# 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

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