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Thread: Evaluating a repeated integral by changing to polar co-ordinates

  1. #1
    Senior Member chella182's Avatar
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    Evaluating a repeated integral by changing to polar co-ordinates

    Okay, so I missed my lectures on this and I haven't a clue I have my friend's assignment here but she's gone from step to step and I totally don't understand how. The question is:

    Evaluate by changing to polar co-ordinates
    $\displaystyle \int_{0}^{1}\int_{0}^{\sqrt{1-x^2}}\sin(\frac{\pi(x^2+y^2)}{2})dydx$

    My friend has then wrote down $\displaystyle x=r\cos\theta$ and $\displaystyle y=r\sin\theta$ then jumped to...

    $\displaystyle \int_{0}^{\pi/2}\int_{0}^{1}\sin(\frac{\pi r^2}{2})r drd\theta$

    ...and I don't know how. Can anyone go through it explaining the steps to me?
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  2. #2
    MHF Contributor
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    Hi

    Have a look to the scheme below
    $\displaystyle \int_{0}^{1}\int_{0}^{\sqrt{1-x^2}}f(x,y)dydx$ means that you integrate f on the domain defined by $\displaystyle 0 \leq x \leq 1$ and for each value of x $\displaystyle 0 \leq y \leq \sqrt{1-x^2}$

    When you plot the function $\displaystyle y = \sqrt{1-x^2}$ you can see that it is a quarter of a circle.
    x between 0 and 1, and y between 0 and $\displaystyle \sqrt{1-x^2}$ defines the quarter of the disk as area of integration (see red line).

    If you want to change to polar co-ordinates, you have to sweep the same domain using $\displaystyle \theta$ and r.

    You can see that you can manage this by choosing $\displaystyle 0 \leq \theta \leq \frac{\pi}{2} $ and for each value of $\displaystyle \theta$ $\displaystyle 0 \leq r \leq 1$

    Now you have to substitute $\displaystyle x=r\cos\theta$ and $\displaystyle y=r\sin\theta$
    $\displaystyle \sin(\frac{\pi(x^2+y^2)}{2}) = \sin(\frac{\pi r^2}{2}) $

    and $\displaystyle dx dy$ by $\displaystyle r drd\theta$

    to get finally

    $\displaystyle \int_{0}^{1}\int_{0}^{\sqrt{1-x^2}}\sin(\frac{\pi(x^2+y^2)}{2})dydx = \int_{0}^{\pi/2}\int_{0}^{1}\sin(\frac{\pi r^2}{2})r drd\theta$

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  3. #3
    Senior Member chella182's Avatar
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    Thankyou it's a lot clearer than it was, anyway. I just didn't want to blindly copy the work without actually knowing what I was copying / if what I was copying was actually right. I don't get how she's managed to get between those two steps with no working out in between like haha!
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  4. #4
    Senior Member chella182's Avatar
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    I get that to change to polar co-ords that $\displaystyle x=r\cos{\theta}$ and $\displaystyle y=r\sin{\theta}$.
    But then, on my assignment I've written...

    $\displaystyle \sqrt{1-x^2}$ makes a quarter circle, so $\displaystyle 0 \leq \theta \leq \frac{\pi}{2}$ and $\displaystyle 0 \leq r \leq 1$

    ...but I totally don't get where I've pulled that from
    Last edited by mr fantastic; Jan 16th 2009 at 04:40 AM.
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  5. #5
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    Quote Originally Posted by chella182 View Post
    I get that to change to polar co-ords that $\displaystyle x=r\cos{\theta}$ and $\displaystyle y=r\sin{\theta}$.
    But then, on my assignment I've written...

    $\displaystyle \sqrt{1-x^2}$ makes a quarter circle, so $\displaystyle 0 \leq \theta \leq \frac{\pi}{2}$ and $\displaystyle 0 \leq r \leq 1$


    ...but I totally don't get where I've pulled that from
    What part of the explanation you've been given in this thread do you not understand?

    Note: $\displaystyle y = \sqrt{1 - x^2} \Rightarrow y^2 = 1 - x^2 \Rightarrow x^2 + y^2 = 1$ which is a circle. You only want the part of the circle given by 0 < x < 1 and y > 0, that is, the part of the circle lying in the first quadrant.
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  6. #6
    Senior Member chella182's Avatar
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    I posted this ages ago. I didn't understand where the 0 and 1 came from on the outer limits. I kind of do now. Flunked this question on the exam still though
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