# Will anyone help me with this line integral problem?

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 $\sqrt{5}+5/3$

• 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$\sqrt{5} + 5/3$ – H_Hassan May 11 '14 at 10:42
• @Ant so any ideas on how to solve this? – H_Hassan May 11 '14 at 10:50
• Post your calculations, so one can check then ;-) – Ant May 11 '14 at 10:56
• @Ant there you go my calculations – H_Hassan May 11 '14 at 11:07

## 1 Answer

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

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

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