Find the flow of vector field through the side of cylindrical surface

Question

Problem is to find the flow of vector field, without Divergence theorem: $$\vec{F}= (x-z) \vec{i}+(y+z)\vec{j}+(x-y)\vec{k}$$ through the outer side of cylindrical surface $x^2+y^2=a^2$, bounded by planes $z=x$ and $z=0$
($z\geq0$)

I tried to do by my own: $$\iint{(x-z)dydz+(y+z)dxdz+(x-y)dxdy},$$ but i think that i messed up with the integrals bounds

Answer 1

This answer concerns itself solely with Part 2 of your problem, as I have not yet learned any differential forms.

The flux of a vector field through a closed surface is given by

$$\Phi=\oint_S\vec{F}\cdot d\vec{S}$$

First, we have to think about the surface that we want to consider. It is above the $xy$-plane for $x>0$ and below the $xy$ for $x<0$ and it intersects itself, so we really have to consider six surfaces here.

Let $x>0$.

First, we consider the part of the surface in the $xy$-plane, which is a semi-circle of radius $a$. We use polar coordinates with $(x,y)\rightarrow(r\cos\phi,r\sin\phi)$ with $r\in(0,a)$ and $\phi\in(-\frac{\pi}{2},\frac{\pi}{2})$. The unit normal vector is $\hat{n}=-\hat{z}$, because the circle lies in the $xy$-plane.

\begin{align*} \Phi_1 &=\int_{S_1}\vec{F}\cdot d\vec{S}\\ &=\int_0^adr\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}d\phi r\vec{F}(r,\phi)\cdot -\hat{z}\\ &=-\int_0^adrr^2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}d\phi(\cos\phi - \sin\phi)\\ &=-2\int_0^adrr^2\\ &=-\frac{2}{3}a^3 \end{align*}

The next surface we consider is the curved surface that is the mantle of the cylinder. Parametrising this surface leads to $(x,y,z)\rightarrow(a\cos\pi, a\sin\phi,z)$ with $\phi\in(-\frac{\pi}{2},\frac{\pi}{2})$ and $z\in(0,x)=(0,a\cos\phi)$. The outward normal is given by $\hat{n}=(\cos\phi,\sin\phi,0)$, and the surface element $d\vec{S}=\hat{n}ad\phi dz$.

Since the bounds of integration on $z$ now depend on the angle $\phi$, we are no longer free to choose the order of integration however we want and set up our integral in the following way:

\begin{align*} \Phi_2&=\int_{S_2}\vec{F}\cdot\ d\vec{S}\\ &=a\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}d\phi\int_0^{a\cos\phi}dz\vec{F}(\phi,z)\cdot(\cos\phi,\sin\phi,0)\\ &=a\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}d\phi\int_0^{a\cos\phi}dz(a\cos\phi-z,a\sin\phi+z,a\cos\phi-a\sin\phi)\cdot(\cos\phi,\sin\phi,0)\\ &=a\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}d\phi\int_0^{a\cos\phi}dz(a\cos^2\phi-z\cos\phi + a\sin^2\phi+z\sin\phi)\\ &=a\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}d\phi\int_0^{a\cos\phi}dz(a+z(\sin\phi-\cos\phi))\\ &=a\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}d\phi\left(a^2\cos\phi+\frac{a^2\cos^2\phi}{2}(\sin\phi-\cos\phi)\right)\\ &=2a^3+\frac{a^3}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}d\phi\cos^2\phi(\sin\phi-\cos\phi)\\ &=2a^3-a^3\int_{0}^{\frac{\pi}{2}}d\phi\cos^3\phi\\ &=2a^3-\frac{2}{3}a^3\\ &=\frac{4}{3}a^3 \end{align*}

Finally, we look at the third surface, the slanted surface on top of the cut cylinder. We parametrise it as follows: $(x,y,z)\rightarrow(r\cos\phi,r\sin\phi,x)=(r\cos\phi,r\sin\phi, r\cos\phi)$ with $r\in(0,a)$, $\phi\in(-\frac{\pi}{2},\frac{\pi}{2})$. The surface element is given by $d\vec{S}=(-1,0,1)rdrd\phi$.

\begin{align*} \Phi_3&=\int_{S_3}\vec{F}\cdot d\vec{S}\\ &=\int_0^adrr\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}d\phi\vec{F}(r,\phi)\cdot(-1,0,1)\\ &=\int_0^adrr\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}d\phi(0,r\sin\phi+r\cos\phi,r\cos\phi-r\sin\phi)\cdot(-1,0,1)\\ &=\int_0^adrr^2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}d\phi(\cos\phi-\sin\phi) \end{align*}

This, however, is just $-\Phi_1$.

Adding this all up, we have the flux through the first part of our intersecting surface:

$$\Phi_A = \Phi_1 + \Phi_2 + \Phi_3 = \frac{4}{3}a^3$$

The procedure for the other part (where $x<0$) is basically the same. Once again, the integrals through the slanted part and the semi-circle in the $xy$-plane should cancel and the flux through the second part of the intersecting surface $\Phi_B$ is just the flux through the mantle of the cylinder.

Adding it all up gives you the total flux $\Phi = \Phi_A + \Phi_B$.

Answer 2

You have to calculate the surface integral

$$\iint\limits_{S} (x-z) dydz + (y+z) dzdx + (x-y)dxdy$$

where $S$ is the part of cylindrical surface $x^2+y^2 = a^2$ between the plane $z=0$ (that is $XY$ plane), and the plane $z=x.$ However the condition $z\ge 0$ means that you should only consider that part of cylindrical surface which lies over the $XY$ plane and under $z=x.$ Also note that for this part of the cylindrical surface you have $x\ge 0.$

To calculate this integral you have to project the surface $S$ onto the $ZY$ plane. Denote the projection by $D$. And you also have to find the coordinates of the normal vector $\vec{n},$ $|\vec{n}|=1$ of $S$.

First find the intersection line of the cylindrical surface $x^2 + y^2 = a^2$ and the plane $z=x$. You get $z^2 + y^2 = a^2.$ This is the first boundary line of the projection $D$. Another boundary line is obviously $z=0$. So you have a semidisk $z^2+y^2 \le a^2,$ $(z\ge 0)$.

Now find the coordinates of the normal vector of $S$. In this situation it is convenient to use $z$ and $y$ as the parameters and $x$ as the function of $z$ and $y$. From $x^2+y^2=a^2$ you get $x = \sqrt{a^2-y^2}.$ Note that you take the $+\sqrt{}$ because of the part of the cylindrical surface you are dealing with.

You probably know that $\vec{n} (\cos \alpha, \cos \beta, \cos \gamma),$ where $\cos \alpha = \pm\frac{A}{\sqrt{A^2+B^2+C^2}},$ $\cos \beta = \pm\frac{B}{\sqrt{A^2+B^2+C^2}},$ $\cos \gamma = \pm\frac{C}{\sqrt{A^2+B^2+C^2}}$ and you have to decide what sign to take $+$ or $-$ using the condition that you are dealing with the outer side of the surface $S$.

Now $A = 1,$ $B = - \frac{\partial x}{\partial y} = \frac{y}{\sqrt{a^2-y^2}},$ $C = - \frac{\partial x}{\partial z}=0.$

The angle $\alpha$ between the normal vector to that part of the surface $S$ you are dealing with, and the $Z$-axis is between $0$ and $\pi/2.$ Because of that you obtain that $\cos \alpha $ should always be $\ge 0$. So you decide to take $+$.

Lets put it all together and write down the integral which we have to calculate

$$\iint\limits_{S} (x-z) dydz + (y+z) dzdx + (x-y)dxdy =$$ $$ = \iint\limits_{D} \left( (\sqrt{a^2-y^2} - z) \cos \alpha + (y+z) \cos \beta + (\sqrt{a^2-y^2} - y) \cos\gamma \right) \times \sqrt{A^2+B^2+C^2}dydz = $$ $$ = \iint\limits_{D} \left( (\sqrt{a^2-y^2} - z) A + (y+z) B + (\sqrt{a^2-y^2} - y) C \right) dydz =$$ $$ = \iint\limits_{D} \left( (\sqrt{a^2-y^2} - z) \cdot 1 + (y+z) \frac{y}{\sqrt{a^2-y^2}} + (\sqrt{a^2-y^2} - y) \cdot 0 \right) dydz =$$ $$ = \iint\limits_{D} \left( (\sqrt{a^2-y^2} - z) + (y+z) \frac{y}{\sqrt{a^2-y^2}} \right) dydz =$$ $$ = \iint\limits_{D} \left( \frac{a^2 + yz}{\sqrt{a^2-y^2}} - z \right) dydz .$$

To calculate the last integral you better go to polar coordinates: $y = r \cos \varphi,$ $z = r \sin \varphi,$ $0\le \varphi \le \pi,$ $0\le r \le a.$

So, the last integral changes to

$$\int\limits_0^{\pi} d\varphi \int\limits_0^a \left( \frac{a^2 + r^2 \sin \varphi \cos \varphi}{\sqrt{a^2 - r^2 \sin^2 \varphi }} - r\cos\varphi \right) rdr =$$ $$ = a^2 \int\limits_0^{\pi} \!\!d\varphi \!\!\int\limits_0^a \frac{r}{\sqrt{a^2 - r^2 \sin^2 \varphi }}dr + \int\limits_0^{\pi}\!\! d\varphi \!\!\int\limits_0^a \frac{r^3 \sin \varphi \cos \varphi}{\sqrt{a^2 - r^2 \sin^2 \varphi }} dr - \int\limits_0^{\pi} \!\!d\varphi \!\!\int\limits_0^a r^2 \cos\varphi dr =$$ $$= I_1 + I_2 + I_3.$$

The last one is obviously 0. $I_1$ and $I_2$ are a bit trickier and involve some improper integrals down the way of calculation.