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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?2012-06-24
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    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 $r$.2012-06-24
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    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 coordinates2012-06-24
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    @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.2012-06-24
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    Arturo, That's a good point it should be from 0 to pi2012-06-24

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