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October 31, 2014

October 31, 2014

Posted by **Saksham** on Tuesday, March 12, 2013 at 9:02am.

- calculus -
**Steve**, Tuesday, March 12, 2013 at 2:29pmThe cylinder is centered at (1/2,0) and has radius 1/2

Seems the best way to handle this one is with cylindrical coordinates (like polar coordinates, plus a z-axis):

As with polar coordinates, the alement of area is

dA = r dr dθ

so, the volume element is

dV = r dr dθ dz

We have z = 1-(x^2+y^2) = 1-r^2

x^2+y^2-x = 0 becomes

r^2 - rcosθ = 0, or

r = 1-cosθ

So,

V = ∫[0,pi/2] ∫[0,cosθ] z r dr dθ

= ∫[0,pi/2] ∫[0,cosθ] r(1-r^2) dr dθ

= ∫[0,pi/2] (1/2 cos^2(θ) - 1/3 cos^3(θ)) dθ

= 1/72 (18θ - 18sinθ + 9sin2θ - 2sin3θ) [0,pi/2]

= pi/8 - 2/9

Hmm. I don't get a*pi/b

Must have messed up. Check it out.

- calculus -
**joshua**, Sunday, March 17, 2013 at 12:34pm12

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