Find the double integral $$\iint_D\frac{1}{(1+2x^2+y^2)^2}\; dx \; dy$$ where $D=\{(x,y)|x \geq 0,y\geq 0\}$, using $ x=ar\cos \theta$, $y=br \sin \theta$.
So i got the jacobian $abr$ but i wasnt so sure about the domain, and the domain widen, and my book said that $2x^2+y^2\leq c^2$, $x\geq 0,y\geq 0$. First why I have to use $c$? and how can I define the domain in problem like this? Is the domain said that it is in first quadrant?
$2x^2+y^2\leq c^2$ how can i get this inequality?
Yes, $D$ is the first quadrant and the jacobian is correct.
Let $a=\frac{1}{\sqrt{2}}$ and $b=1$. Then $x=\frac{r\cos(\theta)}{\sqrt{2}}$ and $y=r\sin(\theta)$, $2x^2+y^2=r^2$, and $$D=\{(x,y):x\geq 0, y\geq 0\}=\left\{\left(\frac{r\cos(\theta)}{\sqrt{2}},r\sin(\theta)\right): r\geq 0 ,\;0\leq \theta\leq \pi/2\right\}.$$ Hence $$\iint_D\frac{1}{(1+2x^2+y^2)^2}\; dx \; dy=\int_{r=0}^{+\infty} \int_{\theta=0}^{\pi/2}\frac{1}{(1+r^2)^2}\cdot\frac{r}{\sqrt{2}}d\theta\,dr.$$ Can you take it from here?