# Double Integration (Stuck at square root step) (Solution Included).

1. Mar 20, 2012

### s3a

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

2. Relevant equations
Double integration.

3. The attempt at a solution
Firstly, I'd like to mention that the additional ρ under the square root is there accidentally and that it should be outside of the square root such that it forms the ending ρ dρ dθ.

I'm stuck at the sqrt(4a^2 - ρ^2) ρ dρ step. I have a feeling that I need to use trigonometry with ρ = 2acosθ or ρ = 2asinθ but firstly, which of the two would I choose and secondly given that I then have to integrate with respect to θ, must I replace ρ with a variable other than θ? Why or why not?

Any help would be greatly appreciated!

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2. Mar 20, 2012

### Karamata

I did it with $\rho=2a \sin t$.

3. Mar 21, 2012

### s3a

Thanks for saying but I also just realized that I don't know why z = √(4a^2 - ρ^2). Could you explain that to me please?

4. Mar 21, 2012

### HallsofIvy

Staff Emeritus
The "sphere of radius 2a", which forms part of the boundary, has equation $x^2+ y^2+ z^2= 4a^2$. In polar or cylindrical coordinates (not spherical coordinates), $\rho=\sqrt{x^2+ y^2}$ so that equation becomes $\rho^2+ z^2= 4a^2$ so that $z^2= 4a^2- \rho^2$ and $z= \pm\sqrt{4a^2- \rho^2}$.

5. Mar 21, 2012

### s3a

How did you go from ρ = √(x^2 + y^2) to ρ^2 + z^2 = 4a^2?

6. Mar 22, 2012

### Karamata

Change $\rho=\sqrt{x^2+y^2}$ in $x^2+y^2+z^2=4a^2$