# Polar Double Integration help

• Nov 11th 2008, 07:15 PM
FalconPUNCH!
Polar Double Integration help
I'm doing polar integration and I'm having some problems.

$\displaystyle 4*\int_{0}^{\frac{\pi}{2}} \int_{0}^{1} ln(r^{2}+1)rdrd\theta$

$\displaystyle \pi(ln(4)-1)$ but I always end up with $\displaystyle \pi(ln(2)-1)$
• Nov 11th 2008, 07:35 PM
mr fantastic
Quote:

Originally Posted by FalconPUNCH!
I'm doing polar integration and I'm having some problems.

$\displaystyle 4*\int_{0}^{\frac{\pi}{2}} \int_{0}^{1} ln(r^{2}+1)rdrd\theta$

$\displaystyle \pi(ln(4)-1)$ but I always end up with $\displaystyle \pi(ln(2)-1)$

If you show your working it will be easier to see your mistake.

The only 'hard' bit is finding $\displaystyle \int_{1}^{2} \ln u \, du = 2 \ln (2) - 1$ where $\displaystyle u = r^2 + 1$. Note the integrals terminals (this is probably what you forgot).
• Nov 11th 2008, 07:36 PM
FalconPUNCH!
Quote:

Originally Posted by mr fantastic
If you show your working it will be easier to see your mistake.

The only 'hard' bit is finding $\displaystyle \int_{1}^{2} \ln u \, du = 2 \ln (2) - 1$ where $\displaystyle u = r^2 + 1$. Note the integrals terminals (this is probably what you forgot).

I would have shown my work but it's two pages long and messy. I think it's because I didn't try substituting. I'll try that thanks.