So i have the following question:
Compute $$\int_{C}(x^2+y)dx + (z+x)dy + (x+2y)dz$$ where $C$ is the intersection of the cylinder $x^2+y^2=4 $ and the plane $x+y=z$
So my thoughts are to parametrise it to make get $$r(t) = 2cos(t) i + 2sin(t)j + (2cos(t) + 2sin(t) ) k$$
and then use this along with $r'(t)$ to calculate
$$\int_{0}^{2\pi} F(r(t)) \cdot r'(t) dt $$
Which when I computed it all, came out to be $$-8\pi$$
I was hoping someone could either verify this method, along with the answer or help me to find out the solution (through tips)
Thank you
Yes you have parametrized the surface correctly but your answer is incorrect.
$r(t) = (2 \cos t, 2 \sin t, 2 \cos t + 2 \sin t)$
So, $dx = -2 \sin t \ dt, \ dy = 2 \cos t \ dt, \ dz = (-2 \sin t + 2 \cos t) \ dt$
$x^2 + y = 4 \cos^2 t + 2 \sin t, x + z = 4 \cos t + 2 \sin t, x + 2y = 2 \cos t + 4 \sin t$
So $\ \displaystyle \int_C \ (x^2+y) \ dx + (z+x) \ dy + (x+2y) \ dz = \int_0^{2 \pi} (12 \cos 2t + 4 \sin 2t - 8 \cos^2t \ \sin t) \ dt = 0$