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integrate $ f(x,y,z) = x+ \sqrt{y} - z^2$ over the path from $(0,0,0)$ to $(1,1,1)$ given by $C_1: r(t)= ti +t^2j , 0\le t\le1$

$C_2: r(t)=i+j+tk, 0\le t \le 1 $

will anyone help me with this problem? Here is what i did: fot the first part $ds= \sqrt{(dx/dt)^2 +(dy/dt)^2}dt$=$\sqrt{1+4t}=\sqrt{5}$

i think the last step is where i went wrong.

$\int^1_0 2\sqrt{5}t dt$=$\sqrt{5}$

for the second part $ds=dt$ so $\int^1_0 (2-t^2)dt$ = $5/3$

so adding them i got $\sqrt{5}+5/3$

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  • $\begingroup$ I tried getting the arclength of both of them and then integrating both over the interval from 0 to 1 and then adding but i always get$ \sqrt{5} + 5/3$ $\endgroup$ – H_Hassan May 11 '14 at 10:42
  • $\begingroup$ @Ant so any ideas on how to solve this? $\endgroup$ – H_Hassan May 11 '14 at 10:50
  • $\begingroup$ Post your calculations, so one can check then ;-) $\endgroup$ – Ant May 11 '14 at 10:56
  • $\begingroup$ @Ant there you go my calculations $\endgroup$ – H_Hassan May 11 '14 at 11:07
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EDIT:

You have just made some mistakes.

You have:

$F(r_1(t)) = 2t$

$ds_1 = ||r'_1(t)||dt = \sqrt{1+4t^2}dt$

$F(r_2(t)) = (2-t^2)$

$ds_2 = ||r_2'(t)||dt = dt$

So all in all your integral is

$$\int_0^1 2t\sqrt{1+4t^2} + (2-t^2) dt = \frac{1}{6} (9 + 5\sqrt{5})$$

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  • $\begingroup$ but it says in the answers that the result is $1/6 (5\sqrt{5} +9)$ $\endgroup$ – H_Hassan May 11 '14 at 11:20
  • $\begingroup$ not really sure where it got that answer from $\endgroup$ – H_Hassan May 11 '14 at 11:26
  • $\begingroup$ @user3333708 I'm sorry about before, I got confused. I edited the answer, it should be okay now :-) $\endgroup$ – Ant May 11 '14 at 11:33
  • $\begingroup$ Yeah thank you very much :D $\endgroup$ – H_Hassan May 11 '14 at 11:35

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