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 $y=3$.

Screenshot of y=3-x^2 from y=0 to y=3


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.

Screenshot of the solid formed via a revolution about the x-axis

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:


When I plug $3-y$ for $u$, the formula reaches it final form:


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!


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.


2 Answers 2


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.

  1. 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).

  2. 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 $\dfrac{48\sqrt{3}\pi}{5}$.

  • $\begingroup$ Thank you for helping debug my mistakes! The changing of the bounds was something I completely forgot to do. If this helps anyone else, I forgot how to simply radical expressions which would have helped me realize what you and Ian pointed out earlier. $\endgroup$ Feb 7, 2021 at 23:44

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.)

  • $\begingroup$ Ian thank you for taking the time to help me conceptualize the problem! You're completely right that my lack of including the left-hand side was causing my volume to be half of what was expected. $\endgroup$ Feb 7, 2021 at 23:43
  • 1
    $\begingroup$ @CristianE.Nuno In general when trying to figure out why your answer differs from the given answer, it is often helpful to finish computing your answer fully and then start doing some simple comparisons between it and the exact answer: subtract, take the ratio etc. You can often find something simple like a factor of two error that you can more easily trace back in your work. $\endgroup$
    – Ian
    Feb 8, 2021 at 0:03

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