## anonymous 5 years ago name the solid obtained by rotating the set of points (x,y) such that (x-4)^2 +y^2 < or =1 around the y-axis. (set up, but do not evaluate, an integral whose value is the volume of this solid. )

1. watchmath
2. anonymous

its kinda confusing to me :(

3. anonymous

there is another way to do this I think. You look at the area of the circle, and multiply that by r*d(theta). That is your differential Volume element. Then you integrate from 0 to 2*Pi. It has the same result as watchmath's method. $\int\limits_{0}^{2\pi}\pi*1^{2}*4d \theta$

4. anonymous

where r is the radius to the center of the circle from the origin, and youre integraing about the y axis

5. watchmath

So we can think of this $$rd\theta$$ as the thickness of this differential volume?

6. anonymous

i think so. it is the arclength of one small piece of the volume which would be the thickness of the differential element. Then adding these all up around the y axis gives the volume.

7. watchmath

Jojo, are you familiar with the cyllindrical shell method?

8. anonymous

watchmath is that method sort of a way to slice the circle into vertical strips and add up each shell over the diameter of the circle? It's been a while since I did these problems.

9. watchmath

Yes, that's correct.

10. anonymous