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Math Help - Area and Arc Length in Polar Coordinates

  1. #1
    TWN
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    Area and Arc Length in Polar Coordinates

    I'm teaching myself calculus and I can't figure out how to do the problems below. Please explain how to do them because my book is confusing me.

    Find the area of the region:

    1.) Inside r = 2acos(theta) and outside r = a


    Find the length of the curve over the given interval:

    2.) r = 2acos(theta)

    3.) r = 8(1+cos(theta))


    Thank you!
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    Re: Area and Arc Length in Polar Coordinates

    Quote Originally Posted by TWN View Post
    I'm teaching myself calculus and I can't figure out how to do the problems below. Please explain how to do them because my book is confusing me.

    Find the area of the region:

    1.) Inside r = 2acos(theta) and outside r = a


    Find the length of the curve over the given interval:

    2.) r = 2acos(theta)

    3.) r = 8(1+cos(theta))


    Thank you!
    It always helps to draw a picture of the graphs, so that you can determine what the terminals of your double integral should be...
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  3. #3
    TWN
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    Re: Area and Arc Length in Polar Coordinates

    How do you determine the terminals? Can it be any two angles within which the figure exists?
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    Re: Area and Arc Length in Polar Coordinates

    Have you drawn the diagrams yet?
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  5. #5
    TWN
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    Re: Area and Arc Length in Polar Coordinates

    Yes I have. Now what do I do?

    Edit: I am going to bed. If someone could have some walkthroughs ready by morning it would be greatly appreaciated.
    Last edited by TWN; October 5th 2012 at 12:59 AM.
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    Re: Area and Arc Length in Polar Coordinates

    Don't be so rude - you don't get given walkthroughs on this site, you get given hints so that you can do your own work!

    If you convert to cartesians and use horizontal strips, you'll see that the x values are bounded between \displaystyle \begin{align*} a - \sqrt{1-y^2} \leq x \leq \sqrt{a^2 - y^2} \end{align*} and the y values are bounded between \displaystyle \begin{align*} -\frac{\sqrt{3}\,a}{2} \leq y \leq \frac{\sqrt{3}\,a}{2} \end{align*}. So the double integral is

    \displaystyle \begin{align*} \int_{-\frac{\sqrt{3}\,a}{2}}^{\frac{\sqrt{3}\,a}{2}}{ \int _{a - \sqrt{1 - y^2}}^{\sqrt{a^2 - y^2}}{\,dx}\,dy} \end{align*}
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