# A Divergence Theorem Exercise

This problem was sent to me recently:

Verify the divergence theorem for the following force field and surface:

$\vec{F} = (4 x, -2 y^2, z^2) \\ S: x^2 + y^2 = 4 \\ 0 \leq z \leq 3$

Divergence theorem:

$\int \int \int \nabla \cdot \vec{F} \, dV = \int \int \vec{F} \cdot \vec{n} \, dS$

Vector field divergence and differential volume:

$\nabla \cdot \vec{F} = 4 - 4 y + 2 z \\ dV = dx \, dy \, dz$

Transforming to polar coordinates:

$\nabla \cdot \vec{F} = 4 - 4 r \, sin \theta + 2 z \\ dV = r \, dr \, d \theta \, dz$

Divergence integral (the triple integral is easy but tedious, so I've simply given the answer):

$\int_0^3 \int_0^{2 \pi} \int_0^2 (4 - 4 r \, sin \theta + 2 z) \, (r \, dr \, d \theta \, dz) \\ \int_0^3 \int_0^{2 \pi} \int_0^2 (4 r - 4 r^2 \, sin \theta + 2 r z) \, dr \, d \theta \, dz = 84 \pi$

As for the right side, it's obvious by inspection that the right-side integral over the $$z = 0$$ face is zero, and that the right-side integral over the $$z = 3$$ face is $$36 \pi$$. If the divergence theorem holds, the right-side integral over the side of the cylinder must equal $$48 \pi$$.

Cylinder side normal vector and vector field:

$\vec{n} = (cos \theta, sin \theta , 0) \\ \vec{F} = (4 r \cos \theta, -2 r^2 sin^2 \theta , z^2)$

Dot product and differential surface area:

$\vec{F} \cdot \vec{n} = 4 r \cos^2 \theta -2 r^2 sin^3 \theta \\ dS = r \, d \theta \, dz$

Over the cylinder side, $$r = 2$$, so:

$\vec{F} \cdot \vec{n} = 8 \cos^2 \theta - 8 sin^3 \theta \\ dS = 2 \, d \theta \, dz \\ \vec{F} \cdot \vec{n} \, dS = 16 (\cos^2 \theta - sin^3 \theta) \, d \theta \, dz$

Right-side integral over cylinder side:

$\int \int \vec{F} \cdot \vec{n} \, dS \\ \int_0^3 \int_0^{2 \pi} 16 (\cos^2 \theta - sin^3 \theta) \, d \theta \, dz \\ = 48 \int_0^{2 \pi} (\cos^2 \theta - sin^3 \theta) \, d \theta \\ = 48 \pi$

Note by Steven Chase
2 weeks, 1 day ago

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We can also solve this by taking a projection on the XZ plane for the surface integral of the cylinder?

- 2 weeks, 1 day ago

In general, you can integrate over a projection instead of the actual surface. However, you have to formulate the integrand in a particular way when doing so. A vector calculus text book should have a description of that.

- 2 weeks, 1 day ago

Oh I see. Thank you!

- 2 weeks, 1 day ago