# Evaluate the line integral $\int_cx^2y\sqrt{z}dz$.

Evaluate the line integral where C is the given curve.

$$\int_cx^2y\sqrt{z}dz\\ C: x = t^3, y = t, z = t^2, 0 \leq t \leq 1$$

I get: $$\int_0^1t^9\sqrt{27t^4+4t^2+1}dt$$

Which is an awfully sloppy integral with no apparent way to simplify it. If I could get help with the appropriate approach to these problems it would be greatly appreciated.

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Do you really want $dz$ in the first displayed integral? –  David Mitra Dec 8 '12 at 19:28
Do you see what David's comment is getting at? You have an integral with respect to $z$, but then expanded it as if it were with respect to arc length. –  alex.jordan Dec 8 '12 at 19:33
$\int_0^1t^9dt = \frac{1}{10}$ This doesn't match the book however. (Sorry, our professor, running out of time at the end of semester, went over 4 sections in our book in a 50 minute session. I'm struggling a bit here.) –  user1405177 Dec 8 '12 at 19:50