Finding the volume of a torus by rotation

[KevinL]

Homework Statement

http://faculty.tcu.edu/richardson/Calc2/H20090323torusVolume.htm This is a link to the homework.

2. The attempt at a solution
I did number 1 by doing 2 * integral from -2 to 2 of sqrt(4-x^2) = 4pi

The second problem is where I am completely confused and don't know how to do it or even conceptualize it. Trying to look at an example of a similar problem on another site, I tried to do 4pi * integral from -2 to 2 of xsqrt(4-x^2) but it equals 0.

I think if I understand this one ill be able to do the rest of the homework.

 

[Dick]

You are close. But the x in the second integral is supposed to be the radius of rotation of the slice located at x and rotated around x=3. How far is x from the center of rotation at x=3?

 

[KevinL]

So it should be 4pi * integral from -2 to 2 of 3sqrt(4-x^2) ?

 

[Dick]

No. If x is -2 the radius of rotation is 5. If x is +2 the radius of rotation is 1. Right? The distance of those points from x=3. What is the radius for a general point x between -2 and 2?

 

[KevinL]

r= 3-x

 

[Dick]

You've got it. Put that into your integral instead of x.