# 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

${\mathbf{F}\left( {x,y,z} \right) } = { \Big( {P\left( {x,y,z} \right),}}\kern0pt{{Q\left( {x,y,z} \right), }}\kern0pt{{R\left( {x,y,z} \right)} \Big)}$

be a vector field whose components have continuous partial derivatives.

The Divergence Theorem states:

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

where

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

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

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

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

${V \text{ = }}\kern0pt{ \frac{1}{3}\Big| {\iint\limits_S {xdydz }}+{{ ydxdz }}+{{ zdxdy} } \Big|}$

## Solved Problems

Click or tap a problem to see the solution.

### 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.

### 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 $$1$$).

### Example 3

Use the Divergence Theorem to evaluate the surface integral $$\iint\limits_S {\mathbf{F} \cdot d\mathbf{S}}$$ of the vector field $$\mathbf{F}\left( {x,y,z} \right) =$$ $$\left( {{x^3},{y^3},{z^3}} \right),$$ where $$S$$ is the surface of a solid bounded by the cone $${x^2} + {y^2} – {z^2}$$ $$= 0$$ and the plane $$z = 1.$$ (Figure $$2$$).

### Example 4

Using the Divergence Theorem calculate the surface integral $$\iint\limits_S {\mathbf{F} \cdot d\mathbf{S}}$$ of the vector field $$\mathbf{F}\left( {x,y,z} \right) =$$ $$\left( {2xy, 8xz, 4yz} \right),$$ where $$S$$ is the surface of tetrahedron with vertices $$O\left( {0,0,0} \right),$$ $$A\left( {1,0,0} \right),$$ $$B\left( {0,1,0} \right),$$ $$C\left( {0,0,1} \right)$$ (Figure $$3$$).

### Example 5

Calculate the surface integral $$\iint\limits_S {\mathbf{F} \cdot d\mathbf{S}}$$ of the vector field $$\mathbf{F}\left( {x,y,z} \right) =$$ $$\left( {2{x^2}y,x{z^2},4yz} \right),$$ where $$S$$ is the surface of the rectangular box bounded by the planes $$x = 0,$$ $$x = 1,$$ $$y = 0,$$ $$y = 2,$$ $$z = 0,$$ $$z = 3$$ (Figure $$4$$).

### Example 6

Find the surface integral $$\iint\limits_S {2xdydz }$$ $${+ \left( {3y + x} \right)dxdz }$$ $${+ \left( {2y + 4z} \right)dxdy} ,$$ where $$S$$ is the outer surface of the pyramid $${\large\frac{x}{a}\normalsize} + {\large\frac{y}{b}\normalsize} + {\large\frac{z}{c}\normalsize} \le 1,$$ $$x \ge 0,$$ $$y \ge 0,$$ $$z \ge 0$$ (see Figure $$5$$).

### 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.

Solution.

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

${I }={ 3\iiint\limits_G {\left( {{x^2} + {y^2} + {z^2}} \right)dxdydz} } = {3\iiint\limits_G {{r^2} \cdot {r^2}\sin \theta drd\psi d\theta } } = {3\int\limits_0^{2\pi } {d\psi } \int\limits_0^\pi {\sin \theta d\theta } \int\limits_0^a {{r^4}dr} } = {3 \cdot 2\pi \cdot \left[ {\left. {\left( { – \cos \theta } \right)} \right|_0^\pi } \right] \cdot}\kern0pt{ \left[ {\left. {\left( {\frac{{{r^5}}}{5}} \right)} \right|_0^a} \right] } = {\frac{{12\pi {a^5}}}{5}.}$

### 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 $$1$$).

Solution.

We apply the Divergence Theorem:

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

By switching to cylindrical coordinates, we have

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

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Problems 1-2
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Problems 3-6