Cylindrical Shell method conceptual question
I am self-teaching myself calculus for the summer to get ready for the actual class. Let us say that we have a region bounded above by the curve $y = 2 - x^2$ and below by the curve $y=x^2$ from $x=0$ to $x=1$. Suppose that the region is revolved around the $y-$axis.
Why do we use this formula instead of the washer method?:
$ 2 \pi \int_a^b x(f(x) - g(x)) dx$
The surface area of a cylinder is $2\pi rh$ (without the top and bottom) and the volume is $\pi r^2 h$, so how can the volume of a cylinder be $2 \pi rh dr$? I don't understand that way of thinking.
I thought that we're adding infinitesimally thin cylinders as we go up and down, so why not use the washer method?
$\endgroup$1 Answer
$\begingroup$If we want, we could use the washer method (in fact, this is the disk method since the $y$-axis is a boundary of the region) and integrate with respect to $y$. However, this would involve two separate integrals, since the radius of each disk is defined by two different curves. Thus, the volume using the disk method is: $$ V= \left[ \pi \int_0^1 (\sqrt{y})^2 dy \right] + \left[ \pi \int_1^2 (\sqrt{2-y})^2 dy \right] = \dfrac{\pi}{2} + \dfrac{\pi}{2} = \boxed{\pi} $$ The computations in this example actually turned out to not be that bad, but sometimes it just isn't feasible this way if the inverse functions of the boundaries can't be easily found.
Instead, it would be easier to add infinitesimally thin cylindrical shells ("hollow pipes"). If "unrolled" into a rectangular prism, the dimensions of an arbitrary shell would be $2\pi r \times h \times dx$, where the radius is $r=x$ and the height is $h=(2-x^2)-x^2$. Note that $2\pi r$ corresponds to the circumference of the original cylindrical shell's circular base and $dx$ corresponds to the shell's infinitesimally thin thickness. This yields the integral: $$ V= \int_0^1 2\pi x ((2-x^2)-x^2)dx = \boxed{\pi} $$
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