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I have a vector field $\vec v=(x^2+y^2+z^2)(x \hat x+y \hat y +z \hat z)$, and I need to computed the integral of $\Delta \cdot \vec v$ over the region $x^2+y^2+z^2 \le 25$. I see that $\vec v=r^2\vec r$ which is $\vec v=r^3 \hat r$. The way I tried to compute the integral was like this:

$$\int r^3\hat r \cdot d\vec a \ ; \ d\vec a = r^2sin(\phi )drd\theta d\phi $$

$$\int_0^5 r^3 \hat r\cdot r^2sin(\phi)d\theta d\phi= \int_0^{{\pi\over 2}}d\theta\int_0^{{\pi\over 2}}sin(\phi)d\phi\int_0^5r^5dr$$

$$=\Big({\pi\over 2} \Big)\Big(1\Big)\Big({1\over 6}5^6\Big)\approx 1302\pi$$

This isn't right because the solution should be $12,500\pi$. However the worked out solution doesn't make much sense to me because it seems like there a lot of steps missing and there isn't an explanation for what was done:

$$\int \vec A \cdot d\vec a= \int \big(r^3 \hat r \big) \cdot d\vec a$$ $$=r^3 4 \pi r^2=4r^5 \pi = 4 \big(5^5 \big) \pi= 12,500 \pi$$ $$where \ \ r=\sqrt{25}=5$$

I don't understand this solution. Why wasn't there any integration?

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You have to compute the integral of the divergence of $\mathbf{v}$ over a closed volume bounded by a sphere of radius 5. The Divergence Theorem states that such volumetric integral is equal to the outward flux of the vector field: $$ \int _{V}\left(\mathbf {\nabla } \cdot \mathbf {v} \right)\,dV= \oint_S (\mathbf {v} \cdot \mathbf {n} )\,dS. $$

As the vector field norm depends only on the radius and it is always normal to the sphere, $\mathbf{v}\cdot\mathbf{n}$ is constant over the sphere and equal to $r^3=125$. Then $$ \oint_S (\mathbf {v} \cdot \mathbf {n} )\,dS=125\oint dS=125\cdot 4\pi r^2=12,500\pi. $$ There was integration, but the symmetries of the problem made the computation very easy.

In practice, you tried to compute the volume integral of the vector field, not its divergence, and you equated a vector area element $d\mathbf{a}$ with a volume element $ r^{2}\sin \theta \,\mathrm {d} r\,\mathrm {d} \theta \,\mathrm {d} \varphi$, obtaining a incorrect result.

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