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Thread: 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). $\displaystyle \int_0^6 \int_0^y x~dx~dy$
    ....

    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 $\displaystyle \theta$ limits are $\displaystyle \frac{\pi}{4} \leq \theta \leq \frac{\pi}{2}$.

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

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

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

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

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

    Clearly the r limits are incorrect!

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

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

    Is this integral the correct one to evaluate?
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  2. #2
    MHF Contributor
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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 $\displaystyle \theta$ limits are $\displaystyle \frac{\pi}{4} \leq \theta \leq \frac{\pi}{2}$.

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



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

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

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

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

    Clearly the r limits are incorrect!

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

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

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

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

    $\displaystyle
    \int_0^{\pi/2} \int_0^{2 \cos \theta} \sin \theta + \cos \theta dr d\theta
    $
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