I want to use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves: $y = 32 − x^2$ and $y = x^2$ about the axis $x=4$.

I know how to find the volume of the rotated region with respect to the $y$-axis which is $x=0$ (the value of the volume here is equal to $\frac{16384\pi }{3}$), but I'm having trouble in the specific case where the axis is $x=4$, I've tried translation but nothing worked.

  • $\begingroup$ No, the $x$-axis is $y=0$. Draw a sketch and draw a cross-section when you rotate about $x=4$. Then think. $\endgroup$ Commented Jul 2, 2020 at 22:52
  • $\begingroup$ @TedShifrin sorry for the typo, I meant "I know how to find the volume rotated with respect to the y-axis" I fixed it $\endgroup$
    – user144435
    Commented Jul 2, 2020 at 22:56
  • $\begingroup$ Write down the integral that you were trying to solve. You say that rotating around $x=0$ gives $0$. I think that's wrong. $\endgroup$
    – Andrei
    Commented Jul 2, 2020 at 23:11

1 Answer 1


So, what the graph is, is a pointed oval like shape from $x = -4$ to $x =4$. And, you rotate it from the rightmost edge, to get sort of a distorted torus. Then, we set up a cylinder with axis along $x = 4$.

No, around this cylinder, there would always be a cylindrical symmetry because of the rotation. Thus we need not worry about the angular part. only the values of $r$ and $z$ matter. And we multiply by $2\pi$ to our integral to account for the angular part of the integral.

now, we place our cylindrical shell such that $r=0$ at $x=4$ (the axis of rotation) and $z = 0$ at $y=16$ (where the two curves meet ie at $(x,y) = (4,16)$).

Then, $r$ extends from 0 to 8. (as the curves meet at $x=-4$ and $x=4$ which are 8 units apart). as $x=4$ at $r=0$, for any value of $r= a$(say), either $x=a+4$ or $x=4-a$. and $f(a+4)=f(4-a)$ for our rotational construction. Then for each $r, z$ extends from $z=(4-r)^2$ to $z= 32-(4-r)^2$. Note, we chose $4-r$ as the original functions are defined to the left of our chosen axis.

thus, we have: $$\int_{r=0}^{8}\int_{z=(4-r)^2}^{32-(4-r)^2} r dr d\theta dz$$ This should be the cylinderical setup you were expecting. $$ V = 2\pi* \int_{r=0}^8 [32 - 2(4-r)^2] r dr$$ $$ V = 4\pi* \int_{r=0}^8 8r^2 - r^3$$ $V = 1365.3$ $\pi$ $units^3$ or $ V = \frac{4096}{3}\pi$ cube units or $V = \frac{8^4}{3}\pi$

I hope the answer matches with what you were expecting.

  • $\begingroup$ There should be a factor of $2$ before the last $r^2$. And for some reason you dropped an extra factor of $r$ (coming from $r dr$). $\endgroup$
    – Andrei
    Commented Jul 3, 2020 at 0:44
  • $\begingroup$ Thanks, I did miss out the r term. However, the 2 was factored out and multiplied with 2pi $\endgroup$
    – Mihir Neve
    Commented Jul 3, 2020 at 8:00
  • $\begingroup$ Then you should have $4r^2-r^3$ $\endgroup$
    – Andrei
    Commented Jul 3, 2020 at 13:41
  • $\begingroup$ Opening the terms gives me 32 - (32 + 2r^2 -16r) = 2(8-r^2) $\endgroup$
    – Mihir Neve
    Commented Jul 3, 2020 at 13:42
  • $\begingroup$ Sorry, my mistake $32-2(16-8r+r^2)=2(8r-r^2)$ $\endgroup$
    – Andrei
    Commented Jul 3, 2020 at 13:44

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