Iterated Intergal

**guess** · March 29th 2010, 09:03 AM

Find the volume of the solid inside the cylinder x^2+y^2 = 2y and bounded above and below by the sphere x^2 + y^2 + z^2 = 4.

Thanks for visiting through my problems. I need to get the solution but i do not have any idea on how to solve it. Please help. Thanks

**HallsofIvy** · March 29th 2010, 11:59 AM

Originally Posted by guess

Find the volume of the solid inside the cylinder x^2+y^2 = 2y and bounded above and below by the sphere x^2 + y^2 + z^2 = 4.

Thanks for visiting through my problems. I need to get the solution but i do not have any idea on how to solve it. Please help. Thanks

By completing the square, that cylinder is $x^2+ (y- 1)^2= 1$ which has axis (0, 1, z) which is slightly of center from the sphere. I think the first thing I would do is "translate" the origin to (0, 1, 0)- replace y by y'= y- 1 so that y= y'+ 1 and the equation of the cylinder is $x^2+ y'^2= 1$ and the equation of the sphere is $x^2+ (y'+ 1)^2+ z^2= 4$ or $x^2+ y'^2+ z^2+ 2y= 3$ .

Now the cylinder has its axis at (0, 0, z) and you can use cylindrical coordinates: z will run from the part of the sphere at the bottom, $z= -\sqrt{3- x^2- y^2- 2y}= -\sqrt{3- r^2- 2rsin(\theta)}$ , to the part at the top, $z= \sqrt{3- x^2- y^2- 2y= \sqrt{3- r^2- 2rsin(\theta)}$ while r and $\theta$ range from 0 to 1 and 0 to $2\pi$ . The volume is given by
$\int_{\theta= 0}^{2\pi}\int_{r= 0}^1\int_{z=-\sqrt{3- r^2- 2rsin(\theta)}}^{\sqrt{3- r^2- 2rsin(\theta)}} r dz dr d\theta$

**shawsend** · March 29th 2010, 02:12 PM

Hi. Here's an alternate way just for fun. Do the complete the square thing, then convert $x^2+(y-1)^2=1$ to polar coordinates to obtain $r=2\sin(t)$ , and the sphere into cylindrical coordinates $z^2=\sqrt{4-r^2}$ , then the integral is:

$4\int_0^{\pi/2}\int_0^{2\sin(t)}\int_0^{\sqrt{4-r^2}} rdzdrdt$ which is the same answer as HallsofIvy above, about 9.6 which is in the ball park since the volume of the whole cylinder of length 4 is $4\pi\approx 12.6$

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