# Evaluate the integral inside domain V, where V is bounded by the planes

1. Apr 11, 2012

### Cloudless

1. Evaluate the integral

VxdV

inside domain V, where V is bounded by the planes x=0, y=x, z=0, and the surface x2+y2+z2=1

Answer given: 1/8 - √2/16 (which is NOT what I got.. :yuck:)
2. The attempt at a solution

Ok, it's a triple integral, I know this.

∫dx runs from 0 to 1

∫dy runs from -√1-x2 to x

∫dz runs from 0 to √1-x2-y2

So the order of integrals go ∫∫∫dz dy dx

Yeah... it's not working. I keep getting stuck with things like x(1-y2-x2)1/2 which I can't integrate onwards...

2. Apr 12, 2012

### tiny-tim

Hi Cloudless!
this is a vertically sliced wedge from θ = 45° to 90°, so it would be much easier to use spherical coordinates

of course, you can use x y z, but …

no

no

3. Apr 12, 2012

### HallsofIvy

In the xy-plane, z= 0 so $x^2+ y^2+ z^2= 1$ projects to the quarter circle $x^2+ y^2= 1$. For each x, y ranges for 0 up to $y= \sqrt{1- x^2}$, NOT $y= -\sqrt{1- x^2}$ because the bounding surface is above the xy-plane.

4. Apr 12, 2012

### Cloudless

I redid it in spherical coordinates. Much easier, but still missing a pi/2. I think the conclusion is the answer key was wrong.

The question was ambiguous in its bounded area. I thought it was from negative pi/2 to pi/4.

5. Apr 12, 2012