How Can I Evaluate This Line Integral Using Curl and Double Integrals?

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SUMMARY

The discussion focuses on evaluating the line integral ∫(y + sin x)dx + (z^2 + cos y)dy + x^3dz along the curve defined by r(t) = for 0 ≤ t ≤ 2π. The participant correctly identifies that the line integral can be expressed as ∫F∙dr, where F = . They compute the curl of F as curl F = <-2z, -3x^2, 1> and the normal vector n for the surface z = 2xy as n = <2x, 2y, -1>. The participant proposes the double integral ∫∫(-14x^2y - 1)dxdy with boundaries 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1, while also considering a polar coordinate representation.

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bodensee9
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Could someone help me with the following? I am asked to evaluate the line integral of ∫(y + sin x)dx + (z^2+cosy)dy +x^3dz where C is the curve r(t) = <sint, cost, sin2t, 0 ≤t≤2π.

Doesn’t this equal to ∫F∙dr where F = <y + sinx, z^2cosy,x^3> and r = <x,y,z>? So wouldn’t ∫F∙dr = ∫∫curlF∙nds where n is the normal vector to the surface z = 2xy (from the parametric equation, z = 2xy).

I got that curl F is -2zi-3x^2j +k, and that n is 2xi + 2yj –k. So if you take the dot product, you would get -4zx – 6x^2y -1, and if you were to want to substitute for z you would get -8x^2y – 6x^2y -1 or -14x^2y -1.

But I’m not sure what the boundaries of integration is other than that x seems to be between 0 and 1, and y seems to be between 0 and 1 as well? So would the double integral be ∫∫-14x^2y-1dxdy where 0≤y≤1 and 0≤x≤1? Many thanks!
 
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Or actually, on further thought, could you represent integral as ∫∫(-14(cosθ)^2(sinθ)-1)rdrdθ where r is between 0 and 1 and θ is between 0 and 2pi?
 

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