# Thread: Line integral

1. ## Line integral

Calculate the line integral on the $\displaystyle s$ along the curve $\displaystyle C$

$\displaystyle \int_C \frac{x}{1+y^2}ds$
$\displaystyle C: x = 1+2t; y = t; (0 \leq t \leq 1)$

2. Originally Posted by Apprentice123
Calculate the line integral on the $\displaystyle s$ along the curve $\displaystyle C$

$\displaystyle \int_C \frac{x}{1+y^2}ds$
$\displaystyle C: x = 1+2t; y = t; (0 \leq t \leq 1)$

Remember

$\displaystyle ds=\sqrt{\left( \frac{dx}{dt} \right)^2 +\left( \frac{dy}{dt} \right)^2 }dt$

$\displaystyle \int_{0}^{1}\frac{1+2t}{1+t^2}\sqrt{(2)^2+1^2}dt$

You should be able to finish from here

3. Thank you.

Please check if these are correct
1)
$\displaystyle F(x,y) = x^2yi +4j$
$\displaystyle C: r(t) = e^t i + e^{-t} j; (0 \leq t \leq 1)$

My solution:

$\displaystyle \int_0^1 (e^t,4).(e^t,-e^{-t})dt$
$\displaystyle = \frac{9}{2}e+\frac{7}{2}$

2)
$\displaystyle F(x,y,z) = zi+xj+yk$
$\displaystyle C: r(t) = senti + 3sentj + sen(t)^2k; (0 \leq t \leq \frac{ \pi}{2})$
My Solution:

$\displaystyle \int_0^{\frac{ \pi}{2}} (sen(t)^2, sent, 3sent).(cost, 3cost, 2sentcost)dt$
$\displaystyle = \frac{23}{6}$