- #1

bodensee9

- 178

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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!