# [Mathematical physics] - Integral problem

## Homework Statement

Calculate the integral

$$\int_{S} (\frac{A}{r^2}\hat{r} + B\hat{z}) \cdot d\vec{S}$$

Where S is the sphere with r = a.

2. The attempt at a solution

I have no clue how to solve this problem. I have thought of introducing spherical coordinates and somehow finding a connection but I don't think that works.

I tried breaking out $$d\vec{S} = \frac{\partial \vec{r}}{\partial u} \cdot \frac{\partial \vec{r}}{\partial v } dudv$$

using the formula above but not sure on how the dot product works. What confuses me with the integrand with the z and r. The answer is $$4πA$$.

Orodruin
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$$d\vec{S} = \frac{\partial \vec{r}}{\partial u} \cdot \frac{\partial \vec{r}}{\partial v } dudv$$
This is not the correct expression for the surface element. The correct expression is
$$d\vec S = \frac{\partial \vec r}{\partial u} \times \frac{\partial \vec r}{\partial v} du\, dv.$$
This is equal to ##\vec n \, dS##, where ##\vec n## is a unit normal and ##dS## the area spanned by the tangent vectors on the sphere. What is the unit normal to the unit sphere?

Also, the second term does not contribute to the integral. Do you know of some integral theorem that you can invoke to come to this conclusion?

Chandra Prayaga
Ask yourself, what is the vector ## d \vec S ##?

This is not the correct expression for the surface element. The correct expression is
$$d\vec S = \frac{\partial \vec r}{\partial u} \times \frac{\partial \vec r}{\partial v} du\, dv.$$
This is equal to ##\vec n \, dS##, where ##\vec n## is a unit normal and ##dS## the area spanned by the tangent vectors on the sphere. What is the unit normal to the unit sphere?

Also, the second term does not contribute to the integral. Do you know of some integral theorem that you can invoke to come to this conclusion?

I might be completely lost but isn't the normal of a surface given by $$\nabla = \frac{\partial}{\partial x} + \frac{\partial}{\partial y} + \frac{\partial}{\partial z}$$? It should be ⊥ to the tangentvector.

Or maybe the position vector $$\vec{r} = x\hat{x} + y\hat{y} + z\hat{z}$$? I think this one simply desribes the coordinates of a sphere as opposed to a normal.

The don't know if the book covered the relevant integral theroem since I read it 4 times.

Orodruin
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The ##\nabla## operator in itself is a differential operator, not a vector. Acting on a scalar field, it becomes the gradient, which is normal to the level surfaces of the field.

The integral theorem is the divergence theorem. You can do without as well, but it helps.

The ##\nabla## operator in itself is a differential operator, not a vector. Acting on a scalar field, it becomes the gradient, which is normal to the level surfaces of the field.

The integral theorem is the divergence theorem. You can do without as well, but it helps.
I don't think I'm supposed to use the divergence theorem since it's not discussed in this chapter.

Orodruin
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