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Evaluate the line integral, where $ C $ is the given curve.
$ \displaystyle \int_C y \, dx + z \, dy + x \, dz $,
$ C: x = \sqrt{t} $, $ y = t $, $ z = t^2 $, $ 1 \leqslant t \leqslant 4 $
$\left(\frac{8}{3}+\frac{64}{3}+\frac{128}{5}\right)-\left(\frac{1}{3}+\frac{1}{3}+\frac{4}{5}\right)$
Vector Calculus
Campbell University
Harvey Mudd College
Baylor University
Boston College
evaluate this line to grow, so we just have tickles. Won't want to. Four wise t X is one over two square root of TD thes t square The wise DT Xs square of t disease to tt so one, two, four This is one over to t to the one over two plus t Square Prostitute to the This should be three over to TT and we find the entire derivative t to the Stree Over to community property to over three. Four t to the Cube over three. Tito the five over to beautify by two over five in front. So for over five, then we probably before we had for the Israel with Who's eight for Cuba's sixty four four to the five over two is to to the fifth Power, which is thirty two thirty two times for his uh, one twenty eight minus one over three plus one over three plus four or five and you're left with fractions. Simplification, which I think I've had. Leave it to you. So I'll have to open another page