Calculate the following integral: double integral over D
Calculate the following integral: $$ \iint_D \frac{dx\,dy}{\sqrt{x}+\sqrt{y}}, $$ where $D$ is the triangle $0\leqslant y\leqslant x\leqslant1$ (vertices $(0,0)$, $(1,0)$, $(1,1)$).
Explanation of Method
Integrating first with respect to $y$ at fixed $x$, the substitution $y=xu^2$ rescales the inner integral into a simple rational form in $u$, producing a clean $\sqrt x\,(1-\ln2)$ factor. The outer integral in $x$ is then a routine power-rule integration.
- Integrate with respect to $y$ using the substitution $y=xu^2$.
- Simplify the resulting rational integral in $u$.
- Integrate the resulting $2\sqrt x(1-\ln2)$ with respect to $x$ over $[0,1]$.
Step 1 · Set up the integral
For each $x\in[0,1]$, $y$ runs from $0$ to $x$:
$$ I = \int_0^1\int_0^x \frac{dy}{\sqrt x+\sqrt y}\,dx. $$
Step 2 · Integrate with respect to $y$
Fix $x>0$ and substitute $y=xu^2$, so $\sqrt y=\sqrt x\,u$ and $dy=2xu\,du$; as $y$ runs from $0$ to $x$, $u$ runs from $0$ to $1$:
$$ \int_0^x \frac{dy}{\sqrt x+\sqrt y} = \int_0^1 \frac{2xu\,du}{\sqrt x(1+u)} = 2\sqrt x\int_0^1 \frac{u}{1+u}\,du. $$
Step 3 · Evaluate the $u$-integral
$$ \int_0^1 \frac{u}{1+u}\,du = \int_0^1\left(1-\frac{1}{1+u}\right)du = \Big[u-\ln(1+u)\Big]_0^1 = 1-\ln2. $$
So the inner integral is:
$$ \int_0^x \frac{dy}{\sqrt x+\sqrt y} = 2\sqrt x\,(1-\ln2). $$
Step 4 · Integrate with respect to $x$
$$ I = 2(1-\ln2)\int_0^1 \sqrt x\,dx = 2(1-\ln2)\left[\frac{2}{3}x^{3/2}\right]_0^1 = 2(1-\ln2)\cdot\frac23 = \frac{4}{3}(1-\ln2). $$
Additional Notes
- Since the integrand is symmetric under $x\leftrightarrow y$, this triangle contributes exactly half of $\iint_{[0,1]^2}\frac{dx\,dy}{\sqrt x+\sqrt y}$ — a useful consistency check, as the full unit square's value works out to $\frac{8}{3}(1-\ln2)$, twice the triangle's value.
- Numerically, $I\approx\frac{4}{3}(1-0.6931)\approx0.4091$.
Result
The solution to the integral is: