# Vector calculus - Divergence Theorem

• joex444
In summary, the problem involves finding the integral of a vector function over a surface defined by a sphere. Using Gauss's theorem and converting between cartesian and spherical coordinates, the solution can be found by evaluating the divergence of the vector function and using the given limits of integration.

## Homework Statement

Find $$\int_{s} \vec{A} \cdot d\vec{a}$$ given $$\vec{A} = ( x\hat{i} + y\hat{j} + z\hat{k} ) ( x^2 + y^2 + z^2 )$$ and the surface S is defined by the sphere $$R^2 = x^2 + y^2 + z^2$$ directly and by Gauss's theorem.

## Homework Equations

$$\int_{s} \vec{A} \cdot d\vec{a} = \int_{V} \nabla \cdot \vec{A} da = \int\int\int \frac{1}{r^2} \frac{\partial}{\partial r}(r^2 A_{r}) r^2 \sin\theta dr d\theta d\phi = \int\int\int \frac{\partial}{\partial r}(r^2 A_{r}) \sin\theta dr d\theta d\phi = 4 \pi R^5$$

## The Attempt at a Solution

Given the answer, I was able to work backwards using limits of integration of 0 to 2pi for phi, 0 to pi for theta and 0 to R on r to find out that A = r^3. Now, the question I have is how can I show that A = r^3 given the A in cartesian coordinates? Since $$r=\sqrt{x^2 + y^2 + z^2}$$ its clear that $$A=r^2(x\hat{i} + y\hat{j} + z\hat{k})$$ but to call the vector valued term r is not agreeing with me. Obviously, since the surface is a sphere, I thought it would be easier to use spherical coordinates...

Edit: Using cartesian coordinates I can find $$\nabla \cdot \vec{A} = 5(x^2 + y^2 + z^2) = 5r^2$$ but using the formula given for spherical coordinates, I would need to show A = r^3 to do a full solution in spherical.

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What do you mean "A= r3"? What is A? There is no A in the statement of the problem.

A is given, it is the vector. It's in the first line... Of course, to solve for A in spherical I worked backwards from the answer (I needed an R^5 in the final answer, and the divergence in spherical gives me an integral of a partial of r^2 * A, so obviously r^2 * A needs to have a 5th power, thus A = r^3, working it through, it does come out correctly).

When I try to convert the given A into spherical directly, I can only get as far as $$\vec{A} = r^2(x\hat{i}+y\hat{j}+z\hat{k})$$

Another question I have is whether it would be easier to show it directly in cartesian or spherical...I'm leaning towards cartesian because A is given in cartesian.

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A = r^3

As Halls pointed out, that's not right.

The conversion from cartesian to spherical, and vice versa, should be found in any vector calculus text.

(or try, http://www.math.montana.edu/frankw/ccp/multiworld/multipleIVP/spherical/body.htm [Broken])

Another question I have is whether it would be easier to show it directly in cartesian or spherical...I'm leaning towards cartesian because A is given in cartesian.

The integral $\int_{s} \vec{A} \cdot d\vec{a}$ over $x^2 + y^2 + z^2=R^2$ should be easier to evaluate in spherical coordinates because of the symmetry (ie, you're integrating over a sphere).

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