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Math Help - Change integral to polar coordinates and evaluate.

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    Change integral to polar coordinates and evaluate.

    Hi, I am having difficulty picturing what to do with this double integral up to a certain point. I have scanned my work and ask that someone take a look at it and see where I got up to and tell me what I should do next. Integrate by parts? Substitution? I'm not sure. I have attached my scans to this thread. Thanks.
    Attached Thumbnails Attached Thumbnails Change integral to polar coordinates and evaluate.-scan0006.jpg   Change integral to polar coordinates and evaluate.-scan0007.jpg  
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    Quote Originally Posted by Undefdisfigure View Post
    Hi, I am having difficulty picturing what to do with this double integral up to a certain point. I have scanned my work and ask that someone take a look at it and see where I got up to and tell me what I should do next. Integrate by parts? Substitution? I'm not sure. I have attached my scans to this thread. Thanks.
    Shouldn't your circle be centered at the origin in image 1?
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    Yes it should, I'll fix that. But what should I do with this integral? Even with the image
    fixed, the integral would be set up the same way.
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    Quote Originally Posted by Undefdisfigure View Post
    Yes it should, I'll fix that. But what should I do with this integral? Even with the image
    fixed, the integral would be set up the same way.
    U-Sub

    u=r\cos(\theta)
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    Quote Originally Posted by dwsmith View Post
    U-Sub

    u=r\cos(\theta)
    Provided you are integrating with respect to \displaystyle \theta first...
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    So do I have to switch the order of integration or is there a way to integrate with respect to r first?
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    You'll have to switch the order...
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    But du would equal -r sin(theta) and it's r^2 in front of the sin(theta). Do I remove the r (1/r) and put it in front of the integral?
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    When integrating with respect to \displaystyle \theta you treat \displaystyle r as a constant.
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    I sent a private message and I don't know if it was received but this is what it was:

    In the question I asked, when you substitute u = r cos(theta), you get
    du = - r sin(theta). When treating r as a constant, does that mean it doesn't matter if it's r^2 sin (theta) and you just integrate e^u du with respect to theta?
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    Quote Originally Posted by Undefdisfigure View Post
    I sent a private message and I don't know if it was received but this is what it was:

    In the question I asked, when you substitute u = r cos(theta), you get
    du = - r sin(theta). When treating r as a constant, does that mean it doesn't matter if it's r^2 sin (theta) and you just integrate e^u du with respect to theta?
    You have this integral

    \iint_{R}ye^{x}dA

    Going to polar gives

    \displaystyle \int_{0}^{5}\int_{0}^{\frac{\pi}{2}} r\sin(\theta)e^{r\cos(\theta)}r d\theta dr

    Then using the recommend substitution gives

    u=r\cos(\theta) \implies du=-r\sin(\theta)d\theta
    u=r\cos(0) \implies u=r and \displaystyle u=r\cos\left(\frac{\pi}{2}\right) \implies u=0

    \displaystyle \int_{0}^{5} \left(-\int_{r}^{0}re^{u}du \right) dr =\int_{0}^{5}\int_{0}^{r}re^{u}dudr
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