# Flux in a rotated cylindrical coordinate system

Kaguro
Homework Statement:
The vector field ##\vec F= 2x^2y \hat i - y^2 \hat j + 4xz^2 \hat k ## is defined over the region in the first octant bounded by ## y^2+z^2=9## and x=2. Find the value of ##\iint_S \vec {F} \cdot \hat n dS##
(a)100
(b)18
(c)0.18
(d)1.8
Relevant Equations:
Gauss's Divergence Theorem.
##\vec F= 2x^2y \hat i - y^2 \hat j + 4xz^2 \hat k ##
## \Rightarrow \vec \nabla \cdot \vec F= 4xy-2y+8xz##
Let's shift to a rotated cylindrical system with axis on x axis:

##x \to h, y \to \rho cos \phi, z \to \rho sin \phi ##
Then our flux, as given by the Divergence theorem is the volume integral of the divergence.

Flux = ##\int_{\phi=0}^{\pi /2} \int_{h=0}^{2} \int_{\rho=0}^{3} (4h \rho cos \phi -2 \rho cos \phi + 8h \rho sin \phi ) \rho d \rho d \phi dh##

=##\int_{\phi=0}^{\pi /2} \int_{h=0}^{2} \int_{\rho=0}^{3} (4h cos \phi -2 cos \phi + 8h sin \phi ) \rho^2 d \rho d \phi dh##

=##\int_{\phi=0}^{\pi /2} \int_{h=0}^{2} (4h cos \phi -2 cos \phi + 8h sin \phi ) [\rho^3 /3]_0^{3} d \phi dh##

=##9\int_{\phi=0}^{\pi /2} \int_{h=0}^{2} (4h cos \phi -2 cos \phi + 8h sin \phi ) d \phi dh##

=##9\int_{\phi=0}^{\pi /2} [2h^2 cos \phi -2h cos \phi + 4h^2 sin \phi ]_0^2 d \phi##

=##9\int_{\phi=0}^{\pi /2} (8cos \phi -4cos \phi + 16 sin \phi)d \phi##

=##36\int_{\phi=0}^{\pi /2} (cos \phi +4 sin \phi)d \phi##

=##36( [sin \phi ]_0^{\pi/2} +4 [cos \phi ]_{\pi/2}^0 )##

=36(1+4) = 36*5 = 180

Which doesn't match anything. The answer given is (a) 100.

Is there something which I did wrong here?

Fred Wright
Why don't you compute the integral in the homework statement? You have$$\vec F=2x^2y\hat i -y^2\hat j +4xz^2\hat k$$and the normal vector to the cylinder in the first quadrant is$$\hat n = \frac{1}{3}(y\hat j +z\hat k)$$Therefore $$\vec F \cdot \hat n =\frac{1}{3}(4xz^3 - y^3)$$ Using the parameterization$$y=3\cos(\theta)$$$$z=3\sin(\theta)$$ The differential element of surface area is$$dS=3dxd\theta$$The integral becomes$$27\int_0^{\frac{\pi}{2}}\int_0^2(4x\sin^3(\theta)-\cos^3(\theta))dxd\theta$$See http://math.mit.edu/~jorloff/suppnotes/suppnotes02/v9.pdf for an explanation on how to do these kind of problems.

Kaguro, Delta2 and etotheipi
Homework Helper
Gold Member
I think your answer is correct. Also, looking at the answer choices, I would bet the first answer is supposed to be ##180##, given the pattern in the answer choices. Probably just a typo.

Kaguro and Delta2
Kaguro
Why don't you compute the integral in the homework statement? You have$$\vec F=2x^2y\hat i -y^2\hat j +4xz^2\hat k$$and the normal vector to the cylinder in the first quadrant is$$\hat n = \frac{1}{3}(y\hat j +z\hat k)$$Therefore $$\vec F \cdot \hat n =\frac{1}{3}(4xz^3 - y^3)$$ Using the parameterization$$y=3\cos(\theta)$$$$z=3\sin(\theta)$$ The differential element of surface area is$$dS=3dxd\theta$$The integral becomes$$27\int_0^{\frac{\pi}{2}}\int_0^2(4x\sin^3(\theta)-\cos^3(\theta))dxd\theta$$See http://math.mit.edu/~jorloff/suppnotes/suppnotes02/v9.pdf for an explanation on how to do these kind of problems.

This integral gives a value of 108.
If I assume my answer is correct, then why doesn't the answer found by using the Divergence Theorem match this one?

Homework Helper
Hello Kaguro,

I second LCKurtz: can't find anything wrong with your calculation.

On a side note some tips: your ##\LaTeX## looks better if you use
• displaystyle for the integrals (So either enclose in $$or use \displaystyle { ...} • \cos and \sin etc. instead of cos and sin, so they look like functions and get correct spacing • enforce some spacing with \, or \; ##\rho \;d \rho\, d \phi\, dh## instead of ##\rho d \rho d \phi dh##$$\int_{\phi=0}^{\pi /2} \int_{h=0}^{2} \int_{\rho=0}^{3} (4h \rho \cos \phi -2 \rho \cos \phi + 8h \rho \sin \phi ) \rho\; d \rho\, d \phi\, dh$$A disputable improvement (I personally don't like it, mainly because I get an ugly font) comes from using ##{\sf d}\ ## ( {\sf d} ) instead of ##d##: ##\rho \;{\sf d} \rho\, {\sf d} \phi\, {\sf d}h## SammyS and Kaguro Kaguro Hello Kaguro, I second LCKurtz: can't find anything wrong with your calculation. On a side note some tips: your ##\LaTeX## looks better if you use • displaystyle for the integrals (So either enclose in$$ or use \displaystyle { ...}
• \cos and \sin etc. instead of cos and sin, so they look like functions and get correct spacing
• enforce some spacing with \, or \; ##\rho \;d \rho\, d \phi\, dh## instead of ##\rho d \rho d \phi dh##
$$\int_{\phi=0}^{\pi /2} \int_{h=0}^{2} \int_{\rho=0}^{3} (4h \rho \cos \phi -2 \rho \cos \phi + 8h \rho \sin \phi ) \rho\; d \rho\, d \phi\, dh$$

A disputable improvement (I personally don't like it, mainly because I get an ugly font) comes from using ##{\sf d}\ ## ( {\sf d} ) instead of ##d##: ##\rho \;{\sf d} \rho\, {\sf d} \phi\, {\sf d}h##
Okay I'll keep these in mind. But could you tell me what is reason that directly evaluating the surface integral gives a different answer?

etotheipi
Homework Helper
Gold Member
Okay I'll keep these in mind. But could you tell me what is reason that directly evaluating the surface integral gives a different answer?

Did you check all the sides of the volume when you did the surface integrals? They each have their own normal and some of them may not be zero.

etotheipi and Kaguro
Kaguro
Did you check all the sides of the volume when you did the surface integrals? They each have their own normal and some of them may not be zero.
Absolutely spectacular observation!

I completely missed this fact!

Gauss's Divergence Theorem converts closed surface integrals into volume integrals!

There are 5 surfaces of the given piece of cylinder.

One the curved surface whose flux is 108, the one done by Fred Wright.

The bottom where x=0 gives 0 flux.
The other two cuts where y=0 and z=0 respectively also give flux 0.

But the last side the top portion where x=2, gives flux 72. ( I checked!)

So now if I add this to the 108, I get 180, the same answer as given by the Divergence Theorem!

Thanks once again for this fantastic observation!

BvU