Solve $\int_{0}^{1}\int_{0}^{1-y} xy\sqrt{(1-x-y)}dxdy$

Question

Calculate the double integral

$$∫∫xy\sqrt{1-x-y}\,dxdy$$

where the domain is $D=\{(x,y):x≥0,y≥0,x+y≤1\}$

I think the range is $0≤x≤1$ and $0≤y≤1−x$ . Is it correct? I understands that this problem needs polar transformation.But cant figure out what to substitute ?

I am getting stuck in this problem. Should I integrate by parts, or is there any other way to solve it? If I have to substitute, then what should I substitute? Please anyone help me solving it. Thanks in advance.

Answer 1

Yes, those are the limits of integration. And there is no need for polar coordinates, since\begin{align}\int_0^1\int_0^{1-x}xy\sqrt{1-x-y}\,\mathrm dy\,\mathrm dx&=\int_0^1x\int_0^{1-x}y\sqrt{1-x-y}\,\mathrm dy\,\mathrm dx\\&=\int_0^1\left[-\frac2{15}(-x-y+1)^{3/2}(-2x+3y+2)\right]_{y=0}^{y=1-x}\,\mathrm dx\\&=\int_0^1\frac4{15}x(1-x)^{5/2}\\&=\frac{16}{945},\end{align}where the second equality comes from the fact that$$\int y\sqrt{a-y}\,\mathrm dy=-\frac2{15}(a-y)^{3/2}(2a+3y).$$

Answer 2

I found another method by using beta function. Putting t=x/(1-y)=>x=t(1-y)=>dx=(1-y)dt

\begin{align}\int_0^1 \int_0^1 t(1-y)y\sqrt{1-t(1-y)-y} (1-y)dtdy&=\int_0^1 \int_0^1 y(1-y)^{5/2} t(1-t)^{1/2} dtdy\\&=\int_0^1y(1-y)^{5/2} dy \int_0^1 t(1-t)^{1/2} dt\\&=\int_0^1 y^{2-1} (1-y)^{7/2-1} dy\int_0^1 t^{2-1} (1-t)^{3/2-1} dt\\&=\beta(2,\frac{7}{2})\beta(2,\frac{3}{2})\\&=\frac{16}{945}\end{align}