# Using the method of cylindrical shells to find the volume generated by rotating a region with respect to a specific axis.

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.

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

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.

• 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 $r dr$). Commented Jul 3, 2020 at 0:44
• Thanks, I did miss out the r term. However, the 2 was factored out and multiplied with 2pi Commented Jul 3, 2020 at 8:00
• Then you should have $4r^2-r^3$ Commented Jul 3, 2020 at 13:41
• Opening the terms gives me 32 - (32 + 2r^2 -16r) = 2(8-r^2) Commented Jul 3, 2020 at 13:42
• Sorry, my mistake $32-2(16-8r+r^2)=2(8r-r^2)$ Commented Jul 3, 2020 at 13:44