Vector Calc 12-24-2020

If we close the surface by placing a disc of unit radius at $z = 1$ , then we can evaluate the flux through the closed surface using the divergence theorem, which states that

$\displaystyle \int_S \vec{F} \cdot \vec{dS} = \int_V \text{div} \vec{F} dV$

We have,

$\text{div} \vec{F} = \dfrac{\partial F_x}{\partial x} + \dfrac{\partial F_y}{\partial y}+ \dfrac{\partial F_x}{\partial x} = 1 + 0 + 2 z + 1 = 2 z + 2$

In cylindrical coordinates, $dV = dz r dr d \phi$

Hence

$\displaystyle \int_V \text{div} \vec{F} dV = \int_{\phi = 0}^{2\pi} \int_{r= 0}^1 \int_{ z = r}^1 (2z + 2) r dz dr d \phi$

$= \displaystyle \int_{\phi = 0}^{2\pi} \int_{r= 0}^1 r (z^2 + 2 z)\Biggr|_{z=r}^{z=1} dr d \phi$

$= \displaystyle \int_{\phi = 0}^{2\pi} \int_{r= 0}^1 r (3 - r^2 - 2 r ) dr d\phi$

$= \displaystyle \int_{\phi = 0}^{2\pi} \int_{r= 0}^1 (3 r - r^3 - 2 r^2 ) dr d\phi$

$= 2 \pi ( \frac{3}{2} - \frac{1}{4} - \frac{2}{3} ) = \dfrac{7\pi}{6}$

Now the flux through the top (which has been added for the purpose of computation) is

$\displaystyle \int_{\phi = 0}^{2\pi} \int_{r= 0}^1 F_z(r, \phi) r dr d\phi$

$= \displaystyle \int_{\phi = 0}^{2\pi} \int_{r= 0}^1 (2 + r sin \phi) r dr d\phi$

$= \displaystyle \int_{\phi = 0}^{2\pi} 1 + \frac{1}{2} sin \phi d \phi$

$= 2 \pi$

Hence the flux through the curved surface is $\dfrac{7 \pi}{6} - 2 \pi$ = $-\dfrac{5 \pi}{ 6}$

and since the question asks for the absolute value then the answer is $\dfrac{5 \pi}{6} \approx \boxed{2.618}$

To verify our answer, we can evaluate the surface integral directly over the lateral surface of the cone.

In cylindral coordinates, the parametrization of the given cone surface is

$p = (r \cos \phi, r \sin \phi, r )$

And the vector differential $\vec{dS}$ is given by $\dfrac{\partial p}{\partial \phi} \times \dfrac{\partial p}{\partial r }$

$\dfrac{\partial p}{\partial \phi} = ( -r \sin \phi, r \cos \phi, 0 )$

$\dfrac{\partial p}{\partial r } = ( \cos \phi, \sin \phi, 1 )$

so that,

$\dfrac{\partial p}{\partial \phi} \times \dfrac{\partial p}{\partial r } = ( r \cos \phi, r \sin \phi, - r )$

The expression for the vector field in cylindrical coordinates is

$\vec{F}(r, \phi) = (r \cos \phi + r^2 \sin^2 \phi -r, r^2 \cos^2 \phi + r, r^2 + r \sin \phi + r )$

Forming the dot product between $\vec{F}$ and $\vec{dS}$ results in the following expression

$\vec{F} \cdot \vec{dS} = (r \cos \phi)( r \cos \phi + r^2 \sin^2 \phi - r ) + (r \sin \phi)( r^2 \cos^2 \phi + r ) - r ( r^2 + r \sin \phi + r )$

Integrating this huge expression with respect to $\phi$ first annihilates (kills) all the terms containing $\cos \phi$ and $\sin \phi$

because their integrals over $[0, 2 \pi ]$ is zero. Hence,

$\displaystyle \int_S \vec{F} \cdot \vec{dS} = \int_{r = 0}^1 \int_{\phi=0}^{2\pi} [ (r \cos \phi)( r \cos \phi + r^2 \sin^2 \phi) + (r \sin \phi)( r^2 \cos^2 \phi ) - r ( r^2 + r ) ] d\phi dr$

$=\displaystyle \int_{r = 0}^1 \pi r^2 - r ( r^2 + r ) (2 \pi) d r$

$= \pi ( \frac{1}{3} - \frac{1}{2} - \frac{2}{3} ) = \dfrac{\pi}{6} (2 - 3 - 4) = -\dfrac{5 \pi}{6}$

Which is in agreement with our previous result.

A point on the surface of the cone is:

$\vec{r} = x \ \hat{i} + y \ \hat{j} + \sqrt{x^2+y^2} \ \hat{k}$

We can write the surface area element as:

$d\vec{S} =\left( \frac{\partial \vec{r}}{\partial x} \times \frac{\partial \vec{r}}{\partial y}\right) dy \ dx$

Plugging in values and evaluating the surface area element vector yields:

$d\vec{S} = \left(-\frac{x}{z}\ \hat{i} - \frac{y}{z} \ \hat{j} + \hat{k}\right) dy \ dx$

Now, the dot product $\vec{F} \cdot d\vec{S}$ is evaluated which after simplification leads to:

$\vec{F} \cdot d\vec{S} = \left(x + x^2+y^2 + \sqrt{x^2+y^2} -\left(\frac{x^2+xy(x+y)}{ \sqrt{x^2+y^2} }\right)\right) dy \ dx$

Calculating the surface integral involves integrating the above function over the unit circle centred at the origin:

$\int \int_{S} \vec{F} \cdot d\vec{S} = \int_{-1}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \left(x + x^2+y^2 + \sqrt{x^2+y^2} -\left(\frac{x^2+xy(x+y)}{ \sqrt{x^2+y^2} }\right)\right) dy \ dx$

This integral can be tackled by converting to polar coordinates. These steps are left out. The transformed integral is:

$\int \int_{S} \vec{F} \cdot d\vec{S} = \int_{0}^{2 \pi} \int_{0}^{1}\left(r\cos{\theta} + r^2+r-r\cos^2{\theta} - r^2\sin{\theta}\cos{\theta}(\sin{\theta}+\cos{\theta})\right) r \ dr \ d\theta$

Evaluating the above is trivial and the closed-form solution is:

$\boxed{\int \int_{S} \vec{F} \cdot d\vec{S} =\frac{5 \pi}{6}}$