Double integral polar cordiantes

[christian0710]

Hi, I need help with this problem

Evaluate the given integral by changing to polar cordinates

∫∫xydA where D is the disc with centre the origin and radius.

My solution so far.

I believe this would give a circle with radius 3 in xy plane. And then x=r*cos(θ) and y=r*sin(θ)

So ∫(∫((r*cos(θ)*r*sin(θ))*r,r,0,3),θ,0,2∏)

But the result is suppose to be zero.
What am I doing wrong?

 

[christian0710]

Ohh never mind, the result is correct :P
Just me who needs to learn how to use my calculator.

I guess it did not make seance that a circle with radius 3 would give 0 as a result, but perhaps it's because z=x*y --> z=0 <> 0=x*y so x=0/y --> x=0 and y=0/x --y=0
Is that a correct interpretation?

 

[Mark44]

Hi, I need help with this problem

Evaluate the given integral by changing to polar cordinates

∫∫xydA where D is the disc with centre the origin and radius.

And radius what?

My solution so far.

I believe this would give a circle with radius 3 in xy plane. And then x=r*cos(θ) and y=r*sin(θ)

So ∫(∫((r*cos(θ)*r*sin(θ))*r,r,0,3),θ,0,2∏)

But the result is suppose to be zero.
What am I doing wrong?

That's what I get.

Ohh never mind, the result is correct :P
Just me who needs to learn how to use my calculator.

I guess it did not make seance that a circle with radius 3 would give 0 as a result, but perhaps it's because z=x*y --> z=0

Why do you think that z = 0?

<> 0=x*y so x=0/y --> x=0 and y=0/x --y=0
Is that a correct interpretation?

No.
Edit: Added the inner integral.
The integral looks like this:
$$ \int_{\theta = 0}^{2\pi}\int_{r=0}^3 rcos(\theta) \cdot rsin(\theta) r~dr~d\theta$$

If you carry out the integration, you get a value of 0.