Anonymous
Anonymous asked in Science & MathematicsPhysics · 9 years ago

Please answer my calculus question below which is about scalar line integrals in (R^3)?

Convert the line integral to an ordinary integral with respect to the parameter & evaluate it.

∫(x+y+z)ds; C is the circle r(t) = <2cost, 0, 2sint>, for 0≤ t ≤ 2π

*THERE IS A C UNDER THE INTEGRAL SYMBOL

*PLEASE GIVE ME THE COMPLETE SOLUTION & NOT JUST THE FINAL ANSWER.

*THE FINAL ANSWER IS 0.

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  • 9 years ago
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    r'(t) = <-2 sin t, 0, 2 cos t>.

    ||r'(t)|| = √(4 sin²t + 4 cos²t) = 2, since sin²t + cos²t = 1.

    We have f(x,y,z) = x+y+z.

    ∴ ∫(x+y+z)ds; C = ∫ (0 to 2π) f(r(t)) ||r'(t)|| dt.

    = 2 ∫ (0 to 2π) (2 cos t + 2 sin t) dt.

    = 4 [sin t - cos t] (0 to 2π).

    = 4(sin 2π - cos 2π - sin 0 + cos 0).

    = 0.

    Of course, it's clear by symmetry that the line integral will equal 0.

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  • 9 years ago

    ds² = dx² + dy² + dz² = 4(sint)²dt² + 4(cost)²dt², therefore ds/dt = 2.

    ∫xds = 2∫cost ds = 2∫cost (ds/dt)dt = 4∫cost dt = 4[sin2π] - 4[sin0] = 0

    Similarly, ∫zds = 0. And, trivially, ∫yds = 0.

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