I have to calculate the following integral:
$$\int_{0}^{1}\int_{\sqrt{3x}}^{\sqrt{4-x^2}}x\sqrt{x^2+y^2}dydx$$
I drew the domain of integration, and I found that :
$$ 0 \leq r \leq 2 $$
$$0 \leq \theta \leq \pi/3$$
Assuming what I found is correct, I how can I rewrite the integral?
I know $ dydx $ becomes $rdrd\theta$, and $\sqrt{x^2+y^2}$ becomes $r$.
Do I convert $x$ to $rcos\theta$ ? This will yield me the answer of $2\sqrt{3}$ after integration, but I cant find any online calculator to verify this.
Thanks in advance!