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

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If you integrating only one variable, why do you need to go to polar coordinates? –  user17762 Jun 24 '12 at 4:30
    
Why are you letting $\theta$ range from $0$ to $2\pi$? The graph of $y=3-x^2$ from $x=-\sqrt{3}$ to $x=\sqrt{3}$ is completely on the top half of the plane, so if you are using polar coordinates, your $\theta$ only varies from $0$ to $\pi$, not to $2\pi$. And the limits of integration for $r$ cannot depend on $r$. –  Arturo Magidin Jun 24 '12 at 4:49
    
I'm just playing around with polar coordinates. I know it's easy to use Cartesian coordinates. But I just want to find out if it's possible to use polar coordinates –  user34369 Jun 24 '12 at 4:50
    
@user34369: But the question is: why are you setting up the limits you are setting up? The description using polar coordinates has to agree with the one in cartesian, and letting $\theta$ range from $0$ to $2\pi$ does not make any sense. –  Arturo Magidin Jun 24 '12 at 4:51
    
Arturo, That's a good point it should be from 0 to pi –  user34369 Jun 24 '12 at 4:51

2 Answers 2

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.)

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Thank you very MUCH I learned a lot from your explanation. It was very clear. I tried to evaluate the integral on Maple and I got a very long function and I wasn't able to get a decimal answer –  user34369 Jun 24 '12 at 5:25

You have established that $\int_{-\sqrt{3}}^{\sqrt{3}} (3 - x^2) dx = 4 \sqrt{3} $ in the standard Cartesian coordinates, and wished to attempt the same integration using polar coordinates. As Arturo Magidin demonstrated, this is no tidy matter if one makes a direct transformation of the curve.

The difficulty is that curves that do not pass through the origin generally do not have simple expressions in polar coordinates, as one observes even with straight lines $ax + by = c $ , with $a, b, c \ne 0$. Perhaps it will be helpful to find a way to produce a parabola which has one "arm" passing through $( 0, 0)$ .

What we want, after all, is the area of a parabola of the same size, bounded by the x-axis. If we translate $y = 3 - x^2$ to the right by $\sqrt{3}$ units, we obtain $y = 3 - (x - \sqrt{3})^2 = 2 \sqrt{3} \cdot x - x^2$ , which now has intercepts at $x = 0$ and $x = 2 \sqrt{3}$ . The transformation of this expression into polar coordinates produces

$$ r \sin \theta = 2 \sqrt{3} \cdot r \cos \theta - r^2 \cos^2 \theta \rightarrow r \cos^2 \theta = 2 \sqrt{3} \ \cos \theta - \sin \theta $$

$$\rightarrow r(\theta) = \frac{2 \sqrt{3} \ \cos \theta - \sin \theta}{\cos^2 \theta} .$$

enter image description here

It is clear that $r(0) = 2 \sqrt{3}$ , so the lower limit of our integration over angle will be $\theta= 0$. We will be able to cover the area under our translated parabola by advancing $\theta$ to the upper limit of integration. Although the radius function appears problematical, we never reach $\theta = \frac{\pi}{2}$, as we have $r(\theta) = 0$ at an angle we will define by $\tan \Theta = 2 \sqrt{3}$ .

We can now set up our "polar area integral" as

$$\int_0^{\Theta} \frac{1}{2} \cdot [r({\theta})]^2 \ d\theta \ = \ \frac{1}{2}\int_0^{\Theta} ( \frac{2 \sqrt{3} }{ \cos \theta} - \frac{\sin \theta}{\cos^2 \theta})^2 \ d\theta $$

$$= \frac{1}{2}\int_0^{\Theta} ( 12 \sec^2 \theta \ - \ 4 \sqrt{3} \cdot \frac{\sin \theta }{ \cos^3 \theta} \ + \ \frac{\sin^2 \theta}{\cos^4 \theta}) \ d\theta $$

$$= \frac{1}{2} [\ 12 \tan\theta \ - \ \ 4 \sqrt{3} \cdot (\frac{\sec^2 \theta }{ 2}) \ + \ (\frac{\tan^3 \theta}{3}) \ ] |_0^{\Theta} $$

$$= \ 6 \cdot (\tan\Theta \ - \ 0 ) \ - \ \sqrt{3} \cdot ( \sec^2 \Theta \ - \ \sec^2 0 ) \ + \ \frac{1}{6} \cdot ( \tan^3 \Theta \ - 0 ) $$

$$= \ 6 \tan\Theta \ - \ \sqrt{3} \cdot ( [ \ \tan^2 \Theta \ + \ 1 \ ] - \ 1 ) \ + \ \frac{1}{6} \cdot \tan^3 \Theta $$

$$= \ 6 \cdot 2\sqrt{3} \ - \ \sqrt{3} \cdot (\ 2\sqrt{3} \ )^2 \ + \ \frac{1}{6} \cdot ( \ 2\sqrt{3} \ )^3 \ = \ 12 \sqrt{3} \ - \ 12 \sqrt{3} \ + \ 4 \sqrt{3} \ = \ 4 \sqrt{3} \ . $$

This method should work for finding the area enclosed by either coordinate axis and a parabola opening toward that axis and with its symmetry axis perpendicular to the coordinate axis. (I just can't imagine why anyone would prefer that to using rectangular coordinates...)

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