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Math Help - Easy question but I forgot how to do it

  1. #1
    s3a
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    Easy question but I forgot how to do it

    Question:
    Find such that the area of the region enclosed by the parabolas and is .

    My problem probably lies with the limits of integration. When I get x^2 = c^2 x = c, I assumed the limits of integration would be -c and c but I can't justify it not to mention the answer is wrong.

    Can someone please help me?

    Any help would be greatly appreciated!
    Thanks in advance!
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  2. #2
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    Quote Originally Posted by s3a View Post
    Question:
    Find such that the area of the region enclosed by the parabolas and is .

    My problem probably lies with the limits of integration. When I get x^2 = c^2 x = c, I assumed the limits of integration would be -c and c but I can't justify it not to mention the answer is wrong.

    Can someone please help me?

    Any help would be greatly appreciated!
    Thanks in advance!
    x^2-c^2 < c^2-x^2 for -c < x < c

    150 = \int_{-c}^c (c^2-x^2) - (x^2-c^2) \, dx

    150 = 4\int_0^c c^2-x^2 \, dx

    \frac{75}{2} = \left[c^2 x - \frac{x^3}{3}\right]_0^c

    \frac{75}{2} = c^3 - \frac{c^3}{3}

    \frac{75}{2} = \frac{2c^3}{3}

    \frac{225}{4} = c^3

    c = \sqrt[3]{\frac{225}{4}}
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  3. #3
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by s3a View Post
    Question:
    Find such that the area of the region enclosed by the parabolas and is .

    My problem probably lies with the limits of integration. When I get x^2 = c^2 x = c, I assumed the limits of integration would be -c and c but I can't justify it not to mention the answer is wrong.

    Can someone please help me?

    Any help would be greatly appreciated!
    Thanks in advance!
    already was answer
    Last edited by 11rdc11; May 19th 2010 at 03:17 PM. Reason: already was answered
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  4. #4
    Senior Member
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    Quote Originally Posted by s3a View Post
    Question:
    Find such that the area of the region enclosed by the parabolas and is .

    My problem probably lies with the limits of integration. When I get x^2 = c^2 x = c, I assumed the limits of integration would be -c and c but I can't justify it not to mention the answer is wrong.

    Can someone please help me?

    Any help would be greatly appreciated!
    Thanks in advance!
    If you are still wondering about why the limits of integration are -c<x<c, you of course know we set the two expressions equal:

    x^2-c^2=c^2-x^2

    \implies 2x^2 = 2c^2

    \implies x^2 = c^2

    \implies \sqrt{x^2} = \sqrt{c^2}

    Now, here is where most people make a mistake (or try to take shortcuts). What is the square root of x^2 ? It's |x|, NOT x. So our next step will be:

    \implies |x| = c

    \implies x = \pm c

    By the way, the reason we can say that \sqrt{c^2}=c is because the problem states that c>0.
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