Calculate the following integral: double integral over D
Calculate the following integral: $$ \iint_D x^2y\sqrt{1-x^3-y^3}\,dx\,dy, $$ where $D$ is the region $x\geqslant0$, $y\geqslant0$, $x^3+y^3\leqslant1$.
Explanation of Method
The substitution $u=x^3$, $v=y^3$ removes all the cube-root structure from the region (turning it into the flat triangle $u+v\leqslant1$) and, remarkably, the extra Jacobian factors combine with $x^2$ to cancel exactly, leaving a clean integrand $v^{-1/3}(1-u-v)^{1/2}$. A further substitution reduces the $u$-integral to a power function, and the remaining $v$-integral is precisely the Beta function integral.
- Substitute $u=x^3$, $v=y^3$ to flatten the region into a triangle.
- Integrate with respect to $u$ using $w=1-u-v$.
- Recognize the remaining $v$-integral as a Beta function integral.
Step 1 · Change of variables
Let $u=x^3$, $v=y^3$, so $x=u^{1/3}$, $y=v^{1/3}$, $dx=\frac13u^{-2/3}\,du$, $dy=\frac13v^{-2/3}\,dv$. The region becomes $u\geqslant0,\,v\geqslant0,\,u+v\leqslant1$, and:
$$ x^2y\sqrt{1-x^3-y^3}\,dx\,dy = u^{2/3}v^{1/3}\sqrt{1-u-v}\cdot\frac19 u^{-2/3}v^{-2/3}\,du\,dv = \frac19 v^{-1/3}\sqrt{1-u-v}\,du\,dv. $$
$$ I = \frac19\int_0^1\int_0^{1-v} v^{-1/3}(1-u-v)^{1/2}\,du\,dv. $$
Step 2 · Integrate with respect to $u$
Fix $v$ and let $w=1-v-u$; when $u=0$, $w=1-v$, and when $u=1-v$, $w=0$:
$$ \int_0^{1-v}(1-u-v)^{1/2}\,du = \int_0^{1-v} w^{1/2}\,dw = \frac23(1-v)^{3/2}. $$
$$ I = \frac{2}{27}\int_0^1 v^{-1/3}(1-v)^{3/2}\,dv. $$
Step 3 · Recognize the Beta function
Using $\displaystyle B(p,q)=\int_0^1 v^{p-1}(1-v)^{q-1}\,dv$ with $p-1=-\frac13$ and $q-1=\frac32$, i.e. $p=\frac23$, $q=\frac52$:
$$ I = \frac{2}{27}\,B\!\left(\frac23,\frac52\right). $$
Step 4 · Express in terms of the Gamma function
$$ B\!\left(\frac23,\frac52\right) = \frac{\Gamma\!\left(\frac23\right)\Gamma\!\left(\frac52\right)}{\Gamma\!\left(\frac{19}{6}\right)}, \qquad \Gamma\!\left(\frac52\right)=\frac34\sqrt\pi, $$
so:
$$ I = \frac{2}{27}\cdot\frac{\frac34\sqrt\pi\,\Gamma\!\left(\frac23\right)}{\Gamma\!\left(\frac{19}{6}\right)} = \frac{\sqrt\pi\,\Gamma\!\left(\frac23\right)}{18\,\Gamma\!\left(\frac{19}{6}\right)}. $$
Additional Notes
- Numerically, $I=\frac{2}{27}B\!\left(\frac23,\frac52\right)\approx0.0569$, confirmed by direct numerical double integration over $D$.
Result
The solution to the integral is: