Solid inside both x^2+y^2+z^2=a^2 and (x-(a/2)^2 +y^2 = (a/2)^2

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- Apr 27th 2010, 10:38 AMasdf122345Need help setting this integral.
Solid inside both x^2+y^2+z^2=a^2 and (x-(a/2)^2 +y^2 = (a/2)^2

- Apr 28th 2010, 10:27 AMhollywood
You need to first visualize what the solid looks like. In this case, the first equation describes a sphere of radius a centered at the origin. For the second, I assume you mean (x-(a/2))^2 +y^2 = (a/2)^2, so it is a cylinder with radius a/2 centered at (a/2,0).

It will be easiest to integrate with respect to z first. The top and bottom surfaces come from the sphere, so solve that equation for z to get the limits of integration.

We now need to integrate this over the circle (x-(a/2))^2 +y^2 = (a/2)^2. So you solve for y to get the limits for y: $\displaystyle \pm\sqrt{\left(\frac{a}{2}\right)^2-\left(x-\frac{a}{2}\right)^2}$.

The limits for x are just the right- and left-most points of the circle, 0 and a.

- Hollywood