Change of variable (involving partial diff.)

[athrun200]

Homework Statement

Homework Equations

The Attempt at a Solution

I am not sure part b and c.

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

 

[Sourabh N]

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.

 

[athrun200]

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 r=\sqrt{x^{2}+y^{2}+z^{2}}, isn't it?

 

[Sourabh N]

Ah, ofcourse! My bad.

 

[athrun200]

Well if I intergrate r at the same time, it becomes \int\frac{x}{({x^{2}+y^{2}+z^{2}})^\frac{3}{2}}

Which is difficult to integrate...

 

[Sourabh N]

Well if I intergrate r at the same time, it becomes \int\frac{x}{({x^{2}+y^{2}+z^{2}})^\frac{3}{2}}

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)

 

[athrun200]

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?

 

[Sourabh N]

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\infty to x² + y² + z².

 

[athrun200]

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

 

[Sourabh N]

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

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 \hat{x} and similarly y and z integral are accompanied with respective unit vectors. Together they form \vec{E} which has to look same as 8.11 (this is part c!)

 

[athrun200]

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 \hat{x} and similarly y and z integral are accompanied with respective unit vectors. Together they form \vec{E} which has to look same as 8.11 (this is part c!)

I don't understand this point.
Do you mean that I miss out the vector?

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

 

[Sourabh N]

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

I cannot possibly give a better explanation. I highly recommend you have a look (Sec 2.3 in Griffith - Intro to electrodynamics)

I don't understand this point.
Do you mean that I miss out the vector?

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

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

 

[athrun200]

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

I would like to know if I get the correct cylindrical form.

 

[Sourabh N]

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

Yes that is correct. You can check that by using \vec{E} = -\nabla \phi and comparing it with 8.11

I would like to know if I get the correct cylindrical form.

To get \phi in cylindrical form, you have to write \vec{E} in cylindrical coordinates and follow the procedure as above. (To know how to go to cylindrical from spherical, look at http://is.gd/JFitBh)

 

[athrun200]

To get \phi in cylindrical form, you have to write \vec{E} 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?