Suppose $A=6z\hat i+(2x+y)\hat j-x\hat k$ .Evaluate $$\iint A.dS$$ Over the entire surface S of the region bounded by the cylinder $x^2+z^2=9,x=0,y=0,z=0$ and $y=8$.
I split it into three surface 1.Upper circle part, $S_1$ 2.Lower circle part, $S_2$ 3.cylindrical part, $S_3$.
I couldn't do surface integral for $S_3$.Since i am familiar with parametrization of a cylinder and cylindrical coordinates but failed to approach to the answer. The solution provided by the book is $18\Pi$.Can anyone help me to explain how to get $\iint A.n dS$ please.
Thanks in Advanced.


If you don't want to use the divergence theorem.

Your volume has 5 surfaces.

$S_1$ is the quarter disk in the plane $y = 0$

$S_2$ is the quarter dis in the plane $y = 8$

$S_3$ is the rectangle in the plane $x = 0$

$S_4$ is the rectangle in the plane $z = 0$

$S_5$ is the surface of the cylinder.

The normal vectors to these respective surfaces are $(0,-1,0), (0,1,0), (-1,0,0),(0,0,-1), (\cos\theta, 0, \sin \theta) $ respectively.

I am going evaluate the surfaces $S_1, S_2$ together for reasons that hopefully become apparent.

$\iint -2x\ dS_1 + \iint 2x + 8 \ dS_2\\ \int_0^3\int_0^\sqrt{9-x^2} -2x\ dz\ dx + \int_0^3\int_0^\sqrt{9-x^2} 2x + 8 \ dz\ dx\\ \int_0^3\int_0^\sqrt{9-x^2} 8\ dz\ dx\\ 18\pi$

$\iint -6z\ dS_3\\ \int_0^3\int_0^8 -6z\ dy\ dz\\ -216$

$\iint x\ dS_4\\ \int_0^3\int_0^8 x\ dy\ dx\\ 36$

$S_5$ We will parameterize the surface.

$x = 3\cos \theta\\ y = y\\ z = 3\sin\theta$

$dS = (3\cos\theta, 0,3\sin\theta)\ dy\ d\theta\\ F(y,\theta)\cdot dS = 45\sin\theta\cos\theta \ dy\ d\theta$

$\int_0^{\frac{\pi}{2}}\int_0^8 45\sin\theta\cos\theta \ dy\ d\theta\\ \int_0^{\frac{\pi}{2}} 180\sin 2\theta d\theta\\ -90\cos 2\theta|_0^{\frac {\pi}2}\\ 180$

Add them together we get $18\pi$

There are some reasons we might deduce that flux over $S_3, S_4$ and $S_5$ will cancel each other out.

  • $\begingroup$ Thanks a lot @Doug M Really i understand the concept clearly $\endgroup$ – emonhossain Aug 30 '18 at 17:40

I don't know if you want to do this way, but in case you are allowed to do by the divergence theorem here it goes (I'm a little bit lazy to do it by surface integrals).

$$\int_{\partial\Omega}{\vec{A}\cdot d\vec{S}}=\int_{\Omega}{\textrm{div}(\vec{A})\,dV}$$ Since $\textrm{div}(\vec{A})=1$ in your case, you must calculate the volume of your body. In this case is a quarter of the volume of a cylinder of radius equal to $R=3$ and a height of $h=8$, giving $$V = \int_{\Omega}{\,dV}=\frac{1}{4}\pi R^2h = 18\pi $$

For the cylindrical part, make a change of coordinates: $$x = R\sin{\theta}\qquad y=y\qquad z = R\cos{\theta}\qquad $$

In the cylindrical surface , $\vec{A}=(6R\cos{\theta},2R\sin{\theta}+y,-R\sin{\theta})$, and $d\vec{S}=(\sin{\theta},0,\cos{\theta})Rd\theta dy$. Multiply and integrate! $\theta\in[0,\pi/2]$ and $y\in[0,8]$

  • $\begingroup$ Thank you @HBR but I have to show that by surface Integral. $\endgroup$ – emonhossain Aug 30 '18 at 13:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.