Triple integral over a sphere in rectangular coordinates

[Batmaniac]

Homework Statement

Evaluate the following integral:

<br /> \iiint \,x\,y\,z\,dV<br />

Where the boundaries are given by a sphere in the first octant with radius 2.

The question asks for this to be done using rectangular, spherical, and cylindrical coordinates.

I did this fairly easily in spherical and rectangular coordinates, except for the fact that I got two different answers and I can't figure out where I went wrong! That's not a problem though because I can fix that.

The Attempt at a Solution

How would I do this problem in rectangular coordinates? My integral would look like this:

<br /> \int_{0}^{{\sqrt{4-x^2-y^2}}}\int_{0}^{{\sqrt{4-x^2-z^2}}}\int_{0}^{{\sqrt{4-z^2-y^2}}}xyz\,dz\,dy\,dx<br />

Which, without some clever transformations and an extremely messy Jacobian calculation, looks unsolvable.

 

[Mystic998]

I think you need to rethink your bounds on that one...

 

[Batmaniac]

How does this look then?

<br /> \int_{0}^{2}\int_{0}^{2}\int_{0}^{{\sqrt{4-z^2-y^2}}}xyz\,dz\,dy\,dx<br />

 

[Batmaniac]

Hmm, MATLAB tells me that's zero.

 

[Dick]

Your dy limit should depend on x.

 

[HallsofIvy]

How does this look then?

<br /> \int_{0}^{2}\int_{0}^{2}\int_{0}^{{\sqrt{4-z^2-y^2}}}xyz\,dz\,dy\,dx<br />

That would be over a square in the xy-plane rising up to the sphere.

Projecting the sphere into the xy-plane gives you the quarter circle x²+ y²= 4, with 0\le x\le 2, 0\le y\le 2. You can let x go from 0 to 2 but then, for each x, y ranges from 0 to \sqrt{4- x^2}.