A Divergence Theorem Exercise

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

F=(4x,2y2,z2)S:x2+y2=40z3\vec{F} = (4 x, -2 y^2, z^2) \\ S: x^2 + y^2 = 4 \\ 0 \leq z \leq 3

Divergence theorem:

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

Vector field divergence and differential volume:

F=44y+2zdV=dxdydz\nabla \cdot \vec{F} = 4 - 4 y + 2 z \\ dV = dx \, dy \, dz

Transforming to polar coordinates:

F=44rsinθ+2zdV=rdrdθdz\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):

0302π02(44rsinθ+2z)(rdrdθdz)0302π02(4r4r2sinθ+2rz)drdθdz=84π\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=0z = 0 face is zero, and that the right-side integral over the z=3z = 3 face is 36π 36 \pi . If the divergence theorem holds, the right-side integral over the side of the cylinder must equal 48π 48 \pi .

Cylinder side normal vector and vector field:

n=(cosθ,sinθ,0)F=(4rcosθ,2r2sin2θ,z2) \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:

Fn=4rcos2θ2r2sin3θdS=rdθdz \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=2r = 2 , so:

Fn=8cos2θ8sin3θdS=2dθdzFndS=16(cos2θsin3θ)dθdz \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:

FndS0302π16(cos2θsin3θ)dθdz=4802π(cos2θsin3θ)dθ=48π \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
1 year 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?

Nashita Rahman - 1 year ago

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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.

Steven Chase - 1 year ago

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Oh I see. Thank you!

Nashita Rahman - 1 year ago

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