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anonymous
 4 years ago
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
anonymous
 4 years ago
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

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amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0you mean the volume of a torus?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0do you know an answer yet to check with? perhaps say 16pi?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1328034607713:dw

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0if 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= 2cos(t) since that would be the top half we would have to double that result \[2*(pi\int_{0}^{pi/2}((2+cos(t))^2(2cos(t))^2)dt\]

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0it makes sense in my head, but other than that .... im sure theres other ways to determine it :)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0that should simplify to: integral 8cos(t), which goes up to 8sin(t) 8sin(pi/2)  8sin(0) = 8; *pi = 8pi doubles to 16pi

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0http://tutorial.math.lamar.edu/Classes/CalcI/MoreVolume.aspx might be a better way to go about it; near the bottom
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