# Thread: double integral help, polar coordinates

1. ## double integral help, polar coordinates

I get to this bit

$\displaystyle \int^{\frac{\pi}{2}}_{0} \int^{1}_{0} \sqrt{r^{2} +1} r dr d\theta$

from here, can I not simplify the expression $\displaystyle \sqrt{r^{2}+1} r$

=

$\displaystyle r^{2} + r$ ??

2. ## Re: double integral help, polar coordinates

The r in "$\displaystyle r drd\theta$" is outside the square root. Use the substitution $\displaystyle u= r^2+ 1$.

3. ## Re: double integral help, polar coordinates

Originally Posted by HallsofIvy
The r in "$\displaystyle r drd\theta$" is outside the square root. Use the substitution $\displaystyle u= r^2+ 1$.
I know that it is outside the square root, but why can i not just take the square root or r^2 and 1 that gives

$\displaystyle \sqrt{r^{2} + 1}$ = r+ 1

than multiply that through by r? why is this wrong ?

4. ## Re: double integral help, polar coordinates

Back to algebra class for you: you can't take the square root of each term like that. For example $\displaystyle \sqrt{3^2+1}$ is $\displaystyle \sqrt{10}$, not 4. Or looking the other way, $\displaystyle (r+1)^2$ is $\displaystyle r^2+2r+1$, not $\displaystyle r^2+1$.

When you make HallsofIvy's substitution, you get

$\displaystyle \int^{\frac{\pi}{2}}_{0} \int^{2}_{1} \frac{1}{2} u^{1/2}\, du \,d\theta$

which you should be able to integrate directly.

- Hollywood

5. ## Re: double integral help, polar coordinates

Originally Posted by Tweety
I know that it is outside the square root, but why can i not just take the square root or r^2 and 1 that gives

$\displaystyle \sqrt{r^{2} + 1}$ = r+ 1

than multiply that through by r? why is this wrong ?
Is $\displaystyle \sqrt{25+ 1}= \sqrt{26}= 6$?

Or $\displaystyle (r+ 1)^2= r^2+ 2r+ 1$ not $\displaystyle r^2+ 1$.