Evaluate $\iint_D (\sqrt{x}+\sqrt{y})^{100}\,dx\,dy$

Question

I'm having difficulty with evaluating the following integral

$\iint_D (\sqrt{x}+\sqrt{y})^{100}\,dx\,dy$

D is the area bounded by $\sqrt{x}+\sqrt{y}=1$ , $x=0$ and $y=0$;

I'm new to the subject of changing variables with double integrals. I'd really appreciate a slow, step by step explanation. Thank you :)

Answer 1

Let $T$ be the triangle whose vertices are $(0,0)$, $(1,0)$, and $(0,1)$ and, for each $(x,y)\in T$, let $g(x,y)=(x^2,y^2)$. Then$$g(T)=\{(x,y)\in\Bbb R^2\mid x,y\geqslant0\wedge\sqrt x+\sqrt y\leqslant1\}.$$Since $g$ is a class $C^1$ bijection from $T$ onto the region of integration of your integral, your integral is equal to$$\iint_T(x+y)^{100}|J(g)_{(x,y)}|\,\mathrm dx\,\mathrm dy.\tag1$$But$$|J(g)_{(x,y)}|=\left|\det\begin{bmatrix}2x&0\\0&2y\end{bmatrix}\right|=4xy.$$So,\begin{align}(1)&=\iint_T(x+y)^{100}4xy\,\mathrm dx\,\mathrm dy\\&=4\int_0^1\int_0^{1-x}(x+y)^{100}xy\,\mathrm dy\,\mathrm dx\\&=4\int_0^1\frac{x \left(x^{102}-102 x+101\right)}{10\,302}\,\mathrm dx\\&=\frac4{624}\\&=\frac1{156}.\end{align}

Answer 2

Use the polar coordinates inspired substitution

$$\begin{cases} x = r\cos^4\theta \\ y = r\sin^4\theta \\ \end{cases} \implies J = 4 r \sin^3\theta \cos^3\theta$$

The bounds become constants giving us the integral

$$\int_0^{\frac{\pi}{2}} \int_0^1 4r^{51}\sin^3\theta\cos^3\theta\:dr\:d\theta = \frac{1}{156}$$