Calculate the following integral: double integral over D
Calculate the following integral: $$ \iint_D x^2\,dx\,dy, $$ where $D$ is the quarter disk $x\geqslant0$, $y\geqslant0$, $x^2+y^2\leqslant a^2$.
Explanation of Method
Rather than integrating $x^2$ directly, note that swapping the roles of $x$ and $y$ leaves the quarter disk unchanged, so $\iint_D x^2\,dA=\iint_D y^2\,dA$. Their sum, $\iint_D(x^2+y^2)\,dA$, is easy to evaluate in polar coordinates, and each individual integral is exactly half of that sum.
- Use symmetry to note $\iint_D x^2\,dA=\iint_D y^2\,dA$.
- Evaluate the sum $\iint_D(x^2+y^2)\,dA$ in polar coordinates.
- Divide by $2$ to isolate $\iint_D x^2\,dA$.
Step 1 · Symmetry argument
The quarter disk $D$ is invariant under the reflection $x\leftrightarrow y$ (this reflection maps the region onto itself, since it is symmetric about the line $y=x$). Since the map only relabels the integration variables:
$$ \iint_D x^2\,dx\,dy = \iint_D y^2\,dx\,dy. $$
Call this common value $I$. Then:
$$ 2I = \iint_D x^2\,dx\,dy + \iint_D y^2\,dx\,dy = \iint_D (x^2+y^2)\,dx\,dy. $$
Step 2 · Evaluate the sum in polar coordinates
With $x=r\cos\theta$, $y=r\sin\theta$, the quarter disk becomes $0\leqslant r\leqslant a$, $0\leqslant\theta\leqslant\frac{\pi}{2}$, and $x^2+y^2=r^2$:
$$ \iint_D (x^2+y^2)\,dx\,dy = \int_0^{\pi/2}\int_0^a r^2\cdot r\,dr\,d\theta = \int_0^{\pi/2}d\theta\int_0^a r^3\,dr = \frac{\pi}{2}\cdot\frac{a^4}{4} = \frac{\pi a^4}{8}. $$
Step 3 · Isolate $I$
$$ 2I = \frac{\pi a^4}{8} \quad\Longrightarrow\quad I = \frac{\pi a^4}{16}. $$
Additional Notes
- The same argument shows $\iint_D y^2\,dx\,dy=\frac{\pi a^4}{16}$ as well, without any further computation.
Result
The solution to the integral is: