Compute the line integral of vector field $F(x,y,z)$ = $⟨x^2,y^2,z^2⟩$ where C is the curve of intersection of $z=x+1$ and $x^2+y^2=1$, from the lowest point on the curve to the highest, traversed counterclockwise when seen from above.
These are a plane tilted 45 degrees and a cylinder of radius 1 centered at the origin, so their intersection is a hoop tilted 45 degrees on the $y$ axis. The hoop's low point is (-1,0,0) and its high point is (1,0,2). I can also do the following conversion to cylindrical coordinates to get the equation of the hoop:
$z=1+x$ $\implies$ $z=1+r\cos\theta$
$x^2+y^2 = 1^2$ $\implies$ $r^2 = 1$ $\implies$ $r=1$
∴ $z=1+1\cos\theta$ $\implies$ $z=1+\cos\theta$
This is all fine. Afterwards, however, I don't know how to convert this into a vector field equation, where everything is supposed to be parameterized entirely into $t$. The formula for the line integral of a vector field is:
$\int^b_aF(x(t),y(t),z(t))\cdot r\prime(t) dt$
And I'm not quite sure how to transform my work so that it fits this formula. I think that I should treat $\theta$ as $t$, and the limits $a$ and $b$ should be $\int^{0}_π$, but I'm stuck beyond that. Would anyone be able to tell me how I should be setting this integral up?
$r(x,y,z) = (\cos t, \sin t, 1+\cos t) \\ r'(t) = (-\sin t, \cos t, -\sin t)$
If you integrate from $(\pi, 0)$ as you suggest above, you will traverse the path in the clockwise direction. In this case, it will prove to be irrelevant.
$\int^b_aF(x(t),y(t),z(t))\cdot r\prime(t) dt\\ \int_{\pi}^{2\pi} (\cos^2 t, \sin^2 t, (1+\cos t)^2)\cdot (-\sin t, \cos t, -\sin t) dt\\ \int_{\pi}^{2\pi} -2\sin t\cos^2 t + \cos t\sin^2 t - \sin t - 2 \sin t \cos t \;dt\\ $
And I will let you take it from here.
It is worth noting that F(x,y,z) is a conservative force.
$F = \nabla \frac 13 (x^3 + y^3 + z^3)$
The integral does not depend on the path, only the endpoints.
$\frac 13 (x^3 + y^3 + z^3) |_{(-1,0,0)}^{(1,0,2)}$