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anonymous
 5 years ago
name the solid obtained by rotating the set of points (x,y) such that (x4)^2 +y^2 < or =1 around the yaxis. (set up, but do not evaluate, an integral whose value is the volume of this solid. )
anonymous
 5 years ago
name the solid obtained by rotating the set of points (x,y) such that (x4)^2 +y^2 < or =1 around the yaxis. (set up, but do not evaluate, an integral whose value is the volume of this solid. )

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watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0Look at here: http://openstudy.com/users/watchmath#/users/watchmath/updates/4dd5dd7ad95c8b0bd2f35cc4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its kinda confusing to me :(

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0there 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
 5 years ago
Best ResponseYou've already chosen the best response.0where r is the radius to the center of the circle from the origin, and youre integraing about the y axis

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0So we can think of this \(rd\theta\) as the thickness of this differential volume?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i 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
 5 years ago
Best ResponseYou've already chosen the best response.0Jojo, are you familiar with the cyllindrical shell method?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0watchmath 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thank you guys! but please help me with other my review questions?? :(
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