MHB NICK's question at Yahoo Answers regarding a solid of revolution

[MarkFL]

Here is the question:

Calculus ii question!?

Using disks or washers, find the volume of the solid obtained by rotating the region bounded by the curves y=x, y=0, x=2, and x=4 about the line x=1.

Volume = ?

Here is a link to the question:

Calculus ii question!? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[MarkFL]

Hello NICK,

Using washers, we find that we have one interval where the inner radius is a function of $y$, and another in which the inner radius is constant. For both intervals, the outer radius is constant.

Thus, we find the volume is:

$$V=\pi\left(\int_0^2 3^2-1^2\,dy+\int_2^4 3^2-(y-1)^2\,dy \right)=$$

$$\pi\left(\left[8y \right]_0^2+\left[9y-\frac{1}{3}(y-1)^3 \right]_2^4 \right)=\pi\left(16+27-\frac{53}{3} \right)=\frac{76\pi}{3}$$

To check our work, let's use the shell method:

$$V=2\pi\int_2^4(x-1)x\,dx=2\pi\left[\frac{x^3}{3}-\frac{x^2}{2} \right]_2^4=$$

$$2\pi\left(\frac{40}{3}-\frac{2}{3} \right)=\frac{76\pi}{3}$$