Calculus

Surface Integrals

The Divergence Theorem

Page 1
Problems 1-2
Page 2
Problems 3-6

Let \(G\) be a three-dimensional solid bounded by a piecewise smooth closed surface \(S\) that has orientation pointing out of \(G.\) 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 on problem description to see 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

\[
{I = 3\iiint\limits_G {dxdydz} }
= {3\int\limits_{ – 1}^1 {dz} \int\limits_0^{2\pi } {d\varphi } \int\limits_0^a {rdr} }
= {3 \cdot 2 \cdot 2\pi \cdot \left[ {\left. {\left( {\frac{{{r^2}}}{2}} \right)} \right|_0^a} \right] }
= {6\pi {a^2}.}
\]
Solid bounded by the cylinder x^2+y^2=a^2 and the planes  z=−1, z=1

Figure 1.

Page 1
Problems 1-2
Page 2
Problems 3-6