Integration in Polar - polar coordinate

[DrunkApple]

Homework Statement

f(x,y) = xy
x ≥ 0, y ≥ 0, x^{2} + y^{2} ≤ 4

Homework Equations

The Attempt at a Solution

f^{pi/2}_{0}f^{2secθ}_{0} rcosθ * rsinθ * r drdθ

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

 

[Dick]

Homework Statement

f(x,y) = xy
x ≥ 0, y ≥ 0, x^{2} + y^{2} ≤ 4

Homework Equations

The Attempt at a Solution

f^{pi/2}_{0}f^{2secθ}_{0} rcosθ * rsinθ * r drdθ

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

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

 

[DrunkApple]

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

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!

 

[Dick]

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!

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

 

[DrunkApple]

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

 

[Dick]

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

That's it.

 

[DrunkApple]

yes so I'll go head and calculate all this because 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

=4f^{pi/2}_{0}cosθsinθ dθ

After u substitution...

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

=4?

 

[Dick]

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

 

[DrunkApple]

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