# Polar coordinate integral

1. Sep 14, 2016

### nysnacc

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

2. Relevant equations
Average (area) = 1/Area * integrate of polar

3. The attempt at a solution
y= r* sin theta
x= r* cos theta
r^2 = x^2+y^2

2. Sep 14, 2016

### BvU

Why the square root ?

3. Sep 14, 2016

### nysnacc

x^2 + y^2 =r^2... OH, so only r not root r! and other than that, its fine?

4. Sep 14, 2016

### Ssnow

Hi, the $Area(R)=\pi a^{2}$ that you must divide, after $z(x,y)=f(r,\theta)=\sqrt{x^{2}+y^{2}}=r$ and not $\sqrt{r}$... so inside there is $r^2$ ...

5. Sep 14, 2016

### nysnacc

so the solution is (1/Area) double integral r dr d(theta) ?

6. Sep 14, 2016

### Ssnow

No, you have another $r$ inside the integral ...

7. Sep 14, 2016

### nysnacc

1/Area* ∫∫ r dr dθ is not correct ...?

8. Sep 14, 2016

### Ssnow

the formula say $\frac{1}{Area(R)}\int\int_{R}f(r,\theta)rdrd\theta$ so you have $f(r,\theta)\cdot r$ inside ...

9. Sep 14, 2016

### nysnacc

so f(r, θ) makes a r,
then 1/Area* ∫∫ r* r dr dθ

10. Sep 14, 2016

### Ssnow

now it is ok,

11. Sep 14, 2016

### nysnacc

thanks buddy!

12. Sep 14, 2016

### BvU

13. Sep 14, 2016

### nysnacc

And I got the answer as ⅓ a

14. Sep 14, 2016

OK it is ...