Shell method vs Disk Method

Question

I have always found using shell method better and easier to grasp. But are there or would there be any situations where only disk method could be used.

Answer 1

Let $R$ be a plane region bounded above and below by function graphs, and to the left and right by vertical lines, and let $S$ be the solid swept out by revolving $R$ about the horizontal axis.

In principle, if the volume of $S$ can be calculated using disks/washers, it can be calculated using shells. In practice, expressing a "disks" volume such as $$ \pi \int_{0}^{\infty} \bigl[e^{-x}(2 + \sin x)\bigr]^{2}\, dx $$ using shells involves breaking the solid $S$ into pieces (perhaps infinitely many) because the "profile" $y = f(x)$ need not be the graph of an invertible function.

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It's a good exercise to show that "shells and disks are equivalent" in the sense that if one integral can be evaluated in closed form (i.e., is elementary), so can the other. A special case is if $R$ is defined by $0 \leq y \leq f(x)$ and $0 \leq x \leq b$. The "shell" and "disk" integrals for the volume of $S$ are, respectively, $$ 2\pi\int_{0}^{f(b)} y (b - f^{-1}(y))\, dy,\qquad \pi\int_{0}^{b} f(x)^{2}\, dx. $$ Making the substitution $y = f(x)$ in the "shells" integral and integrating by parts gives the "disks" integral: \begin{align*} 2\pi\int_{0}^{f(b)} y (b - f^{-1}(y))\, dy &= 2\pi\int_{0}^{b} f(x) (b - x) f'(x)\, dx \\ &= \pi\int_{0}^{b} (b - x) (f^{2})'(x)\, dx \\ &= \pi(b - x)f(x)^{2} \Big|_{x=0}^{b} + \pi\int_{0}^{b} f^{2}(x)\, dx \\ &= \pi\int_{0}^{b} f^{2}(x)\, dx. \end{align*}