How do I set up the integral to find the volume of the disc bounded by the circle x^2+y^2 = 1 rotated about the line y = 2

Question

amistre64 · Answer

you mean the volume of a torus?

anonymous · Answer

yes

amistre64 · Answer

do you know an answer yet to check with?  perhaps say 16pi?

amistre64 · Answer

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amistre64 · Answer

if we do a washer type method, and if im right ... big if!!

pi (R^2 - r^2) gives the area for any given washer cross section of this from angles 0 to pi/2

R = 2+cos(t)
r= 2-cos(t)

since that would be the top half we would have to double that result

$$2*(pi\int_{0}^{pi/2}((2+cos(t))^2-(2-cos(t))^2)dt$$

amistre64 · Answer

it makes sense in my head, but other than that .... im sure theres other ways to determine it :)

amistre64 · Answer

that should simplify to:  integral  8cos(t), which goes up to 8sin(t)

8sin(pi/2) - 8sin(0) = 8;  *pi = 8pi

doubles to 16pi

amistre64 · Answer

http://tutorial.math.lamar.edu/Classes/CalcI/MoreVolume.aspx might be a better way to go about it; near the bottom