Integrate function $f(x,y) = y^2$ in $x^2+4y^2 \leq a^2$

Question

I am asked to integrate function $f(x,y) = y^2$ in $x^2+4y^2 \leq a^2$

To do that using polar coordinates, how may I find the boundaries for $r$? Is there a procedure that always works (for ellipses and circles)?

I know that the boundaries for $\theta$ are $0$ and $2 \pi$ therefore I must set up an integral

$$\int_{0}^{2\pi} \int_{\cdots}^{\cdots} \cdots r dr d\theta$$

Is there a procedure that always works (for ellipses and circles)?

Answer 1

Divide both sides of the inequality by $4$ to get $$\dfrac{x^2}{4} + y^2 \le \frac{a^2}{4}.$$

Then let $x = 2r\cos\theta$, $y = r\sin\theta$. So now you have: \begin{align} \frac{x^2}{4} + y^2 &\le \frac{a^2}{4}\\[0.3cm] \frac{4r^2\cos^2\theta}{4} + r^2\sin^2\theta &\le \frac{a^2}{4}\\[0.3cm] r^2\left(\cos^2\theta + \sin^2\theta\right) &\le \frac{a^2}{4}\\[0.3cm] r^2 &\le \frac{a^2}{4} \end{align}

This means $$ r \le \frac{a}{2}. $$

Because we also have $r \ge 0$ in general, and because $y^2 = r^2\sin^2\theta$, and because Carl Gustav Jacob Jacobi, the integral is now $$ \int_0^{2\pi} \int_0^{a/2} r^2\sin^2\theta \cdot 2r \, dr \, d\theta.$$

Let me know if you require further assistance.