# Thread: Double integral..ln

1. ## Double integral..ln

Hi all,

I need help with the following task:

Calculate
∫∫ ln(1+x^2+y^2) dxdy
D

where D is: (x,y); 1 <=x^2+y^2<=2

(<= means bigger than or equal to)

I think I should use polar coordinates where

x= r cos §
y= r sin §

Then I get:

∫∫ ln(1+r^2)

∫ dr goes from 1 to 2
∫ d§ goes from 0 to 2 pi

But what is the integral of ln(1+r^2)?

I really need help,

thanks!

2. Remember that the transformation from cartesians to polars is $\displaystyle dy\,dx \to r\,dr\,d\theta$.

So your double integral should be

$\displaystyle \int_0^{2\pi}{\int_1^2{r\ln{(1 + r^2)}\,dr}\,d\theta}$.

You can solve this using a $\displaystyle u$ substitution.

3. Thank you, but I dont understand how to use u substitution.

The thing that troubles me is that ln(1+r^2)*r has two "steps". one is that the anti-derivative is 1/(1+r^2) but then there is the inner r^2 to consider and also the r from the Jacobian...

4. Originally Posted by Prove It
$\displaystyle \int_0^{2\pi}{\int_1^2{r\ln{(1 + r^2)}\,dr}\,d\theta}$.
look at the radius of the second circle.

5. Originally Posted by tinyone
Thank you, but I dont understand how to use u substitution.

The thing that troubles me is that ln(1+r^2)*r has two "steps". one is that the anti-derivative is 1/(1+r^2) but then there is the inner r^2 to consider and also the r from the Jacobian...
How could you possibly be doing double integrals in polar coordinates and not know how to use "substitution"?

Let $u= 1+ r^2$. What is "du"?

(And, as Krizalid suggests, the radii of the two circles are 1 and $\sqrt{2}$, not 1 and 2.)