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Math Help - Line Integration

  1. #1
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    Line Integration

     \int _c xy^4ds C is the right half of the circle x^2+y^2=16

    Here is what i did. Since C is the right half of the circle my bounds are from  -\pi/2 \leq  t \leq \pi/2 . I got x = 4cost and y = 4sint. I then take the arclength of that, and multiple that by the original function with the polar coordinates substituted into it which is \int (4cost)(4sint)^4\sqrt{(-4sint^2)+(4cost^2)}dt

    Im kind of confused on how to simplify the algebra before i integrate. Any step by step guidance would be appreciated .
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  2. #2
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    Quote Originally Posted by purplerain View Post
     \int _c xy^4ds C is the right half of the circle x^2+y^2=16

    Here is what i did. Since C is the right half of the circle my bounds are from  -\pi/2 \leq t \leq \pi/2 . I got x = 4cost and y = 4sint. I then take the arclength of that, and multiple that by the original function with the polar coordinates substituted into it which is \int (4cost)(4sint)^4\sqrt{(-4sint^2)+(4cost^2)}dt

    Im kind of confused on how to simplify the algebra before i integrate. Any step by step guidance would be appreciated .
    So you have

    4^6\int_{-\pi/2}^{\pi/2} \cos t \sin^4 t \,dt then let u = \sin t.
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  3. #3
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    Quote Originally Posted by Danny View Post
    So you have

    4^6\int_{-\pi/2}^{\pi/2} \cos t \sin^4 t \,dt then let u = \sin t.

    I understand the integration, im just confused on how to simplfy the function before i integrate. The book has  4^5\int costsin^4t 4dt

    I have no idea how they got the 4dt, any help would be appreciated, my algebra is fuzzy.
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  4. #4
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    Quote Originally Posted by purplerain View Post
    I understand the integration, im just confused on how to simplfy the function before i integrate. The book has  4^5\int costsin^4t 4dt

    I have no idea how they got the 4dt, any help would be appreciated, my algebra is fuzzy.
     <br />
x = 4 \cos t, y = 4\sin t \; \text{so}\; dx = -4 \sin t\, dt \; \text{and }\;dy = 4 \cos t\, dt<br />
    so
     <br />
ds = \sqrt{dx^2+dy^2}\,dt = \sqrt{(-4 \sin t)^2 + (4 \cos t)^2}\, dt = \sqrt{4^2(\sin^2 t+ \cos^2t)}\,dt = \sqrt{4^2}\,dt = 4 dt<br />
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  5. #5
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    Quote Originally Posted by Danny View Post
     <br />
x = 4 \cos t, y = 4\sin t \; \text{so}\; dx = -4 \sin t\, dt \; \text{and }\;dy = 4 \cos t\, dt<br />
    so
     <br />
ds = \sqrt{dx^2+dy^2}\,dt = \sqrt{(-4 \sin t)^2 + (4 \cos t)^2}\, dt = \sqrt{4^2(\sin^2 t+ \cos^2t)}\,dt = \sqrt{4^2}\,dt = 4 dt<br />
    Thank you!
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