Evaluate integral using Stokes' theorem

Evaluate the integral $\int_C \vec{F} \cdot d\vec{r}$ with $\vec{F}$ and $C$ as given and the direction integration along $C$ being clockwise as seen by a person standing at the origin. $\vec{F}=[-z, 5x, -y]$ and $C$ is the ellipse $x^2+y^2=4, z=x+2$.

The problem wants us to use Stokes' theorem, which says $$\int_C \vec{F} \cdot d\vec{r} = \int\int \text{curl} \ \vec{F} \cdot d\vec{S}$$

I know exactly what I have to do, but I'm having trouble coming up with the unit normal vector to evaluate this surface integral. Thanks in advance.

• which do you want to evaluate, left side or right side or both? – Santosh Linkha Sep 25 '13 at 9:14
• The question is asking to calculate the line integral without actually calculating the line integral (ie. apply Stokes' Theorem and calculate the surface integral instead) – Lefty Sep 25 '13 at 9:15
• then calculate the right side. – Santosh Linkha Sep 25 '13 at 9:17
• Like I explain in the post, I'm not sure how you come up with $d\vec{S}=\hat{n}dS$. – Lefty Sep 25 '13 at 9:18

The curl is the curl is $(-1,-1,5)$. The surface of that vertical plane inclined along x-axis in between that cylinder is an ellipse like this. The parametric equation of this ellipse is given by $$\Phi( r, \theta ) = ( 2 r \cos(\theta), 2 r\sin(\theta), 2 r \cos(\theta)+2)$$
To calculate surface integral, (possibly $-$ve since you are going clockwise loop) $$\int_0^{2\pi}\int_{0}^{2\pi } (-1,-1,5) \cdot \left( \frac{\partial \Phi}{ \partial r} \times \frac{\partial \Phi}{ \partial \theta} \right ) d\theta d\phi = 24 \pi$$
Another way to calculate it, note that ellipse lies in surface $S \implies z - x = 2$ bound below by circle at $(0,0)$ of unit length. You can use another formula. $$\int_{-2}^2 \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2} } (-1, -1, 5) \cdot \vec \nabla S dy dx = 4 \int_0^2 \int_0^{\sqrt{4-x^2}} (-1, -1, 5) \cdot (-1, 0 , 1) dy dx = 24 \pi$$
• Doesn't $\frac{\partial \Phi}{\partial r} \times \frac{\partial \Phi}{\partial \theta}$ have to be a unit normal vector? – Lefty Sep 25 '13 at 10:12
• @Lefty yes that thing has to be unit vector, but again you are evaluating on the surface ... so it get's multiplied again by $|\frac{\partial \Phi}{\partial r} \times \frac{\partial \Phi}{\partial \theta}|$. note that $$\iint |\frac{\partial \Phi}{\partial r} \times \frac{\partial \Phi}{\partial \theta}| dr d\theta$$ should give you area of that ellipse. – Santosh Linkha Sep 25 '13 at 10:14