Evaluating $\iint_R\big(x^2+y^2\big)\,dA$

Question

Evaluate the following double integral: $$\iint_R\big(x^2+y^2\big)\,dA,$$ where $R$ is the region given by plane $x^2+y^2\leq a^2$.

My attempts:

\begin{align} \iint_{R}\big(x^2+y^2\big)\,dA &=\int_{-a}^{a}\int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\big(x^2+y^2\big)\,dy\,dx\\ &=\int_{-a}^{a}\left(x^2y+\dfrac{y^3}{3}\right)_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\,dx\\ &=\dfrac{2}{3}\int_{-a}^{a}\sqrt{a^2-x^2}\cdot\left(2x^2+a^2\right)dx. \end{align}

I can't go further from here, please help.

Answer 1

Hint: Try $x=a\sin(t)$, but really polar coordinates would be the move here.

Answer 2

Whenever you have $\sqrt{a^2-x^2}$ in an integral, use the substitution $x=a \sin(\theta )$.

For information :

In case of having $\sqrt{x^2-a^2}$ use the substitution $x=a \sec(\theta )$.

In case of having $\sqrt{x^2+a^2}$ use the substitution $x=a \tan(\theta )$

Answer 3

$$\iint_R x^2+y^2 dA=\int_0^{2\pi}\int_0^a r^3drd\theta=2\pi\int_0^a r^3 dr$$

Answer 4

As has been hinted, polar coordinates are the way to go. But if you really need to continue, the first thing I would note is that your integrand is an even function of $x$ over a symmetric integral. Therefore, you can simplify a tad to $$\dfrac{4}{3}\int_{0}^{a}\sqrt{a^2-x^2}\cdot\left(2x^2+a^2\right)dx. $$ Now, whenever I see something like $\sqrt{a^2-x^2},$ I think trig substitution. In this case, you draw a triangle with $a$ as the hypotenuse, and one of the other sides as $x$. The third side will be $\sqrt{a^2-x^2}.$ I'll just put $\theta$ as the angle opposite $x,$ so that we have the following: \begin{align*} \frac{x}{a}&=\sin(\theta)\\ x&=a\sin(\theta)\\ dx&=a\cos(\theta)\,d\theta\\ \frac{\sqrt{a^2-x^2}}{a}&=\cos(\theta). \end{align*} For $x$ going from $0$ to $a,$ we can write that $0\le\theta\le\pi/2.$ Our integral becomes \begin{align*} \dfrac{4}{3}\int_{0}^{a}\sqrt{a^2-x^2}\cdot\left(2x^2+a^2\right)dx &=\dfrac{4}{3}\int_{0}^{\pi/2}a\cos(\theta)\cdot\left(2a^2\sin^2(\theta)+a^2\right)a\cos(\theta)\,d\theta\\ &=\frac{4a^4}{3}\int_0^{\pi/2}\cos^2(\theta)\left(2\sin^2(\theta)+1\right)d\theta. \end{align*} This integral breaks up into two pieces, each of which succumbs to the usual methods for integrating products of $\sin(\theta)$ and $\cos(\theta).$ Can you take it from here? The result should be $$\frac{4a^4}{3}\cdot\frac{3\pi}{8}=\frac{\pi a^4}{2}.$$ Check with polar version: \begin{align*} \iint_R\big(x^2+y^2\big)\,dA &=\int_0^{2\pi}\int_0^a r^2\cdot r\,dr\,d\theta\\ &=2\pi\,\frac{r^4}{4}\bigg|_0^a\\ &=\frac{\pi a^4}{2}, \end{align*} as required.