What is the average area using polar coordinates?

[nysnacc]

Homework Statement

Homework Equations

Average (area) = 1/Area * integrate of polar

The Attempt at a Solution

y= r* sin theta
x= r* cos theta
r^2 = x^2+y^2

 

[BvU]

Why the square root ?

 

[nysnacc]

Why the square root ?

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

 

[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 $r^2$ ...

 

[nysnacc]

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

 

[Ssnow]

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

 

[nysnacc]

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

 

[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 ...

 

[nysnacc]

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

 

[Ssnow]

now it is ok,

 

[nysnacc]

thanks buddy!

 

[BvU]

You can check your answer with the expression for the volume of a cone

 

[nysnacc]

now it is ok,

And I got the answer as ⅓ a

 

[BvU]

OK it is ...