# Triple integral over a sphere in rectangular coordinates

1. Apr 1, 2008

### Batmaniac

1. The problem statement, all variables and given/known data

Evaluate the following integral:

$$\iiint \,x\,y\,z\,dV$$

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.

3. The attempt at a solution

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

$$\int_{0}^{{\sqrt{4-x^2-y^2}}}\int_{0}^{{\sqrt{4-x^2-z^2}}}\int_{0}^{{\sqrt{4-z^2-y^2}}}xyz\,dz\,dy\,dx$$

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

2. Apr 1, 2008

### Mystic998

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

3. Apr 1, 2008

### Batmaniac

How does this look then?

$$\int_{0}^{2}\int_{0}^{2}\int_{0}^{{\sqrt{4-z^2-y^2}}}xyz\,dz\,dy\,dx$$

4. Apr 1, 2008

### Batmaniac

Hmm, MATLAB tells me that's zero.

5. Apr 1, 2008

### Dick

Your dy limit should depend on x.

6. Apr 2, 2008

### HallsofIvy

Staff Emeritus
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 $\sqrt{4- x^2}$.