Let $R$ be the region enclosed by $x^2+4y^2\geq 1$ and $x^2+y^2\leq 1$. Then the value of $\int \int_R |xy|dx dy$

Question

Let $R$ be the region enclosed by $x^2+4y^2\geq 1$ and $x^2+y^2\leq 1$. Then the value of $$\int \int_R |xy|dx dy$$ is _________

My attempt

I am getting $5/24$. But answer given is $.375$ Where is my mistake? Is there any short cut to solve this question?

Answer 1

Because the integrand is $|xy|$, I would just calculate the region in the first quadrant then multiply by $4$.

$4\int_0^1 \int_{\frac{1}{2}{\sqrt{1-x^2}}}^{\sqrt{1-x^2}} |xy| \; dy \;dx=4\int_0^1 \frac{x}{2} y^2 \bigg\rvert_{\frac{1}{2}{\sqrt{1-x^2}}}^{\sqrt{1-x^2}} \;dx=\frac{3}{2}\int_0^1 x-x^3 \;dx =\frac{3}{2}\cdot\frac{1}{4}=\boxed{\frac{3}{8}}$

Answer 2

The last step in your computation of part A has an error.

The integrand is $x(-\tfrac34+\tfrac34 x^2)$, which you incorrectly simplified to $-\tfrac34x +\tfrac34x^2$.

I haven't looked any further than that, so you should carefully check part B as well.

Answer 3

You can also consider converting to polar coordinates:

$$x^2+y^2=1\implies r^2=1\implies r=1$$

$$x^2+4y^2=1\implies r^2(1+3\sin^2\theta)=1\implies r=\frac1{\sqrt{1+3\sin^2\theta}}$$

Then

$$\begin{align*} \iint_R|xy|\,\mathrm dx\,\mathrm dy&=\int_0^{2\pi}\int_{\frac1{\sqrt{1+3\sin^2\theta}}}^1|r^2\cos\theta\sin\theta|r\,\mathrm dr\,\mathrm d\theta\\[1ex] &=\int_0^{2\pi}\int_{\frac1{\sqrt{1+3\sin^2\theta}}}^1r^3|\cos\theta\sin\theta|\,\mathrm dr\,\mathrm d\theta\\[1ex] &=\frac14\int_0^{2\pi}|\cos\theta\sin\theta|\left(1-\frac1{(1+3\sin^2\theta)^2}\right)\,\mathrm d\theta \end{align*}$$

We have $\cos\theta\sin\theta>0$ for $\theta\in\left(0,\frac\pi2\right)\cup\left(\pi,\frac{3\pi}2\right)$, and negative for $\theta\in\left(\frac\pi2,\pi\right)\cup\left(\frac{3\pi}2,2\pi\right)$.

Rewrite the integrand as

$$\cos\theta\sin\theta\frac{(1+3\sin^2\theta)^2-1}{(1+3\sin^2\theta)^2}=\frac{6\sin^3\theta+9\sin^5\theta}{(1+3\sin^2\theta)^2}\cos\theta$$

The integral over $\left(0,\frac\pi2\right)$ is

$$\frac14\int_0^{\frac\pi2}\frac{6\sin^3\theta+9\sin^5\theta}{(1+3\sin^2\theta)^2}\cos\theta\,\mathrm d\theta=\frac14\int_0^1\frac{9s^5+6s^3}{(1+3s^2)^2}\,\mathrm ds=\frac3{32}$$

(where we substitute $s=\sin\theta$)

You would find the same value over the other intervals, so the integral has a total value of $4\times\frac3{32}=\frac38$.