## 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

1. amistre64

you mean the volume of a torus?

2. anonymous

yes

3. amistre64

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

4. amistre64

|dw:1328034607713:dw|

5. amistre64

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 $2*(pi\int_{0}^{pi/2}((2+cos(t))^2-(2-cos(t))^2)dt$

6. amistre64

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

7. amistre64

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

8. amistre64

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