The problem:
Line integral 鈭玞 (y^2 dx + 6xy dy), where C is the boundary of the region bounded by y=sqrt(x), y=0 and x=4, and oriented in the clockwise direction.|||Well, you have a dx and a dy, which suggests that it is a vector line integral. Parametrize C as
y = t
x = t^2
and now t goes from 0 to 2.
So the integral is over the curve C(t) = t^2i + tj and C'(t) = 2ti + j.
Let F(x,y) = y^2i + 6xyj. Then F(C(t)) = t^2i + 6t^3 and
F(C(t)) dot C'(t) = 2t^3 + 6t^3 = 8t^3
So we integrate 8t^3 dt from 0 to 2. This ends up being 32.
I'm not really sure how you would have even begun to do this as a scalar line integral...|||Don't quote me on that; I did it pretty quickly.
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