name the solid obtained by rotating the set of points (x,y) such that (x-4)^2 +y^2

Question

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. )

watchmath · Answer

Look at here: http://openstudy.com/users/watchmath#/users/watchmath/updates/4dd5dd7ad95c8b0bd2f35cc4

anonymous · Answer

its kinda confusing to me :(

anonymous · Answer

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.

$$\int\limits_{0}^{2\pi}\pi*1^{2}*4d  \theta$$

anonymous · Answer

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

watchmath · Answer

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

anonymous · Answer

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.

watchmath · Answer

Jojo, are you familiar with the cyllindrical shell method?

anonymous · Answer

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.

watchmath · Answer

Yes, that's correct.

anonymous · Answer

thank you guys! but please help me with other my review questions?? :(