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.

  • $\begingroup$ which do you want to evaluate, left side or right side or both? $\endgroup$ – Santosh Linkha Sep 25 '13 at 9:14
  • $\begingroup$ 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) $\endgroup$ – Lefty Sep 25 '13 at 9:15
  • $\begingroup$ then calculate the right side. $\endgroup$ – Santosh Linkha Sep 25 '13 at 9:17
  • $\begingroup$ Like I explain in the post, I'm not sure how you come up with $d\vec{S}=\hat{n}dS$. $\endgroup$ – 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. enter image description here

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. $$\int_{-2}^2 \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2} } (-1, -1, 5) \cdot \vec \nabla S dy dx = 4 \int_0^2 \int_0^{\sqrt{4-x^2}} (-1, -1, 5) \cdot (-1, 0 , 1) dy dx = 24 \pi $$

  • $\begingroup$ Doesn't $\frac{\partial \Phi}{\partial r} \times \frac{\partial \Phi}{\partial \theta}$ have to be a unit normal vector? $\endgroup$ – Lefty Sep 25 '13 at 10:12
  • $\begingroup$ @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. $\endgroup$ – Santosh Linkha Sep 25 '13 at 10:14

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