1. ## Polar

Hi just want to check if my solution to this question is correct:

Evaluate $\displaystyle \int_{R}\int \frac{1}{(x^2+y^2+1)^\frac{7}{2}} dA$ where $\displaystyle R$ is the region bounded by the y-axis, $\displaystyle y = x$ and $\displaystyle x^2+y^2 = 4$.

edit: sorry the title is supposed to be Polar coordinates + double integral

My solution

By using polar coordinates:
$\displaystyle x = r cos\theta$

$\displaystyle y = r sin\theta$

Thus drawing the region R on the x-y plane, we deduce
$\displaystyle R = \{(r,\theta): 0\leq r\leq 2 , \frac{\pi}{4} \leq \theta \leq \frac{\pi}{2}\}$

Then
$\displaystyle \int_{R}\int \frac{1}{(x^2+y^2+1)^\frac{7}{2}} dA$

$\displaystyle = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\int_{0}^{2} \frac{r}{(r^2+1)^\frac{7}{2}} dr d\theta$

$\displaystyle = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{1}{5}(1 - \frac{1}{25\sqrt{5}}) d\theta$

$\displaystyle = \frac{\pi}{20}(1 - \frac{1}{25\sqrt{5}})$

2. Originally Posted by shinn
Hi just want to check if my solution to this question is correct:

Evaluate $\displaystyle \int_{R}\int \frac{1}{(x^2+y^2+1)^\frac{7}{2}} dA$ where $\displaystyle R$ is the region bounded by the y-axis, $\displaystyle y = x$ and $\displaystyle x^2+y^2 = 4$.

edit: sorry the title is supposed to be Polar coordinates + double integral

My solution

By using polar coordinates:
$\displaystyle x = r cos\theta$

$\displaystyle y = r sin\theta$

Thus drawing the region R on the x-y plane, we deduce
$\displaystyle R = \{(r,\theta): 0\leq r\leq 2 , \frac{\pi}{4} \leq \theta \leq \frac{\pi}{2}\}$

Then
$\displaystyle \int_{R}\int \frac{1}{(x^2+y^2+1)^\frac{7}{2}} dA$

$\displaystyle = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\int_{0}^{2} \frac{r}{(r^2+1)^\frac{7}{2}} dr d\theta$

$\displaystyle = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{1}{5}(1 - \frac{1}{25\sqrt{5}}) d\theta$

$\displaystyle = \frac{\pi}{20}(1 - \frac{1}{25\sqrt{5}})$
It looks OK.