Calculate the following integral: double integral over D
Calculate the following integral: $$ \iint_D \sqrt{\sqrt{x}+\sqrt{y}}\,dx\,dy, $$ where $D$ is the region bounded by the coordinate axes and the curve $\sqrt{x}+\sqrt{y}=1$.
Explanation of Method
The substitution $u=\sqrt x$, $v=\sqrt y$ turns the awkward boundary $\sqrt x+\sqrt y=1$ into the straight line $u+v=1$, at the cost of picking up a Jacobian factor of $4uv$. The resulting polynomial-times-square-root integrand is then handled with an inner substitution $w=u+v$.
- Substitute $u=\sqrt x$, $v=\sqrt y$, turning the region into the triangle $u+v\leqslant1$.
- Integrate with respect to $v$ using $w=u+v$.
- Integrate the resulting polynomial in $u$.
Step 1 · Substitution
Let $u=\sqrt x$, $v=\sqrt y$, so $x=u^2$, $y=v^2$, and the Jacobian is:
$$ \frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix}2u&0\\0&2v\end{vmatrix} = 4uv, \qquad dx\,dy = 4uv\,du\,dv. $$
The region becomes the triangle $u\geqslant0$, $v\geqslant0$, $u+v\leqslant1$, and $\sqrt{\sqrt x+\sqrt y}=\sqrt{u+v}$:
$$ I = 4\iint_{u+v\leqslant1} uv\sqrt{u+v}\,du\,dv. $$
Step 2 · Set up and integrate with respect to $v$
$$ I = 4\int_0^1\int_0^{1-u} uv\sqrt{u+v}\,dv\,du. $$
Fix $u$ and let $w=u+v$, so $v=w-u$, $dv=dw$; as $v$ runs $0\to1-u$, $w$ runs $u\to1$:
$$ \int_0^{1-u} v\sqrt{u+v}\,dv = \int_u^1 (w-u)\sqrt w\,dw = \int_u^1\left(w^{3/2}-uw^{1/2}\right)dw. $$
$$ = \left[\frac{2w^{5/2}}{5}-\frac{2uw^{3/2}}{3}\right]_u^1 = \left(\frac25-\frac{2u}{3}\right) - \left(\frac{2u^{5/2}}{5}-\frac{2u^{5/2}}{3}\right). $$
Since $\frac13-\frac15=\frac{2}{15}$:
$$ \int_0^{1-u} v\sqrt{u+v}\,dv = \frac25-\frac{2u}{3}+\frac{4u^{5/2}}{15}. $$
So:
$$ \int_0^{1-u} uv\sqrt{u+v}\,dv = \frac{2u}{5}-\frac{2u^2}{3}+\frac{4u^{7/2}}{15}. $$
Step 3 · Integrate with respect to $u$
$$ I = 4\int_0^1\left(\frac{2u}{5}-\frac{2u^2}{3}+\frac{4u^{7/2}}{15}\right)du = 4\left[\frac{u^2}{5}-\frac{2u^3}{9}+\frac{8u^{9/2}}{135}\right]_0^1. $$
With common denominator $135$: $\frac15=\frac{27}{135}$, $\frac29=\frac{30}{135}$, $\frac{8}{135}=\frac{8}{135}$:
$$ I = 4\left(\frac{27}{135}-\frac{30}{135}+\frac{8}{135}\right) = 4\cdot\frac{5}{135} = \frac{20}{135} = \frac{4}{27}. $$
Additional Notes
- Verified numerically: direct double integration gives $I\approx0.1481$, matching $\frac{4}{27}\approx0.1481$.
Result
The solution to the integral is: