Question
Via the Shell Method, what is the volume of the solid of revolution formed by revolving a portion of $y=3-x^2$ about the x-axis bounded by $y=0$ and $y=3$.
Context
The answer from Wolfram Alpha leads me to believe the solution is $V=\frac{48\sqrt{3}\pi}{5}$. While I believe that this problem can be solved via the Washer Method, I am interested in how to do solve this problem via the Shell Method.
Work So Far
Shell method formula for finding volume
The formula for volume via the shell method is $V=2\pi\int_a^{b}[r(x)h(x)]\mathrm dx$, where:
- $V$ represents volume;
- $a$ represents the lower bound;
- $b$ represents the upper bound;
- $r(x)$ represent the distance from the axis of rotation to $x$; and
- $h(x)$ represents the height of the solid at $x$.
Given that the revolution occurs around the x-axis, I know I need to set the given equation from terms of $x$ into terms of $y$: $V=2\pi\int_a^{b}[r(y)h(y)]\mathrm dy$. When I rewrite the formula to account for this problem, I end up with $V=2\pi\int_0^{3}[y*\sqrt{3-y}]\mathrm dy$.
Use of u-substitution
Via u-substitution, I would set $u=3-y$ and determine two things: $du=-dy$ and $y=3-u$. Rewriting the volume formula with this information reveals:
$$V=2\pi\int_0^{3}[(u-3)*\sqrt{u}]\mathrm du$$
This simplifies to:
$$V=2\pi\int_0^{3}[u^\frac{3}{2}-3u^\frac{1}{2}]\mathrm du$$
Which further simplifies to:
$$V=2\pi*[\frac{2u^\frac{5}{2}}{5}-2u^\frac{3}{2}]\Big|_0^3$$
When I plug $3-y$ for $u$, the formula reaches it final form:
$$V=2\pi*[\frac{2(3-y)^\frac{5}{2}}{5}-2(3-y)^\frac{3}{2}]\Big|_0^3$$
At this point, I'm getting a volume that is nowhere close to $V=\frac{48\sqrt{3}\pi}{5}$. Any help would be much appreciated!
Appendix
This is my first time encountering an equation where $x=\pm\sqrt(3-y)$ with these types of volume problems. For now, I'm blindly ignoring the negative and am not sure if that's a good idea.
Best Answer
There are of course several easier ways to solve this question, but I'll stick to your shell method. You have close to the right answer, but I found two mistakes.
When you change variables from $y$ to $u$, you need to change your bounds (so you would have the integral from 3 to 0, rather than from 0 to 3).
Your height function is only half of what it should be. This is because you ignored the negative square root; you are only counting the height (width in this problem) from the y-axis to the right, rather than from the left side of the rotated surface (the negative square root) to the right.
After those changes, I got the same answer $\dfrac{48\sqrt{3}\pi}{5}$.