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Math Help - Double intergral using polar

  1. #1
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    Double intergral using polar

    i'm sorry if this look messy and uni computers does not have latex installed, kindly please help to convert it to latex if possible and thank you in advance

    Evaluate the following interral using geometrical interpretation
    I= intergral( intergral( (a^2 -x^2 -y^2)^1/2 dxdy , where R is the disk given by
    R = {(x,y):x^2+y^2 (smaller or equal than) a^2

    the next step that i have done is to change it to intergral( intergral( (a^2 -r^2)^1/2 r.dr.d(digt)

    after that i'm lost can any kind soul enlighten me on further steps?

    and thank in advance to the guy if he manage to change this question to lastex format.
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  2. #2
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    So as I see it, you've got yourself the integral

    \iint\sqrt{a^{2}-r^{2}}\,r\,dr\,d\theta.

    Is that correct? If so, I would definitely try to figure out the limits for the integrals next. What are those?
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  3. #3
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    the domain given is R = {(x,y):x^2+y^2 (smaller or equal than) a^2
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  4. #4
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    Yes, but since you've got the integral defined in terms of polar coordinates (definitely the way to go here), you'd like to be able to describe your domain in polar coordinates. If you could do that, then you should be able to read off the limits for your integral. So how would you describe your domain in polar coordinates?
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  5. #5
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    i have no idea on how describe the domain in polar coordinates or how to do it, could u mind showing me how?
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  6. #6
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    Well, first describe the region in words. What does this look like? What's its shape?
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  7. #7
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    hemisphere
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  8. #8
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    Quote Originally Posted by salohcinseah View Post
    hemisphere
    No, I don't think you've quite got it. The word "hemisphere" implies three dimensions, but you're only working in two dimensions. What does the region

    \{(x,y):x^{2}+y^{2}\le 1\} look like? How about

    \{(x,y):x^{2}+y^{2}\le 4\}? And what about

    \{(x,y):x^{2}+y^{2}\le 9\}?
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  9. #9
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    x^2+y^2 = a circle
    r= radius

    since i know this , what is the next step?
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  10. #10
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    Technically, it's a disk (the circle is the boundary). That is, because you have an inequality in the sets of my last post, you're including all the points on a circle of radius a, as well as all the points inside that circle.

    So let's take our disk of radius a. What are the limits on r? What are the limits on theta?
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  11. #11
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    thus r cannot be more or equal to 4?
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