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The Divergence Theorem |
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The Divergence Theorem is the three-dimensional version of the flux form of Green's Theorem and it relates the flow (or flux) through the boundary of a closed surface S to the divergence of the vector field through the volume Q. The surface S encloses the volume Q. |
Divergence Theorem |
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If \(\vec{N}\) is the unit normal vector directed outward from a solid region Q, then |
\(\displaystyle{ \iint\limits_S { \vec{F} \cdot \vec{N} ~dS = \iiint\limits_Q { ( \nabla \vec{F} ) ~ dV } } }\) |
Divergence Theorem in Detail |
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When you studied surface integrals, you learned how to calculate the flux integral \(\displaystyle{\iint_S{\vec{F} \cdot \vec{N} ~ dS} }\) which is basically the flow through a surface S.
When you have a solid region bounded by a closed surface S, there is an easier way to calculate this integral, using the Divergence Theorem.
Basically what this Divergence Theorem says is that the flow or flux of the vector field \(\vec{F}\) through the closed surface S, which is given by the integral \(\displaystyle{\iint_S{\vec{F} \cdot \vec{N} ~ dS} }\) is equal to the divergence of \(\vec{F}\) through the volume Q. This is given by the integral \(\displaystyle{ \iiint\limits_Q { ( \nabla \vec{F} ) ~ dV } }\).
The required relationship between the surface S in the first integral and the volume Q in the second integral is that the volume Q must be bounded by the surface S.
Here is a great video explaining the Divergence Theorem.
Dr Chris Tisdell - Divergence theorem of Gauss [12min-20secs] | |
Usually one of the integrals is easier to set up and evaluate than the other. Most of the time the easier integral is the volume integral since surface integrals are more difficult to set up in general. But you will get a feel for this as you get more experience.
Before we go on, let's look at an example.
Divergence Theorem Example |
Example 1 - Verify the Divergence Theorem for \(\vec{F}(x,y,z)=xy\hat{i}+z\hat{j}+(x+y)\hat{k}\) where the surface consists of the bounding planes \(y=4\), \(z=4-x\) and the coordinate planes. | |
Divergence Theorem in Depth |
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Here are several full length video lectures that explain the divergence theorem in more depth. These are important to watch in order to understand what is really going on with this theorem and how to use it properly.
Evans Lawrence - Lecture 34 - Divergence Theorem [22min-22secs] | |
MIT OCW - Lec 28-29 | MIT 18.02 Multivariable Calculus, Fall 2007 [2 videos 99min-27secs total] | |
Divergence Theorem Proof |
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Although not completely necessary to the use of the divergence theorem, going through the proof will help you understand in more detail how this proof works.
Khan Academy - Divergence Theorem Proof [5 videos] | |
Okay, you are now ready for some practice problems. |
Stokes' Theorem → |
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Practice Problems |
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Instructions - - Unless otherwise instructed, solve these problems using The Divergence Theorem.
- If you are asked to calculate the surface integral, you will usually evaluate the volume integral.
- Similarly, if you are asked to calculate the volume integral, you will usually evaluate the surface integral.
Level A - Basic |
Practice A01 | |
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Determine the flux of the vector field \(\vec{F}=\langle x^2,y^2,z^3\rangle\) across the surface bounded by the planes \(x=0\), \(x=2\), \(y=0\), \(y=2\), \(z=0\), \(z=4\). | |
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Practice A02 | |
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Determine the flux of the vector field \(\vec{F}=\langle x^2/2,yz,-xz\rangle\) across the surface \(z=\sqrt{4-x^2-y^2}\), \(z=0\). | |
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Practice A03 | |
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Compute the flux of \(\vec{F}=\langle x^4y, -2x^3y^2,z^2\rangle\) through the surface of the solid bounded by \(z=0\), \(z=h\) and \(x^2+y^2=R^2\). | |
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Practice A04 | |
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Calculate the surface integral for \(\vec{F}=\langle 4x^3z,4y^3z,3z^4\rangle\) where the surface of the solid is bounded by the hemispheres \(z=\sqrt{4-x^2-y^2}\), \(z=\sqrt{1-x^2-y^2}\) and the plane \(z=0\). | |
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Practice A05 | |
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Calculate the surface integral for \(\vec{F}=\langle 3xy^2,xe^z,z^3\rangle\) where the solid is bounded by the cylinder \(y^2+z^2=25\) and the planes \(x=-4\) and \(x=3\). | |
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