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Math Help - Area of a region enclosed by polar equation

  1. #1
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    Area of a region enclosed by polar equation

    I have this problem that reads "Find the area enclosed by r^2 = 4cos(\theta)<br />
."

    so, what I ended up with

    2 <br />
\int_a^b cos(\theta) d\theta

    I may have messed up in finding the correct integral, but I think the problem is that I don't know what to choose for a or b. Can anyone help explain how I would go about choosing proper values for a and b?

    P.S. I seem to be having trouble with the LaTex, is anyone else having trouble?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by auslmar View Post
    I have this problem that reads "Find the area enclosed by r^2 = 4cos(\theta)<br />
."

    so, what I ended up with

    2 <br />
\int_a^b cos(\theta) d\theta

    I may have messed up in finding the correct integral, but I think the problem is that I don't know what to choose for a or b. Can anyone help explain how I would go about choosing proper values for a and b?

    P.S. I seem to be having trouble with the LaTex, is anyone else having trouble?
    Have you tried to graph the Lemniscate? The limits are a little tricky, but its not that complex...take into account that 4 \cos\theta can't be negative...

    Just wondering, what did you get for your limits of integration? You're integral set up is correct.

    --Chris
    Last edited by Chris L T521; July 27th 2008 at 07:17 PM. Reason: added something
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by auslmar View Post
    I have this problem that reads "Find the area enclosed by r^2 = 4cos(\theta)<br />
."

    so, what I ended up with

    2 <br />
\int_a^b cos(\theta) d\theta

    I may have messed up in finding the correct integral, but I think the problem is that I don't know what to choose for a or b. Can anyone help explain how I would go about choosing proper values for a and b?

    P.S. I seem to be having trouble with the LaTex, is anyone else having trouble?


    Does this make sense?

    --Chris
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  4. #4
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    Hello, auslmar!

    By the way, this is not a lemniscate.
    . . It could be called "the square root of a circle."


    I have this problem that reads: Find the area enclosed by r^2 = 4 \cos\theta

    so, what I ended up with: .2 <br />
\int_a^b cos(\theta) d\theta

    I may have messed up in finding the correct integral.

    Set r = 0 and solve for θ.

    . . 4 cos θ .= .0 . . θ .= .π

    and there are our limits . . .

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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, auslmar!

    By the way, this is not a lemniscate.
    . . It could be called "the square root of a circle."
    Um...





    It might not be a lemniscate per se, but it has the form of one.

    --Chris
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