Originally Posted by

**mathemagister** Okay, I did some researching on the almighty Internet and found this...

Draw a 2-dimensional picture of a cross-section through the middle. Look at the right hand side. You have a 2-dimensional object bounded on the right by the curve $\displaystyle x^2 + y^2 = 4$ and on the left by the line $\displaystyle x = \sqrt {3}$. The points where these curves intersect are $\displaystyle (\sqrt {3},1)$ and $\displaystyle (\sqrt {3}, - 1)$. We want the volume of the solid of revolution formed by rotating this about the y axis.

Method 1 (the shell method):

Slice the 2-dimensional object into vertical strips. Each such vertical strip is located at position x, has width dx, and height $\displaystyle 2\sqrt {4 - x^2}$. Rotating this strip about the vertical axis gives a thin cylindrical shell. Slice this open and lay it out flat. The result is a rectangular slap of width $\displaystyle 2\pi x$ (the circumference of the shell), height $\displaystyle 2\sqrt {4 - x^2}$, and thickness dx so $\displaystyle dV = 4\pi x\sqrt {4 - x^2}\,dx$ and

$\displaystyle V = \int_{\sqrt {3}}^24\pi x\sqrt {4 - x^2}\,dx $

Method 2 (the "washer" method):

Slice the 2-dimensional object into horizontal strips. Each such horizontal strip is located at position y and has width dy. Its left endpoint is at $\displaystyle x = \sqrt {3}$ and its right endpoint is at $\displaystyle x = \sqrt {4 - y^2}$. Rotating this strip about the vertical axis gives a thin flat "washer" of thickness dy. The inner radius of this (annulus) is $\displaystyle \sqrt {3}$ and the outer radius is $\displaystyle \sqrt {4 - y^2}$. The volume is

$\displaystyle dV = \pi\left(\left(\sqrt {4 - y^2}\right)^2 - \left(\sqrt {3}\right)^2\right)dy = \pi(1 - y^2)dy$

$\displaystyle V = \int_{ - 1}^1 \pi(1 - y^2)dy$

I'm sure you know how to solve either of these integrals.