Change of variable (involving partial diff.)

  • Thread starter athrun200
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  • #1
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Homework Statement



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Homework Equations



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The Attempt at a Solution


I am not sure part b and c.

Also, I don't know how to answer the last question.

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Answers and Replies

  • #2
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In your solution for part 3, you have a factor of r3 in the denominator outside the integral (in each of the three integrals). Remember r = x + y + z, so you cannot take it outside the integral.

Your approach for part c looks confusing. We'll talk about part c, let's get through part b first.
 
  • #3
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In your solution for part 3, you have a factor of r3 in the denominator outside the integral (in each of the three integrals). Remember r = x + y + z, so you cannot take it outside the integral.

Your approach for part c looks confusing. We'll talk about part c, let's get through part b first.

It seems [itex]r=\sqrt{x^{2}+y^{2}+z^{2}}[/itex], isn't it?
 
  • #4
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Ah, ofcourse! My bad.
 
  • #5
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Well if I intergrate r at the same time, it becomes [itex]\int\frac{x}{({x^{2}+y^{2}+z^{2}})^\frac{3}{2}}[/itex]

Which is difficult to integrate...
 
  • #6
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Well if I intergrate r at the same time, it becomes [itex]\int\frac{x}{({x^{2}+y^{2}+z^{2}})^\frac{3}{2}}[/itex]

Which is difficult to integrate...
Not really. For example, for x integral take x2 + y2 + z2 = t, then 2xdx = dt and your integral is simply dt/2*(t^1.5)
 
  • #7
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Not really. For example, for x integral take x2 + y2 + z2 = t, then 2xdx = dt and your integral is simply dt/2*(t^1.5)

Oh. I forget that I can use substitution.
But how about the new interval?
From 0 to x^2+y^2+z^2?
 
  • #8
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Oh. I forget that I can use substitution.
But how about the new interval?
From 0 to x^2+y^2+z^2?
Common convention is to express potential relative to potential at infinity, so limits would be[itex]\infty[/itex] to x2 + y2 + z2.
 
  • #9
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Now, I find that the integral for x, y and z are the same.
Is that correct now?


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  • #10
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Now, I find that the integral for x, y and z are the same.
Is that correct now?


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2 things. First, although you wrote 0 as lower limit, you realize it's actually infinity (as I mentioned in previous post). Having the correct upper and lower limit will help you fix the sign of final value for x integral you get.
Second, note it's the x integral; it is accompanied with a [itex]\hat{x}[/itex] and similarly y and z integral are accompanied with respective unit vectors. Together they form [itex]\vec{E}[/itex] which has to look same as 8.11 (this is part c!)
 
  • #11
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2 things. First, although you wrote 0 as lower limit, you realize it's actually infinity (as I mentioned in previous post). Having the correct upper and lower limit will help you fix the sign of final value for x integral you get.
Do you mean I should put infinty as the upper limit and x^+y^+z^2 as lower limit?
What is the physical meaning if I put infinty as the lower limit and x^+y^+z^2 as the upper limit?


Second, note it's the x integral; it is accompanied with a [itex]\hat{x}[/itex] and similarly y and z integral are accompanied with respective unit vectors. Together they form [itex]\vec{E}[/itex] which has to look same as 8.11 (this is part c!)

I dont understand this point.
Do you mean that I miss out the vector?

However, inside the integral is a dot product. i.e.[itex]F\bullet dr[/itex]
Since dot product produce scalar only, there should be not any vector left.
 
  • #12
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Do you mean I should put infinty as the upper limit and x^+y^+z^2 as lower limit?
What is the physical meaning if I put infinty as the lower limit and x^+y^+z^2 as the upper limit?
Look at it this way - x2 + y2 + z2 = t (where y and z and not constant anymore). So, 2xdx + 2ydy + 2zdz = dt. This substitution makes more sense than taking x2 + y2 + z2 = t with y and z constant as I had suggested previously. Because now the limit of integral is infinity and x2 + y2 + z2. I am not sure about limits of the integral if y and z are kept constant.

On the meaning of upper lower limits - Griffith defines potential like
potential.jpg
I cannot possibly give a better explanation. I highly recommend you have a look (Sec 2.3 in Griffith - Intro to electrodynamics)

I dont understand this point.
Do you mean that I miss out the vector?

However, inside the integral is a dot product. i.e.[itex]F\bullet dr[/itex]
Since dot product produce scalar only, there should be not any vector left.

Ignore this. I jumped on part c in my head.
 
Last edited:
  • #13
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Look at it this way - x2 + y2 + z2 = t (where y and z and not constant anymore). So, 2xdx + 2ydy + 2zdz = dt. This substitution makes more sense than taking x2 + y2 + z2 = t with y and z constant as I had suggested previously. Because now the limit of integral is infinity and x2 + y2 + z2. I am not sure about limits of the integral if y and z are kept constant.

On the meaning of upper lower limits - Griffith defines potential like View attachment 37100 I cannot possibly give a better explanation. I highly recommend you have a look (Sec 2.3 in Griffith - Intro to electrodynamics)



Ignore this. I jumped on part c in my head.

So [itex]\phi=\frac{q}{r}[/itex]?
If it is correct, let's move to part c.

I would like to know if I get the correct cylindrical form.
 
  • #14
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So [itex]\phi=\frac{q}{r}[/itex]?
If it is correct, let's move to part c.
Yes that is correct. You can check that by using [itex]\vec{E}[/itex] = -[itex]\nabla[/itex] [itex]\phi[/itex] and comparing it with 8.11
I would like to know if I get the correct cylindrical form.
To get [itex]\phi[/itex] in cylindrical form, you have to write [itex]\vec{E}[/itex] in cylindrical coordinates and follow the procedure as above. (To know how to go to cylindrical from spherical, look at http://is.gd/JFitBh)
 
  • #15
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To get [itex]\phi[/itex] in cylindrical form, you have to write [itex]\vec{E}[/itex] in cylindrical coordinates and follow the procedure as above. (To know how to go to cylindrical from spherical, look at http://is.gd/JFitBh)

Do you mean that my approach above can't get the answer?
If so, why?
 

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