# Thread: Evaluating the line integral....

1. ## Evaluating the line integral....

...where curve C is given by the integral of (x^2)*y*sqrt(z)dz
C: x=t^3, y=t, z=t^2, t is between 0 and 1

I came up with the integral (from 1 to 0) (2t^12)*sqrt(1+9t^4+4t^2)

is this integral correct? if yes, how do I integrate it?

2. I'm not sure what you're doing but you want the line integral

of (x^2)*y*sqrt(z)dz on the curve x=t^3, y=t, z=t^2 ?

F = (x^2)*y*sqrt(z) k

Since

x=t^3, y=t, z=t^2

F = t^6*t*t k = t^8 k

r = t^3 i + t j + t^2 k

r ' = 3t^2 i + j + 2t k

F*r ' = 2t^9

Which is easy to integrate over the interval 0 to 1