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## Homework Statement

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

## Homework Equations

[tex]\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[/tex]

## 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 [tex]r=\sqrt{x^2 + y^2 + z^2}[/tex] its clear that [tex]A=r^2(x\hat{i} + y\hat{j} + z\hat{k})[/tex] 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 [tex]\nabla \cdot \vec{A} = 5(x^2 + y^2 + z^2) = 5r^2[/tex] 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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