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Math Help - [SOLVED] Line integral

  1. #1
    MHF Contributor arbolis's Avatar
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    [SOLVED] Line integral

    I must calculate \int _{C} xy^4 ds where C is the right side of x^2+y^2=16.
    My attempt : It's my first exercise involving such integrals and I don't understand well something.
    So C is the right side of a circle whose radius is 4.
    \vec{r}(t)=(4\cos t, 4\sin t)
    If f(x,y)=xy^4 then \int _{C} xy^4 ds=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} f(\vec{r}(t))|\vec {r}'(t)|dt.
    How do I interpret f(\vec{r}(t))?
    Is it \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} f(\vec{r}(t))|\vec {r}'(t)|dt=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (4\cos t \cdot (4\sin(t))^4) \cdot 4 dt?
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  2. #2
    Super Member Random Variable's Avatar
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    Yes, that is correct.

    Another way to write your parametrization is

    x = 4cost
    y = 4sint
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