Polar Double Integration help

**FalconPUNCH!** · November 11th 2008, 07:15 PM

I'm doing polar integration and I'm having some problems.

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

The answer comes out to

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

**mr fantastic** · November 11th 2008, 07:35 PM

Originally Posted by FalconPUNCH!

I'm doing polar integration and I'm having some problems.

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

The answer comes out to

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

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

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

**FalconPUNCH!** · November 11th 2008, 07:36 PM

Originally Posted by mr fantastic

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

The only 'hard' bit is finding $\int_{1}^{2} \ln u \, du = 2 \ln (2) - 1$ where $u = r^2 + 1$ . 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.

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