Let $m,n,p\in (0,+\infty)$ be constants and let $D$ be the region in the plane enclosed by lines $x=0,y=0$ and $x+y=1$. Calculate the following integral:
$$\iint\limits_{D}x^{m-1}y^{n-1}（1-x-y)^{p-1}\mathrm{d}x\mathrm{d}y$$
Note: I know how to calculate it by the routine procedure, but the answer computed it in another way: It turned the simplex $D$ into the region $\Omega$ in $\mathbb{R}^3$ which is enclosed by planes $x=0, y=0, z=0$ and $x+y+z=1$, it then claimed that the original integral is equal to $$(p-1)\iiint\limits_{\Omega}x^{m-1}y^{n-1}z^{p-2}\mathrm{d}x\mathrm{d}y\mathrm{d}z $$
But I could not get why the triple integral is equivalent to the original double integral?(I know how to compute the triple integral, but I just can't see how it's equivalent to the double integral.)
Best Answer
Let $\Omega = \{ (x,y,z) \;|\; (x,y) \in D \mbox{ and } 0 \leq z \leq 1-x-y \}$. Then turn the triple integral into a double $+$ a single integral...
$$(p-1)\iiint\limits_{\Omega} x^{m-1}y^{n-1}z^{p-2} \,dV = \iint\limits_D \left(\int_0^{1-x-y} (p-1)x^{m-1}y^{n-1}z^{p-2}\,dz\right)\,dA$$ $$= \iint_D \left. x^{m-1}y^{n-1}z^{p-1} \right|_0^{1-x-y}\,dA = \iint_D x^{m-1}y^{n-1}(1-x-y)^{p-1}\,dA$$