Flux in a rotated cylindrical coordinate system

In summary: 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...
  • #1
Kaguro
221
57
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?
 
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  • #2
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.
 
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  • #3
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.
 
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  • #4
Fred Wright said:
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?
 
  • #5
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##
 
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  • #6
BvU said:
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?
 
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  • #7
Kaguro said:
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.
 
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  • #8
LCKurtz said:
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!
 
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Related to Flux in a rotated cylindrical coordinate system

What is flux in a rotated cylindrical coordinate system?

Flux in a rotated cylindrical coordinate system refers to the flow of a vector field through a surface that is oriented in a rotated cylindrical coordinate system. This is a mathematical concept that is often used in physics and engineering to analyze the behavior of vector fields.

How is flux calculated in a rotated cylindrical coordinate system?

The calculation of flux in a rotated cylindrical coordinate system involves using a mathematical formula that takes into account the vector field, the surface area, and the orientation of the surface. This formula is derived from the dot product of the vector field and the unit normal vector to the surface.

What are some real-world applications of flux in a rotated cylindrical coordinate system?

Flux in a rotated cylindrical coordinate system has many practical applications, such as in fluid dynamics, electromagnetism, and heat transfer. For example, it can be used to analyze the flow of fluids through pipes or the distribution of heat in a cylindrical object.

What is the significance of using a rotated cylindrical coordinate system in flux calculations?

Using a rotated cylindrical coordinate system allows for a more accurate and efficient calculation of flux in certain situations. It can simplify the equations and make them easier to solve, especially in cases where the vector field has a cylindrical symmetry.

Are there any limitations to using a rotated cylindrical coordinate system in flux calculations?

While using a rotated cylindrical coordinate system can be beneficial in certain cases, it may not be the most appropriate coordinate system for all situations. For example, if the vector field does not have a cylindrical symmetry, using a rotated cylindrical coordinate system may complicate the calculations and make them less accurate.

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