Convert to polar $\int_{-\sqrt{3}}^{\sqrt{3}}(-x^2+3) \, \mathrm dx$

Question

I took calculus about 2 semester ago, and I'm trying to brush up on polar coordinates. $$\int_{-\sqrt{3}}^{\sqrt{3}}-x^2+3 \, \mathrm dx$$ from and I got $6.93$

Now I tried to convert it to polar coordinates, but I'm having trouble setting up the integral. This what I did.

$-x^2+3$ => polar coordinates => $-(r\cos\theta)^2+3$

then I did, $\int_0^{2\pi} \int_0^\sqrt{3} ( (-(r\cos(\theta))^2+3)r\space drd\theta$ and when I evaluate this I get a different answer than the Cartesian coordinate integral.

I also tried this $$\int_0^{2\pi} \int_0^{(-(r\cos\theta)^2+3)*r}1\space drd\theta = ?? $$

Any ideas? sorry about typing the integral, I don't know the syntax for laTex.

Thank you

Answer 1

The region you are integrating over is the part of $y=3-x^2$ that lies above the $x$-axis. If you are going to describe it using polar coordinates, then $\theta$ should only range from $\theta=0$ (to consider the line from $(0,0)$ to $(\sqrt{3},0)$) to $\theta=\pi$ (which gives you the line from $(0,0)$ to $(-\sqrt{3},0)$.

So $\theta$ will only range from $0$ to $\pi$, not from $0$ to $2\pi$. What about $r$?

You want to express $r$ as a function of $\theta$. Your graph is $y=3-x^2$ (you seem to have forgotten the $y$...) So the curve you are trying to express would correspond to: $$\begin{align*} y &= 3-x^2\\ r\sin\theta &= 3-r^2\cos^2\theta \end{align*}$$ So you want to express $r$ as a function of $\theta$, so that you can have that in the "inner" integral (the limits of integration for $r$ may depend on $\theta$, but they should not depend on $r$). So we need to solve for $r$; this gives you a quadratic in $r$: $$(\cos^2\theta) r^2 + (\sin\theta)r - 3 = 0.$$ Thus, the graph corresponds to $$r = \frac{-\sin\theta \pm \sqrt{\sin^2\theta +12\cos^2\theta}}{2\cos^2\theta} = \frac{-\sin\theta \pm \sqrt{1 + 11\cos^2\theta}}{2\cos^2\theta}.$$ Since $r$ is positive for the region you want, you would use the $+$ sign. So the region you want would correspond to $$\begin{align*} 0 &\leq r \leq \frac{-\sin\theta + \sqrt{1+11\cos^2\theta}}{2\cos^2\theta}\\ 0 &\leq \theta \leq \pi. \end{align*}$$ And your integral would be $$\int_0^{\pi}\int_0^{\frac{-\sin\theta + \sqrt{1+11\cos^2\theta}}{2\cos^2\theta}} 1r\,dr\,d\theta$$ ... at which point I would grimace in disgust and switch back to cartesian coordinates, because this is definitely not a nice way to go...

(The issue, of course, is that while circles can be described nicely with polar coordinates, parabolas in general are somewhat nasty.)