Surface Integrals

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The Divergence Theorem

Let G be a three-dimensional solid bounded by a piecewise smooth closed surface S that has orientation pointing out of G and let

\[\mathbf{F}\left( {x,y,z} \right) = \Big( {P\left( {x,y,z} \right), Q\left( {x,y,z} \right), R\left( {x,y,z} \right)} \Big)\]

be a vector field whose components have continuous partial derivatives.

The Divergence Theorem states:

\[\iint\limits_S {\mathbf{F} \cdot d\mathbf{S}} = \iiint\limits_G {\left( {\nabla \cdot \mathbf{F}} \right)dV} ,\]


\[\nabla \cdot \mathbf{F} = \frac{{\partial P}}{{\partial x}} + \frac{{\partial Q}}{{\partial y}} + \frac{{\partial R}}{{\partial z}}\]

is the divergence of the vector field \(\mathbf{F}\) (it's also denoted \(\text{div}\,\mathbf{F}\)) and the surface integral is taken over a closed surface.

The Divergence Theorem relates surface integrals of vector fields to volume integrals.

The Divergence Theorem can be also written in coordinate form as

\[\iint\limits_S {Pdydz + Qdxdz + Rdxdy} = \iiint\limits_G {\left( {\frac{{\partial P}}{{\partial x}} + \frac{{\partial Q}}{{\partial y}} + \frac{{\partial R}}{{\partial z}}} \right)dxdydz}\]

In a particular case, by setting \(P = x,\) \(Q = y,\) \(R = z,\) we obtain a formula for the volume of solid \(G:\)

\[V = \frac{1}{3}\left| {\iint\limits_S {xdydz + ydxdz + zdxdy} } \right|.\]

Solved Problems

Example 1.

Evaluate the surface integral \[\iint\limits_S {{x^3}dydz + {y^3}dxdz + {z^3}dxdy} ,\] where \(S\) is the surface of the sphere \[{x^2} + {y^2} + {z^2} = {a^2}\] that has upward orientation.


Using the Divergence Theorem, we can write:

\[I = \iint\limits_S {{x^3}dydz + {y^3}dxdz + {z^3}dxdy} = \iiint\limits_G {\left( {3{x^2} + 3{y^2} + 3{z^2}} \right)dxdydz} = 3\iiint\limits_G {\left( {{x^2} + {y^2} + {z^2}} \right)dxdydz}.\]

By changing to spherical coordinates, we have

\[I = 3\iiint\limits_G {\left( {{x^2} + {y^2} + {z^2}} \right)dxdydz} = 3\iiint\limits_G {{r^2} \cdot {r^2}\sin \theta drd\psi d\theta } = 3\int\limits_0^{2\pi } {d\psi } \int\limits_0^\pi {\sin \theta d\theta } \int\limits_0^a {{r^4}dr} = 3 \cdot 2\pi \cdot \left[ {\left. {\left( { - \cos \theta } \right)} \right|_0^\pi } \right] \cdot \left[ {\left. {\left( {\frac{{{r^5}}}{5}} \right)} \right|_0^a} \right] = \frac{{12\pi {a^5}}}{5}.\]

Example 2.

Use the Divergence Theorem to evaluate the surface integral \(\iint\limits_S {\mathbf{F} \cdot d\mathbf{S}} \) of the vector field \[\mathbf{F}\left( {x,y,z} \right) = \left( {x,y,z} \right),\] where \(S\) is the surface of the solid bounded by the cylinder \({x^2} + {y^2} = {a^2}\) and the planes \(z = -1,\) \(z = 1\) (Figure \(1\)).


Solid bounded by the cylinder x^2+y^2=a^2 and the planes  z=−1, z=1
Figure 1.

We apply the Divergence Theorem:

\[\iint\limits_S {\mathbf{F} \cdot d\mathbf{S}} = \iiint\limits_G {\left( {\nabla \cdot \mathbf{F}} \right)dV} = \iiint\limits_G {\left[ {\frac{\partial }{{\partial x}}\left( x \right) + \frac{\partial }{{\partial y}}\left( y \right) + \frac{\partial }{{\partial z}}\left( z \right)} \right]dxdydz} = \iiint\limits_G {\left( {1 + 1 + 1} \right)dxdydz} = 3\iiint\limits_G {dxdydz} .\]

By switching to cylindrical coordinates, we have

\[I = 3\iiint\limits_G {dxdydz} = 3\int\limits_{ - 1}^1 {dz} \int\limits_0^{2\pi } {d\varphi } \int\limits_0^a {rdr} = 3 \cdot 2 \cdot 2\pi \cdot \left[ {\left. {\left( {\frac{{{r^2}}}{2}} \right)} \right|_0^a} \right] = 6\pi {a^2}.\]

See more problems on Page 2.

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