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Math Help - Parametric Loop Area

  1. #1
    Del
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    Parametric Loop Area

    Notice that the curve given by the parametric equations

    is symetric about the x-axis.
    The curve makes a loop which lies along the x-axis. What is the total area inside the loop?

    What integral do I need to integrate, or what better method can I use, to find this? Please help!
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by Del View Post
    Notice that the curve given by the parametric equations




    is symetric about the x-axis.
    The curve makes a loop which lies along the x-axis. What is the total area inside the loop?

    What integral do I need to integrate, or what better method can I use, to find this? Please help!
    Always draw a graph it really helps

    As you noted the graph is symmetric with respect to the x axis. So we can find the area above the x axis and double it.

    y=t^3-25=t(t-5)(t+5)
    so the y intercepts are at t= -5,0,5

    \int_{-5}^{0}ydx=\int_{-5}^{0}(t^3-25t)(-2tdt)=\int_{-5}^{0}-2t^4+50t^2dt

    \int_{-5}^{0}-2t^4+50t^2dt=-\frac{2}{5}t^5+\frac{50}{3}t^3|_{-5}^{0}= \frac{2}{5}(0)^5+\frac{50}{3}(0)^3- \left( \frac{2}{5}(-5)^5+\frac{50}{3}(-5)^3 \right)=\frac{10000}{3}

    This is half of the area(The part above the x axis)

    So multipling by 2 we get \frac{20000}{3}
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  3. #3
    Del
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    That makes sense all the way, but the program I am inputting my answers into is saying that this is wrong. Thanks though!
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  4. #4
    Behold, the power of SARDINES!
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    Pesky minus sign Dang. I still can't add

    <br />
\int_{-5}^{0}-2t^4+50t^2dt=-\frac{2}{5}t^5+\frac{50}{3}t^3|_{-5}^{0}= \frac{2}{5}(0)^5+\frac{50}{3}(0)^3- \left( \frac{2}{5}(-5)^5+\frac{50}{3}(-5)^3 \right)=\frac{10000}{3}

    Notice the missing negative after the large parenthsis before the 2/5.

    <br />
-\frac{2}{5}t^5+\frac{50}{3}t^3|_{-5}^{0}= -\frac{2}{5}(0)^5+\frac{50}{3}(0)^3- \left(- \frac{2}{5}(-5)^5+\frac{50}{3}(-5)^3 \right)=\frac{2500}{3}<br />

    mult by 2 to get \frac{5000}{3}
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  5. #5
    Del
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    Thanks a ton, you really cleared this up!
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