Stokes' Thm. $\vec F = <2y,xz,x+y>$ through intersection of $x^2+y^2=1$ and $z=y+2$

Question

I am trying to evaluate $\int_C \vec F \cdot d \vec r$ where

$\vec F = <2y,xz,x+y>$ and $C$ is the intersection of $x^2+y^2=1$ and $z=y+2$.

I believe that $$curl \vec F = <1-x, -1, z-2> $$

The surface is an ellipse, on the plane $z=y+2$ so the unit normal vector should be

$$\vec n = \frac{1}{\sqrt{2}}<0,-1,1>$$

because the dot product is $\frac{1}{\sqrt 2} \left(z-1\right)$ I am thinking that we are to evaluate

$$\int \int_S \frac{1}{\sqrt 2} \left(z-1\right) dS$$

I used $x= \cos\theta, \space y= \sin\theta, z=z$ to evaluate this and got the expression

$$\frac{1}{\sqrt2} \int_0^{2\pi} \int_1^3 z-1 \space dz d\theta$$

where I got $4\sqrt{2}\pi$.

Supposedly the answer should be $\pi$.

Can any one help with what my error is?

Answer 1

Your curl is correct, as is your normal vector and your dot product. The problem I believe is in the integral; I'm not sure where you got that integral or those limits.
The area below the ellipse is the unit circle. With the factor of $\sqrt{2}$ converting $dS$ to $dA$, I would have expected $$ \int_{0}^{2\pi} \int_{0}^{1}(1 + r sin {\theta})r dr d\theta$$ which gives $$ \int_{0}^{2\pi} \frac{1}{2} + \frac{1}{6}sin {\theta} d\theta = \pi$$

Answer 2

Stokes Theorem says $\displaystyle \int_C \vec F \cdot dr = \iint_S curl \vec{F} \cdot dS$

where $S$ is a surface with the boundary and $C$ is the boundary curve of the surface.

You have been asked to find the line integral over the elliptic curve which is intersection of cylinder $x^2+y^2= 1$ and plane $z = y + 2$.

You are applying Stokes' Theorem and instead finding the right hand side of the equation. The easiest way to do so is to consider a disk which has the same boundary as in the question and find flux of the curl of the vector field over the disk.

$\vec{n} = (0,-1,1)$

If you normalize the vector then you have to note that $dS = \sqrt2 dA$ where $dA$ is the projection of $dS$ in XY plane.

Parametrization of the disk surface:

$x = r \cos \theta, y = r \sin\theta, z = y + 2 = 2 + r \sin \theta$.

So the integral becomes,

$\displaystyle \int_0^{2\pi} \int_0^1 (r^2 \sin \theta + r) \ dr \ d\theta = \pi$