Integration in Polar - polar coordinate

1. Nov 13, 2011

DrunkApple

1. The problem statement, all variables and given/known data
f(x,y) = xy
x ≥ 0, y ≥ 0, $x^{2}$ + $y^{2}$ ≤ 4

2. Relevant equations

3. The attempt at a solution
$f^{pi/2}_{0}$$f^{2secθ}_{0}$ rcosθ * rsinθ * r drdθ

I just wanted to check... is this right?? cuz I really don't think it is

2. Nov 13, 2011

Dick

Why are you putting the upper limit for r to be 2*sec(theta)? The outer boundary is just a circle, isn't it?

3. Nov 13, 2011

DrunkApple

right! in terms of x, the bound should be from 0 to 2, but since x = 2 is rcos = 2, doesn't r becomes 2sec? or am I really really wrong and fault?
OH WAIT
is it just bounded by 0 to 2?

OH OH WAIT I JUST HAD AN EPIC EPIPHANY!!! THIS IS U SUBSTITUTION!!!!

Last edited: Nov 13, 2011
4. Nov 13, 2011

Dick

Uh, the boundary isn't x=2. It's x^2+y^2=4. u-substitution?? I think you are overcomplicating this.

5. Nov 13, 2011

DrunkApple

$f^{pi/2}_{0}$$f^{2}_{0}$ rcosθ * rsinθ * r drdθ
So it's not this?

6. Nov 13, 2011

Dick

That's it.

7. Nov 13, 2011

DrunkApple

yes so I'll go head and calculate all this cuz there seems to be a problem and I don't know if it's calculation error or book error

$f^{pi/2}_{0}$$f^{2secθ}_{0}$ $r^{3}$cosθ * rsinθ * drdθ

=$f^{pi/2}_{0}$($r^{4}$cosθsinθ)/4 * drdθ

=4$f^{pi/2}_{0}$cosθsinθ dθ

After u substitution...

=4$f^{1}_{0}$u du

=4?

8. Nov 13, 2011

Dick

The integral from 0 to 1 of u*du isn't 1.

9. Nov 13, 2011

DrunkApple

oh oh wow... How could I miss that...
Thank you very much ;D