Evaluating surface integral.

Question

The question is:

Let $S$ be that part of the surface of the paraboloid $z=x^2+y^2$ between the planes $z=1$ and $z=4$. Given $\vec{V}=x^3j+z^3k$, I want to evaluate the surface intergal

$$\iint_s\nabla\times V.\hat n dS$$

would the paramatisation be something like $x=r\cos\theta,y=r\sin\theta,z=r^2 $ or $ z=r$

$z=x^2+y^2$ = constant when z=4 or z=2 does this mean we can calculate $\hat ndS$

Answer 1

Let $\vec V=\hat y x^3+\hat z z^3$. Then, $\nabla \times \vec V=3\hat z x^2$.

A vector point on the surface can be written as $\vec r=\hat \rho\rho +\hat z\rho^2$, where $\hat \rho=\hat x\cos(\phi)+\hat y\sin(\phi)$.

So, the surface element is $\hat n\,dS=\left(\frac{\partial \vec r}{\partial \rho}\times\frac{\partial \vec r}{\partial \phi}\right)\,d\rho\,d\phi=\left(-2\hat\rho\rho^2+ \hat z \rho\right)\,d\rho\,d\phi$.

Therefore,

$$\int_S\nabla\times \vec V\cdot \hat n\,dS=\int_0^{2\pi}\int_1^2 3\rho^3\cos^2(\phi)\,d\rho\,d\phi=45\pi/4$$

Answer 2

alternative

Green's / Stokes theorem:

$\iint_s\nabla\times V.\hat n dS = \oint_c V\cdot dr_2 - \oint_c V \cdot dr_1$

$r_1 = (\cos\theta, \sin\theta, 0), r_2 = (2\cos\theta, 2\sin\theta, 0)$

$\int_0^{2pi} 2^4\cos^4\theta\ d\theta - \int_0^{2pi} \cos^4\theta\ d\theta\\ 15\int cos^4 \theta\ d\theta\\ \frac {15}{2}\int (\frac 12 (1+cos2\theta))^2 d\theta\\ \frac {15}{4}\int 1+ 2cos2\theta + \frac 12 + \frac 12 cos 4\theta d\theta\\ \frac {45}{4}\pi$

Alternative number 2 Divergence theorem:

$\iint_s\nabla\times V.\hat n dS + \iint_{D_4}\nabla\times V \cdot (0,0,1) dA + \iint_{D_1}\nabla\times V \cdot (0,0,-1) dA = \iiint \nabla \cdot (\nabla \times V) dV = 0$

$\int_0^{2\pi} \int_0^2 (3r^3 cos^2\theta)\ dr\ d\theta - \int_0^{2\pi} \int_0^1 (3r^3 cos^2\theta)\ dr\ d\theta$

$\int_0^{2\pi} \int_1^2 3r^3 cos^2\theta\ dr\ d\theta\\ \int_0^{2\pi} \frac {3}{4}(15) cos^2\theta\ d\theta\\ \frac {45}{4} \pi$