Using shell method, find the volume of the solid generated when $y=3-x^2$ is revolved around x-axis from $y=0$ to $y=3$ 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.
 A: Approach 1: the total volume is the sum of the volume of revolution of the region to the right of $x=0$ and the volume of revolution of the region to the left of $x=0$. By symmetry, the two volumes are the same. On the right, you have $x=\sqrt{3-y}$, so the volume of the part on the right is $2 \pi \int_0^3 y \sqrt{3-y} \, dy$ and then the total volume is twice that.
Approach 2: in general the height $h(y)$ is really the distance from the left side of the region to the right side of the region for each fixed value of $y$, so $h(y)=\sqrt{3-y}-(-\sqrt{3-y})=2\sqrt{3-y}$. (Note that this needs to be adjusted if you have a curve determined by a function that isn't even two-to-one, like the region located above the $x$ axis and below $y=\sin(x)$ for $x$ between $0$ and $4\pi$. But there is no issue here.)
A: 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 $\dfrac{48\sqrt{3}\pi}{5}$.
