Calculate the following integral: double integral over D
Calculate the following integral: $$ \iint_D x^2y^2\sqrt{1-(x^3+y^3)}\,dx\,dy, $$ where $D$ is the region $x^3+y^3<1$, $x>0$, $y>0$.
Explanation of Method
The substitution $u=x^3$, $v=y^3$ flattens the cubic boundary into the flat triangle $u+v\leqslant1$. As in the closely related Beta-function integral seen elsewhere in this collection, the Jacobian factors $u^{-2/3}v^{-2/3}$ cancel exactly against the $x^2y^2=u^{2/3}v^{2/3}$ from the integrand, leaving a clean integral of $\sqrt{1-u-v}$ over the triangle.
- Substitute $u=x^3$, $v=y^3$ to flatten the region into a triangle.
- Simplify the integrand and Jacobian together.
- Integrate $\sqrt{1-u-v}$ over the triangle.
Step 1 · Substitution
Let $u=x^3$, $v=y^3$, so $x=u^{1/3}$, $y=v^{1/3}$, and:
$$ \frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix}\frac13u^{-2/3}&0\\[2pt]0&\frac13v^{-2/3}\end{vmatrix} = \frac19u^{-2/3}v^{-2/3}, \qquad dx\,dy=\frac{1}{9u^{2/3}v^{2/3}}\,du\,dv. $$
The region becomes the triangle $u,v>0$, $u+v<1$, and $x^2y^2=u^{2/3}v^{2/3}$, so:
$$ I = \iint_{u+v<1} u^{2/3}v^{2/3}\sqrt{1-u-v}\cdot\frac{1}{9u^{2/3}v^{2/3}}\,du\,dv = \frac19\iint_{u+v<1}\sqrt{1-u-v}\,du\,dv. $$
Step 2 · Set up the integral
$$ I = \frac19\int_0^1\int_0^{1-u}\sqrt{1-u-v}\,dv\,du. $$
Step 3 · Integrate with respect to $v$
With $w=1-u-v$:
$$ \int_0^{1-u}\sqrt{1-u-v}\,dv = \int_0^{1-u}\sqrt w\,dw = \frac23(1-u)^{3/2}. $$
Step 4 · Integrate with respect to $u$
$$ I = \frac19\cdot\frac23\int_0^1(1-u)^{3/2}\,du = \frac{2}{27}\int_0^1 t^{3/2}\,dt = \frac{2}{27}\cdot\frac25 = \frac{4}{135}. $$
Additional Notes
- Verified numerically: direct double integration over the original region gives $I\approx0.02963$, matching $\frac{4}{135}\approx0.02963$.
Result
The solution to the integral is: