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Math Help - Integrating using polar co-ordinates.

  1. #1
    Super Member Showcase_22's Avatar
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    Integrating using polar co-ordinates.

    I ran into some trouble with some integrals that use polar coordinates.

    Change the following cartesian integrals into an equivalent polar integral. Then evaluate the polar integral.
    Sounds fair!

    1). \int_0^6 \int_0^y x~dx~dy
    ....

    The region for integration is a triangle which I thought was a little weird. I think the r limits start from 0 to something (I called it b in the integral below), but I think the \theta limits are \frac{\pi}{4} \leq \theta \leq \frac{\pi}{2}.

    So, without the final r limit, I have \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \int_0^b r^2cos(\theta)~dr~d \theta.

    2). \int_0^2 \int_0^{\sqrt{1-(x-1)^2}}\frac{x+y}{x^2+y^2}~dy~dx
    My problem with this one is that the region of integration is not a circle centred at the origin but rather the point (1,0). I am aware that the region is a semi circle. For this reason, are the \theta limits 0 \leq \theta \leq \pi?

    Are the r limits 0 \leq r \leq \sqrt{1-(rcos(\theta)-1)^2}?

    However, the integrand is going to be \frac{r(rcos(\theta)+rsin(\theta))}{r^2}=cos(\thet  a)+sin(\theta).

    So overall, I get that the integral should be \int_0^{\sqrt{1-(rcos(\theta)-1)^2}} \int_0^{\pi} cos(\theta)+sin(\theta)~dr~d \theta.

    Clearly the r limits are incorrect!

    From here, I decided to put \theta=\pi into the upper limit of the double integral:

    \int_0^{\sqrt{r^2+2r}} \int_0^{\pi} cos(\theta)+sin(\theta)~dr~d \theta=\int_0^{\pi} \int_0^{\sqrt{r^2+2r}} cos(\theta)+sin(\theta)~d \theta dr.

    Is this integral the correct one to evaluate?
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  2. #2
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    Quote Originally Posted by Showcase_22 View Post
    I ran into some trouble with some integrals that use polar coordinates.



    Sounds fair!



    ....

    The region for integration is a triangle which I thought was a little weird. I think the r limits start from 0 to something (I called it b in the integral below), but I think the \theta limits are \frac{\pi}{4} \leq \theta \leq \frac{\pi}{2}.

    So, without the final r limit, I have \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \int_0^b r^2cos(\theta)~dr~d \theta.



    My problem with this one is that the region of integration is not a circle centred at the origin but rather the point (1,0). I am aware that the region is a semi circle. For this reason, are the \theta limits 0 \leq \theta \leq \pi?

    Are the r limits 0 \leq r \leq \sqrt{1-(rcos(\theta)-1)^2}?

    However, the integrand is going to be \frac{r(rcos(\theta)+rsin(\theta))}{r^2}=cos(\thet  a)+sin(\theta).

    So overall, I get that the integral should be \int_0^{\sqrt{1-(rcos(\theta)-1)^2}} \int_0^{\pi} cos(\theta)+sin(\theta)~dr~d \theta.

    Clearly the r limits are incorrect!

    From here, I decided to put \theta=\pi into the upper limit of the double integral:

    \int_0^{\sqrt{r^2+2r}} \int_0^{\pi} cos(\theta)+sin(\theta)~dr~d \theta=\int_0^{\pi} \int_0^{\sqrt{r^2+2r}} cos(\theta)+sin(\theta)~d \theta dr.

    Is this integral the correct one to evaluate?
    Problem 1, r will go from the origin to the line y = 6 so r = 0 to r \sin \theta = 6 or r = 6 \csc \theta

    Problem 2. It's best to draw a picture of the region. It a half circle (upper half) centered at x =1 with radius r=1. The polar equation for this is r = 2 \cos \theta. To sweep our the area \theta = 0 \to \pi/2 and so you have

     <br />
\int_0^{\pi/2} \int_0^{2 \cos \theta} \sin \theta + \cos \theta dr d\theta<br />
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