})dx\\ & =2\sqrt{a^{2}-x^{2}}dx\end{aligned} and the volume of the shell is $dV=2\pi\rho\,dA$ where $\rho=\text{distance of the thin rectangle from the axis of rotation}=x.$ Therefore, $dV=2\pi x\left(2\sqrt{a^{2}-x^{2}}\right)dx.$ Because $$x$$ (= the distance of the thin rectangle from the $$y$$-axis) varies between $$0$$ and $$a$$, the total volume is: \begin{aligned} V & =\int_{0}^{a}dV\\ & =\int_{0}^{a}2\pi x\left(2\sqrt{a^{2}-x^{2}}\right)dx\\ & =2\pi\int_{0}^{a}2x\sqrt{a^{2}-x^{2}}dx\end{aligned} Let $$u=a^{2}-x^{2}$$. Then $$du=-2x\,dx$$ and \begin{aligned} V & =-2\pi\int_{-a^{2}}^{0}\sqrt{u}\,du\qquad\small{(u=0 \text{ when } x=a \text{ and } u=-a^{2} \text{ when } x=0)}\\ & =2\pi\int_{0}^{a^{2}}\sqrt{u}\,du\\ & =2\pi\left[\frac{2}{3}u^{\frac{3}{2}}\right]_{0}^{a^{2}}\\ & =\frac{4\pi}{3}a^{3}\end{aligned}

Example 3
The region inside the curve $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is revolved about the $$y$$-axis. Find the volume of the resulting ellipsoid.

Solution 3

Figure 6

If we solve the equation of the ellipse for $$y$$, we get $y_{1}=b\sqrt{1-\frac{x^{2}}{a^{2}}},\quad y_{2}=-b\sqrt{1-\frac{x^{2}}{a^{2}}}.$ Then \begin{aligned} dA & =\text{area of the thin rectangle}\\ & =(y_{1}-y_{2})dx\\ & =2b\sqrt{1-\frac{x^{2}}{a^{2}}}dx\end{aligned} and the volume of the shell is \begin{aligned} dV & =2\pi xdA\\ & =2\pi x\left(2b\sqrt{1-\frac{x^{2}}{a^{2}}}\right)\,dx\end{aligned} Because $$x$$ (= the distance between the thin rectangle and the $$y$$-axis) varies between $$0$$ and $$a$$, the total volume is \begin{aligned} V & =\int_{0}^{a}dV\\ & =\int_{0}^{a}4b\pi x\sqrt{1-\frac{x^{2}}{a^{2}}}\,dx\end{aligned} Let $$u=1-\frac{x^{2}}{a^{2}}$$. Then $du=\frac{-2}{a^{2}}x\,dx.$ We know that $$u=1$$ when $$x=0$$ and $$u=0$$ when $$x=a$$. Therefore \begin{aligned} V & =\int_{0}^{a}4b\pi x\sqrt{1-\frac{x^{2}}{b^{2}}}dx\\ & =4b\pi\int_{1}^{0}\sqrt{u}\underbrace{\left(-\frac{a^{2}}{2}du\right)}_{xdx}\\ & =-2a^{2}b\pi\left[\frac{2}{3}u^{\frac{3}{2}}\right]_{1}^{0}\\ & =\frac{4}{3}\pi a^{2}b\end{aligned}

Example 4
Given that a disk with radius $$1$$ and center $$(5,0)$$ is revolved about the $$y$$-axis, compute the volume of the resulting doughnut-shaped solid.

Solution 4

Figure 7

The equation of a circle with radius $$1$$ and center $$(5,0)$$ is $(x-5)^{2}+y^{2}=1$ This equation gives us two functions $y_{1}=\sqrt{1-(x-5)^{2}}\quad\text{and}\quad y_{2}=-\sqrt{1-(x-5)^{2}}$ Therefore, the height of the typical thin rectangle is \begin{aligned} MN & =y_{1}-y_{2}\\ & =\sqrt{1-(x-5)^{2}}-\left(-1\sqrt{1-(x-5)^{2}}\right)\\ & =2\sqrt{1-(x-5)^{2}},\end{aligned} and the area of a thin rectangle is \begin{aligned} dA & =(MN)\,dx\\ & =2\sqrt{1-(x-5)^{2}}dx,\end{aligned} and the volume of the shell is \begin{aligned} dV & =2\pi\rho\,dA\\ & =2\pi x\left(2\sqrt{1-(x-5)^{2}}\right)dx\\ & =4\pi x\sqrt{1-(x-5)^{2}}dx.\end{aligned} Because $$\rho=x$$ ($$=$$ the distance between the thin rectangle and the origin) varies between $$4$$ and $$6$$, the total volume is $V=\int_{4}^{6}dV=\int_{4}^{6}4\pi x\sqrt{1-(x-5)^{2}}dx$ Let $$u=x-5$$. Then \begin{aligned} V & =\int_{-1}^{1}4\pi({\color{red}u}+{\color{blue}5})\sqrt{1-u^{2}}du\\ & =4\pi\int_{-1}^{1}{\color{red}u}\sqrt{1-u^{2}}du+4\pi\int_{-1}^{2}{\color{blue}5}\sqrt{1-u^{2}}du\end{aligned} Because the function $$f(u)=u\sqrt{1-u^{2}}$$ is odd, that is $f(-u)=-u\sqrt{1-(-u)^{2}}=-f(u)$ we know $\int_{-1}^{1}u\sqrt{1-u^{2}}du=0$ On the other hand, $$\int_{-1}^{1}\sqrt{1-u^{2}}du$$ is equal to the area of a semicircle of radius $$1$$ (Figure 8). Therefore $\int_{-1}^{1}\sqrt{1-u^{2}}du=\frac{\pi}{2}(1)^{2}=\frac{\pi}{2}$ Figure 8: The integral $$\int_{-1}^{1}\sqrt{1-u^{2}}du$$ is equal to the area of a semicircle of radius 1.

Therefore, the volume of the doughnut-shaped solid is \begin{aligned} V & =4\pi\underbrace{\int_{-1}^{1}u\sqrt{1-u^{2}}du}_{=0}+20\pi\underbrace{\int_{-1}^{1}\sqrt{1-u^{2}}du}_{=\frac{\pi}{2}}\\ & =10\pi^{2}.\end{aligned}

If a circle is revolved about an axis, the doughnut-shaped solid is called a torus. If the radius of the circle is $$r$$ and the distance between the center of the circle and the axis is $$a$$ then the volume of the torus is $V=\left(\pi r^{2}\right)(2\pi a)=2\pi^{2}r^{2}a.$

Figure 9: : A torus and its volume.