I have a joint pdf:
$c\sqrt{1 - x^2 - y^2}, x^2 + y^2 \le 1$
I need to find c and I know this can be done by doing the double integration over the possible of values of y and x and equating the result to 1. However, that seems a tedious approach involving trigonometric substitutions. Hence, I'm looking for other suggestions/ approaches to find the constant. Any suggestions?
Passing in polar coordinates you have only to integrate over the unit disk with radius $\leq1$ and angle $\theta$, say $x^2+y^2=\rho^2$ with jacobian =$\rho$ you get immediately
$$\int_{0}^{2\pi} d\theta \int_{0}^{1} \rho\sqrt{1-\rho^2} d\rho=\frac{2}{3}\pi$$
the first integral is $2\pi$ while the second is immediate when you set $1-\rho^2=u$
your constant is the reciprocal of this quantity
Summary to use polar coordinates:
the problem is to integrate the function in the domain $x^2+y^2<1$. This domain is the entire unit disk, say a disk with radius 1 and $\theta \in (0;2\pi)$. To pass in polar coordinates you have to make the following substitution;
$$ \begin{cases} x=\rho cos\theta \\ y=\rho sin\theta \end{cases}$$
It is very easy to calculate the jacobian writing the matrix of the partial derivatives that gives $|J|=\rho$
$$det \begin{bmatrix} cos\theta &-\rho sin\theta \\ sin\theta &\rho cos\theta \\ \end{bmatrix}=\rho(\underbrace{cos^2\theta+sin^2\theta}_{=1})=\rho$$
so, observing that $x^2+y^2=\rho^2$ you integral becomes
$$\underbrace{\int_{0}^{2\pi}d \theta}_{\text{angle from 0 to $2\pi$}}\underbrace{\int_{0}^{1}\rho\sqrt{1-\rho^2} d\rho}_{\text{radius $0<\rho<1$}}$$