## Cylindrical Shells Method |

Consider the following problem:

Find the volume of the solid obtained by rotating the region bounded by \(y=x^2\sin(x^2)\), \(x=0\),
\(x=\sqrt{\pi}\), and \(y=0\) about the \(y\)-axis.

First note that the cross-section of the solid at \(y\) perpendicular to the \(y\)-axis is a washer. To apply
the washer method, we need to find the inner radius and the outer radius which will require to solve for \(x\)
in terms of \(y\) from \(y=x^2\sin(x^2)\). It is complicated and sometimes analytically impossible to solve for
\(x\) in terms of \(y\) from \(y=f(x)\). We have a better method for this kind of problems:

** Cylindrical Shells Method:**

Consider the region bounded by \(y=f(x)\), \(x=a\), \(x=b\), and the \(x\)-axis. If \(S\) is the solid obtained
by rotating the region about the \(y\)-axis, then the volume of \(S\) is
\[V(S)= \int_a^b 2\pi x f(x)\;dx.\]
We can remember the preceding formula by associating \(2\pi x,\; f(x)\), and \(dx\) with the circumference,
height, and thickness of the cylindrical shell with the radius \(x\) respectively.

Break \([a,b]\) into \(n\) subintervals \([x_0,x_1],[x_1,x_2],\ldots,[x_{n-1},x_n]\) where \(x_i=a+i \Delta x\)
and \(\Delta x=(b-a)/n\). Partition \(S\) into \(n\) solids \(S_1,S_2,\ldots,S_n\) by intersecting \(S\) by
the vertical planes at \(x_1,x_2,\ldots,x_{n-1}\). Then the volume of \(S_i\) is approximately that of the
cylindrical shell with the inner radius \(x_{i-1}\), the outer radius \(x_i\), and height \(f(x_i^*)\) where
\(x_i^*=\frac{x_{i-1}+x_i}{2}\), \(i=1,2,\ldots,n\). Therefore
\[\begin{align*}
V(S_i) &\approx \pi x_i^2 f(x_i^*)-\pi x_{i-1}^2 f(x_i^*)\\
&=\pi f(x_i^*)\left( x_i^2 -x_{i-1}^2\right)\\
&=\pi f(x_i^*) (x_i+x_{i-1})(x_i-x_{i-1})\\
&=2\pi f(x_i^*) \frac{x_{i-1}+x_i}{2} (x_i-x_{i-1})\\
&=2\pi f(x_i^*) x_i^* \Delta x.
\end{align*}\]
Thus the volume of \(S\) is
\[V(S)\approx \sum_{i=1}^n 2\pi x_i^* f(x_i^*) \Delta x.\]
This approximation of \(V(S)\) gets better as \(n\to \infty\). Thus
\[V(S)=\lim_{n\to \infty} \sum_{i=1}^n 2\pi x_i^* f(x_i^*) \Delta x=\int_a^b 2\pi x f(x)\;dx.\]

**Example.**
Find the volume of the solid obtained by rotating the region bounded by \(y=x^2\sin(x^2)\), \(x=0\), \(x=\sqrt{\pi}\),
and \(y=0\) about the \(y\)-axis.

*Solution.*

The volume of the solid is \[\begin{align*} & \int_0^{\sqrt{\pi}} 2\pi x\cdot x^2\sin(x^2)\;dx \\ =\;& \pi\int_0^{\sqrt{\pi}} x^2\sin(x^2) \cdot 2x\;dx \\ =\;& \pi\int_0^{\pi} u\sin u\;du \;\;\left( \text{Let } u=x^2. \text{ Then } du=2x\;dx\right) \\ =\;& \pi \left.\left( -u\cos u +\sin u \right)\right\vert_0^{\pi} \;\;\left( \text{Integrating by parts} \right) \\ =\;& \pi \left( -\pi\cos \pi +\sin \pi \right) -\pi \left( -0\cos 0 +\sin 0 \right) \\ =\;& \pi^2. \end{align*}\]

For some problems, the formula of Cylindrical Shells Method \(V(S)= \int_a^b 2\pi x f(x)\;dx\) is modified by replacing \(x\) and \(f(x)\) by the radius and the height of the cylindrical shell at \(x\) respectively.

**Example.**
Find the volume of the solid obtained by rotating the region bounded by \(y=x\) and \(y=\frac{2x}{x^3+1}\)
about the line \(x=-1\).

*Solution.*

First note that \(y=x\) and \(y=\frac{2x}{x^3+1}\) intersect at \(x=0,1\). The cylindrical shell at \(x\) has radius \(x-(-1)\) and height \(\frac{2x}{x^3+1}-x\). The volume of the solid is \[\begin{align*} & \int_0^1 2\pi (x+1) \left( \frac{2x}{x^3+1} -x\right)\;dx \\ =\;& 2\pi\int_0^1 \left[-(x+1)x+(x+1)\frac{2x}{x^3+1} \right]\;dx\\ =\;& 2\pi\int_0^1 \left[-x^2-x+\frac{2x}{x^2-x+1} \right]\;dx\\ =\;& 2\pi\int_0^1 \left[-x^2-x+\frac{2x-1}{x^2-x+1} +\frac{1}{\left(x-\frac{1}{2}\right)^2+\left( \frac{\sqrt{3}}{2} \right)^2} \right]\;dx\\ =\;& 2\pi \left.\left( -\frac{x^3}{3}-\frac{x^2}{2}+\ln|x^2-x+1|+\frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{x-\frac{1}{2}}{\frac{\sqrt{3}}{2}} \right) \right)\right\vert_0^1 \\ =\;& 2\pi\left( -\frac{5}{6}+\frac{4}{\sqrt{3}} \tan^{-1}\left( \frac{1}{\sqrt{3}} \right) \right)\\ =\;& 2\pi\left( -\frac{5}{6}+\frac{2\pi}{3\sqrt{3}} \right). \end{align*}\]

The roles of \(x\) and \(y\) in the Cylindrical Shells Method may be switched for some problems.

**Example.**
Find the volume of the solid obtained by rotating the region bounded by \(x=y(3-e^y)\) and the \(y\)-axis
about the \(x\)-axis.

*Solution.*
First note that at the points of intersection of \(x=y(3-e^y)\) and \(x=0\), we have
\[ y(3-e^y)=0\implies y=0,\ln 3. \]
The cylindrical shell at \(y\) has radius \(y\) and height \(y(3-e^y)\). The volume of the solid is
\[\begin{align*}
& \int_0^{\ln 3} 2\pi y \cdot y(3-e^y) \;dy\\
=\;& 2\pi\int_0^{\ln 3} \left(3y^2-y^2e^y \right) \;dy\\
=\;& 2\pi \left.\left(y^3-e^y(y^2-2y+2) \right) \right\vert_0^{\ln 3} \;\;\left( \text{Integrating by parts} \right) \\
=\;& 2\pi \left((\ln 3)^3-3((\ln 3)^2-2\ln 3+2) \right)+4\pi\\
=\;& 2\pi \left((\ln 3)^3-3(\ln 3)^2+6\ln 3-4) \right).
\end{align*}\]

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