Some Help With Calculus Please?

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cepheid
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#1 We were given the following information involving Gauss' law in a math assignment, just as an application of the surface integral of a vector field:

[tex] Q = \epsilon_{0} \iint_{S} \vec{E} \cdot d \vec{S} [/tex]

[tex] \vec{E}(x,y,z) = x \hat{i} + y \hat{j} + z \hat{k} [/tex]

[itex] S [/itex] is the cube with vertices
[itex](\pm 1, \pm 1, \pm 1)[/itex]

I approached this problem by finding an expression for each of the faces (e.g. z = 1, within the appropriate bounds), calculating the surface integral over each one, and adding them together. I have two problems with this:

1. The textbook says that the integral given represents the charge [itex] Q [/itex] enclosed by a closed surface. So what does it mean to calculate this integral for a surface that isn't closed, like the face of a cube?

2. I got an answer of [itex] Q = 0 [/itex]. Is this correct?

________________________________________________________________
 
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  • #2
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Originally posted by cepheid

I approached this problem by finding an expression for each of the faces (e.g. z = 1, within the appropriate bounds), calculating the surface integral over each one, and adding them together. I have two problems with this:

1. The textbook says that the integral given represents the charge [itex] Q [/itex] enclosed by a closed surface. So what does it mean to calculate this integral for a surface that isn't closed, like the face of a cube?
that quantity is called flux.

2. I got an answer of [itex] Q = 0 [/itex]. Is this correct?__________________________________________


i don t believe so. i got Q=24
 
  • #3
cepheid
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It looks like you got the answer four for each of the six integrals. But when I computed them, since the normal vectors for opposite faces had opposite signs, half of my integrals evaluated to negative four. This must be a conceptual error...where did I go wrong?
 
  • #4
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for the faces that are parallel to the x-y plane, [itex]\mathbf{E}\cdot d\mathbf{A}=zdxdy[/itex]. so there is a minus sign for the bottom face from the normal vector, and another minus sign from the fact that z=-1 on the bottom, they cancel out and the flux is positive.

this way, all faces give a positive contribution.
 
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  • #5
cepheid
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Thank you!

# 2 We were to make use of Stokes' theorem to solve the following problem:

Evaluate:

[tex] \int_{C} (y + \sin x)dx + (z^{2} + \cos y)dy + x^{3}dz [/tex]

where [itex] C [/itex] is the curve:

[tex]\vec{r}(t) = <\sin t, \cos t, \sin 2t> [/tex]
[tex] 0 \leq t \leq 2\pi [/tex]

Hint : observe that [itex] C [/itex] lies on [itex] z = 2xy [/itex].

I made an honest attempt, but what I ended up with is far too cumbersome to repeat here. Could someone please point me in the right direction?
 
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  • #6
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lets see.... by stokes theorem, this integral is equal to the integral over some region whose boundary is C of the following monkey:

[tex]-dx\wedge dy-2zdy\wedge dz+3x^2dx\wedge dz[/tex]

if that doesn t look familiar to you, its just a way of writing the curl of the thing you gave me. so now all we have to do is choose a convenient area to integrate over. how about the area of the surface [itex]z=2xy[/itex] over the region [itex]x^2+y^2\leq 1[/itex]?
 
  • #7
cepheid
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Hmm...I'm not sure I follow what the significance of the z = 2xy is (I can't even figure out what that surface is), or how you came up with that region.

#3 (the last one). This one completely stumped me as well.

Use the Divergence Theorem to evaluate:

[tex] \iint_{S} \vec{F} \cdot d \vec{S} [/tex]

That is, calculate the flux of [itex] \vec{F} [/itex] across [itex] S [/itex] .

[tex] \vec{F}(x,y,z) = x^{2}y \hat{i} + xy^{2} \hat{j} + 2xyz \hat{k} [/tex]

[itex] S [/itex] is the surface of the tetrahedron bounded by the planes [itex] x = 0, y = 0, z = 0, \textrm{and} x + 2y + z = 2 [/itex].

Obviously, I need to evaluate:

[tex] \iiint_{E} (\nabla \cdot \vec{F}) dV [/tex]

But I'm not sure what to do with that darn tetrahedron.
 
  • #8
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it sounds to me like you need to some refreshers on multiple integrations. i would say a bunch of stuff here, but it is time for me to go to sleep, i think.

maybe tomorrow.

sorry
 
  • #9
cepheid
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No problem! I am doing multiple integration for the first time (the course is multivariable and vector calculus squeezed into one), so I did double and triple integrals scarcely a month ago, and haven't had time to become proficient. I appreciate the help that I did get. A demain...
 
  • #10
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Originally posted by cepheid

But I'm not sure what to do with that darn tetrahedron.

so for the divergence, i got [itex]\nabla\cdot\mathbf{F}=6xy[/itex]

to take care of the tetrahedron, you should choose the order you want to do your integrations. how about x, then y, then z.... for a generic z, and y, x ranges from 0 to 2-2y-z, according to the equation you gave me. then once i have integrated over x, for some generic y, that y can range from 0 to 1-z/2, and once you have tallied all the area over x and y for a generic z, let z run from 0 to 2. if its not clear where i got those numbers from, well, they are the vertices of your tetrahedron.

for problems like this, a picture is immensly helpful

doing all those integrations is a pain in the ass, but my final answer is 2/5... it should start out looking something like this:

[tex]
\int^2_0\left(\int^{1-z/2}_0\left(\int^{2-2y-z}_06xy\ dx\right)dy\right)dz
[/tex]
 

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