Calculate the integral over the half ellipse

Question

Calculate the integral $\displaystyle \iint_E x^2 dA$ where E is the half ellipse given by the inequalities $x^2 + 2y^2 \le 2$ and $y \ge 0$.

This is what I did: $$\begin {align}\iint_E x^2 dA &=\iint r^3 \cos^2(\theta) dr d\theta\\ &= \int_1^{\sqrt2}r^3 \times \int_0^{\pi} \frac12(\cos(2\theta)+1) dr d \theta\\ &= \left[\frac{r^4}4\right]_1^{\sqrt2}\left[\frac{\sin(2\theta)}2+ \frac{\theta}2\right]_0^{\pi}\\ &= \left(0+\frac{\pi}2-0\right)\left (\frac44 - \frac14\right)\\ &= \frac{3\pi}4\end{align}$$

This is wrong and the correct answer is $\dfrac{\sqrt{2\pi}}4$

Answer 1

The correct and obviously best possible way to evaluate the integral is given in the preceding answer. I just want to show the error in your calculation.

You had obviously decided to compute the integral in the polar coordinate system with $x=r\cos\theta$, and $y=r\sin\theta$. In this coordinate system the ellipse has the equation: $$ \frac{r^2\cos^2\theta}{a^2}+\frac{r^2\sin^2\theta}{b^2}=1 $$ or $$ r=\frac1{\sqrt{\frac{\cos^2\theta}{a^2}+\frac{\sin^2\theta}{b^2}}}. $$ Therefore the correct way to set the limits in the integral is: $$ \int\limits_0^\pi\cos^2\theta\,d\theta\int\limits_0^{\frac1{\sqrt{\frac{\cos^2\theta}{a^2}+\frac{\sin^2\theta}{b^2}}}}r^3\, dr =\frac14\int\limits_0^\pi\frac{\cos^2\theta\,d\theta}{\left(\frac{\cos^2\theta}{a^2}+\frac{\sin^2\theta}{b^2}\right)^2}=\frac18\pi a^3b, $$ where I skipped rather boring steps for evaluation of the last integral.

Answer 2

Your limits for $r$ are wrong. You are actually evaluating the integral over a half ring with inner radius $1$ and outer radius $\sqrt{2}$.

Note that by letting $x=\sqrt{2}r\cos(\theta)$ and $y=r\sin(\theta)$ then the half ellipse $E$ is obtained for $(r,\theta)\in [0,1]\times [0,\pi]$. Hence $$\iint_E x^2 dA=\int_{r=0}^1\int_{\theta=0}^{\pi}(\sqrt{2}r\cos(\theta))^2\sqrt{2}r drd\theta=2\sqrt{2}\left[\frac{r^4}4\right]_0^1\frac{\pi}{2}=\boxed{\frac{\sqrt{2}\pi}4}.$$ Are you sure that in the correct answer there is a square-root of $\pi$?

P.S. In general, for an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}\leq 1$ you should apply the change of coordinates $x=ar\cos(\theta)$ and $y=br\sin(\theta)$. Then $$\frac{x^2}{a^2}+\frac{y^2}{b^2}\leq 1\Leftrightarrow \frac{a^2r^2\cos^2(\theta)}{a^2}+\frac{b^2r^2\sin^2(\theta)}{b^2}\leq 1\Leftrightarrow r^2\leq 1$$ that is $r\in [0,1]$. Moreover $dxdy$ is replaced by $ab\,rdr\,d\theta$.