Let C be the intersection curve between surfaces $z = 3x$ and $x^2+y^2 = 1$, oriented counterclockwise as seen from above. Calculate $$\int_c (1-4x) dx + 2x dy + (1-5y) dz$$ I want to calculate this using Stoke's theorem
I found the unit normal vector from the intersection plane to be $n = (-3i + k)/\sqrt10$
Then I found $∇×F = -5i-4j+2k$
I found the dot product and then evaluated $\int^{2\pi}_0\int^1_0 {17\over\sqrt10}$ and got $34\pi\over\sqrt10$
However the correct answer is simply $17\pi$
Please help me to find and understand my error here
Edit:
Ok so from the intersection curve I found $z^2 = 9x^2$ Using this, I found ${z^2\over9} + y^2 = 1$ and the area of this ellipse would be $\sqrt9*\sqrt1*\pi$. So the integrand multiplied by this area would be $(17*3\pi)\over\sqrt10$. Confirmation of this would be greatly appreciated.
As always, these kinds of problems are long, so let's do it!
First, plot the figure:
As you may see, the intersection is an ellipse. From the integral you show, we have $\mathbf F = (1 -4x) \boldsymbol{\hat \imath} + 2x \boldsymbol{\hat \jmath} + (1 - 5y) \boldsymbol{\hat k}$. Since you want to use the Stokes' theorem:
$$\int_C \mathbf F \cdot d\mathbf r = \iint_S (\nabla \times \mathbf F) \cdot d \mathbf S,$$
we have to find $\nabla \times \mathbf F$:
$$\nabla \times \mathbf F = -5 \boldsymbol{\hat \imath} + 2 \boldsymbol{\hat k}.$$
Now, we have to find the normal vector of that ellipse (which is just the normal vector of the plane). You have already found it:
$$\boldsymbol{\hat{n}} = \sqrt{10}\, \dfrac{-3 \boldsymbol{\hat \imath} + \boldsymbol{\hat k}}{10},$$
so:
$$\iint_S (\nabla \times \mathbf F) \cdot d \mathbf S = \iint_S (\nabla \times \mathbf F) \cdot \boldsymbol{\hat{n}} \, dS = \iint_S (-5 \boldsymbol{\hat \imath} + 2 \boldsymbol{\hat k}) \cdot \left(\sqrt{10}\, \dfrac{-3 \boldsymbol{\hat \imath} + \boldsymbol{\hat k}}{10} \right) dS$$
$$=\dfrac{9 \sqrt{10}}{5} \iint_S dS.$$
The last step is to find the area of that ellipse. Now, the hard part is the following: how do you parametrize the surface of the ellipse? One idea for that may come from the parametrization of curve of the ellipse:
$$\mathbf \Gamma (t) = \cos t \, \boldsymbol{\hat \imath} + \sin t \, \boldsymbol{\hat \jmath} + 3 \sin t \, \boldsymbol{\hat k}, \quad 0 \le t \le 2 \pi. $$
After thinking a lot, a good parametrization of the surface is:
$$\mathbf r (t, s) = \sin t \cos s \,\boldsymbol{\hat \imath} + \sin t \cos t \,\boldsymbol{\hat \jmath} + 3 \sin t \cos s \,\boldsymbol{\hat k}, \quad 0 \le t \le \pi/2,\quad 0 \le s \le 2\pi.$$
I show the plot of the curve and the surface just to be sure:
The upper limit of $t$ comes from:
$$ \begin{align} \sin^2 t \, \cos^2 s + \sin^2 t \, \sin^2 s &= 1, \\ \sin^2 t &= 1, \\ t &= \pi /2. \end{align} $$
We are doing this because we need:
$$\iint_S dS = \iint_D \lvert \mathbf r_s \times \mathbf r_t \rvert \, dA.$$
After finding the derivatives and the cross product:
$$\lvert \mathbf r_s \times \mathbf r_t \rvert = \sqrt{10} \cos t \sin t.$$
With that, you can find the area:
$$\iint_D \lvert \mathbf r_s \times \mathbf r_t \rvert \, dA = \int_0^{2\pi} \! \int_0^{\pi /2} \sqrt{10} \cos t \sin t \, dt \, ds = \sqrt{10} \pi.$$
Finally:
$$\iint_S (\nabla \times \mathbf F) \cdot d \mathbf S = \dfrac{9 \sqrt{10}}{5} \iint_S dS = \dfrac{9 \sqrt{10}}{5} \sqrt{10} \pi = 18 \pi.$$