1. ## Polar

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

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

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

My solution

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

$y = r sin\theta$

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

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

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

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

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

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

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

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

My solution

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

$y = r sin\theta$

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

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

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

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

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