Calculus

Surface Integrals

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Surface Integrals of Vector Fields

We consider a vector field \(\mathbf{F}\left( {x,y,z} \right)\) and a surface \(S,\) which is defined by the position vector

\[\mathbf{r}\left( {u,v} \right) = x\left( {u,v} \right) \cdot \mathbf{i} + y\left( {u,v} \right) \cdot \mathbf{j} + z\left( {u,v} \right) \cdot \mathbf{k}.\]

Suppose that the functions \(x\left( {u,v} \right),\) \(y\left( {u,v} \right),\) \(z\left( {u,v} \right)\) are continuously differentiable in some domain \(D\left( {u,v} \right)\) and the rank of the matrix

\[\left[ {\begin{array}{*{20}{c}} {\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial y}}{{\partial u}}}&{\frac{{\partial z}}{{\partial u}}}\\ {\frac{{\partial x}}{{\partial v}}}&{\frac{{\partial y}}{{\partial v}}}&{\frac{{\partial z}}{{\partial v}}} \end{array}} \right]\]

is equal to \(2.\)

We denote by \(\mathbf{n}\left( {x,y,z} \right)\) a unit normal vector to the surface \(S\) at the point \(\left( {x,y,z} \right).\) If the surface \(S\) is smooth and the vector function \(\mathbf{n}\left( {x,y,z} \right)\) is continuous, there are only two possible choices for the unit normal vector:

\[\mathbf{n}\left( {x,y,z} \right)\;\;\text{or}\;\; - \mathbf{n}\left( {x,y,z} \right).\]

If the choice of the vector is done, the surface \(S\) is called oriented.

If \(S\) is a closed surface, by convention, we choose the normal vector to point outward from the surface.

The surface integral of the vector field \(\mathbf{F}\) over the oriented surface \(S\) (or the flux of the vector field \(\mathbf{F}\) across the surface \(S\)) can be written in one of the following forms:

Here \(d\mathbf{S} = \mathbf{n}dS\) is called the vector element of the surface. Dot means the scalar product of the appropriate vectors. The partial derivatives in the formulas are calculated in the following way:

\[\frac{{\partial \mathbf{r}}}{{\partial u}} = \frac{{\partial x}}{{\partial u}}\left( {u,v} \right) \cdot \mathbf{i} + \frac{{\partial y}}{{\partial u}}\left( {u,v} \right) \cdot \mathbf{j} + \frac{{\partial z}}{{\partial u}}\left( {u,v} \right) \cdot \mathbf{k},\]
\[\frac{{\partial \mathbf{r}}}{{\partial v}} = \frac{{\partial x}}{{\partial v}}\left( {u,v} \right) \cdot \mathbf{i} + \frac{{\partial y}}{{\partial v}}\left( {u,v} \right) \cdot \mathbf{j} + \frac{{\partial z}}{{\partial v}}\left( {u,v} \right) \cdot \mathbf{k}.\]

If the surface \(S\) is given explicitly by the equation \(z = z\left( {x,y} \right),\) where \(z\left( {x,y} \right)\) is a differentiable function in the domain \(D\left( {x,y} \right),\) then the surface integral of the vector field \(\mathbf{F}\) over the surface \(S\) is defined in one of the following forms:

We can also write the surface integral of vector fields in the coordinate form.

Let \(P\left( {x,y,z} \right),\) \(Q\left( {x,y,z} \right),\) \(R\left( {x,y,z} \right)\) be the components of the vector field \(\mathbf{F}.\) Suppose that \(\cos \alpha,\) \(\cos \beta,\) \(\cos \gamma\) are the angles between the outer unit normal vector \(\mathbf{n}\) and the \(x\)-axis, \(y\)-axis, and \(z\)-axis, respectively. Then the scalar product \(\mathbf{F} \cdot \mathbf{n}\) is

\[\mathbf{F} \cdot \mathbf{n} = \mathbf{F}\left( {P,Q,R} \right) \mathbf{n}\left( {\cos \alpha ,\cos \beta ,\cos \gamma } \right) = P\cos \alpha + Q\cos \beta + R\cos \gamma .\]

Consequently, the surface integral can be written as

\[\iint\limits_S {\left( {\mathbf{F} \cdot \mathbf{n}} \right)dS} = \iint\limits_S {\left( {P\cos \alpha + Q\cos \beta + R\cos \gamma } \right)dS} .\]

As \(\cos \alpha \cdot dS = dydz\) (Figure \(1\)), and, similarly, \(\cos \beta \cdot dS = dzdx,\) \(\cos \gamma \cdot dS = dxdy,\) we obtain the following formula for calculating the surface integral:

\[\iint\limits_S {\left( {\mathbf{F} \cdot \mathbf{n}} \right)dS} = \iint\limits_S {\left( {P\cos \alpha + Q\cos \beta + R\cos \gamma } \right)dS} = \iint\limits_S {Pdydz + Qdzdx + Rdxdy}.\]
Relationship cos(alpha)*dS=dydz
Figure 1.

If the surface \(S\) is given in parametric form by the vector \(\mathbf{r}\big( {x\left( {u,v} \right),y\left( {u,v} \right),}\) \({z\left( {u,v} \right)} \big),\) the latter formula can be written as

\[ \iint\limits_S {\left( {\mathbf{F} \cdot \mathbf{n}} \right)dS} = \iint\limits_S {Pdydz + Qdzdx + Rdxdy} = \iint\limits_{D\left( {u,v} \right)} {\left| {\begin{array}{*{20}{c}} P&Q&R\\ {\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial y}}{{\partial u}}}&{\frac{{\partial z}}{{\partial u}}}\\ {\frac{{\partial x}}{{\partial v}}}&{\frac{{\partial y}}{{\partial v}}}&{\frac{{\partial z}}{{\partial v}}} \end{array}} \right|dudv} ,\]

where the coordinates \(\left( {u,v} \right)\) range over some domain \(D\left( {u,v} \right).\)

Solved Problems

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Example 1

Evaluate the flux of the vector field \[\mathbf{F}\left( {x,y,z} \right) = \left( {x, - 1,z} \right)\] across the surface \(S\) that has downward orientation and is given by the equation

\[z = x\cos y, 0 \le x \le 1, \frac{\pi }{4} \le y \le \frac{\pi }{3}.\]

Example 2

Find the flux of the vector field \[\mathbf{F}\left( {x,y,z} \right) = \left( {y,x,z} \right)\] through the surface \(S,\) parameterized by the vector

\[\mathbf{r}\left( {u,v} \right) = \left( {\cos v,\sin v,u} \right), 0 \le u \le 2, \frac{\pi }{2} \le v \le \pi .\]

Example 1.

Evaluate the flux of the vector field \[\mathbf{F}\left( {x,y,z} \right) = \left( {x, - 1,z} \right)\] across the surface \(S\) that has downward orientation and is given by the equation

\[z = x\cos y, 0 \le x \le 1, \frac{\pi }{4} \le y \le \frac{\pi }{3}.\]

Solution.

We apply the formula

\[\iint\limits_S {\mathbf{F} \cdot d\mathbf{S}} = \iint\limits_{D\left( {x,y} \right)} {\mathbf{F} \left( {\frac{{\partial z}}{{\partial x}}\mathbf{i} + \frac{{\partial z}}{{\partial y}}\mathbf{j} - \mathbf{k}} \right)dxdy}.\]

Since

\[\frac{{\partial z}}{{\partial x}} = \frac{\partial }{{\partial x}}\left( {x\cos y} \right) = \cos y,\;\;\; \frac{{\partial z}}{{\partial y}} = \frac{\partial }{{\partial y}}\left( {x\cos y} \right) = - x\sin y,\]

the flux of the vector field can be written as

\[\iint\limits_S {\mathbf{F} \cdot d\mathbf{S}} = \iint\limits_{D\left( {x,y} \right)} {\left[ {x \cdot \cos y + \left( { - 1} \right) \cdot \left( { - x\sin y} \right) + z \cdot \left( { - 1} \right)} \right]dxdy} = \iint\limits_{D\left( {x,y} \right)} {\left( {\cancel{x\cos y} + x\sin y - \cancel{x\cos y}} \right)dxdy} = \iint\limits_{D\left( {x,y} \right)} {x\sin ydxdy} .\]

After some algebra we find the answer:

\[ \iint\limits_S {\mathbf{F} \cdot d\mathbf{S}} = \int\limits_0^1 {xdx} \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{3}} {\sin ydy} = \left[ {\left. {\left( {\frac{{{x^2}}}{2}} \right)} \right|_0^1} \right] \cdot \left[ {\left. {\left( { - \cos y} \right)} \right|_{\frac{\pi }{4}}^{\frac{\pi }{3}}} \right] = \frac{1}{2}\left( { - \cos \frac{\pi }{3} + \cos \frac{\pi }{4}} \right) = \frac{1}{2}\left( { - \frac{1}{2} + \frac{{\sqrt 2 }}{2}} \right) = \frac{{\sqrt 2 - 1}}{4}.\]

Example 2.

Find the flux of the vector field \[\mathbf{F}\left( {x,y,z} \right) = \left( {y,x,z} \right)\] through the surface \(S,\) parameterized by the vector

\[\mathbf{r}\left( {u,v} \right) = \left( {\cos v,\sin v,u} \right), 0 \le u \le 2, \frac{\pi }{2} \le v \le \pi .\]

Solution.

First we calculate the partial derivatives:

\[\frac{{\partial \mathbf{r}}}{{\partial u}} = \left( {0,0,1} \right),\;\; \frac{{\partial \mathbf{r}}}{{\partial v}} = \left( { - \sin v,\cos v,0} \right).\]

It follows that

\[ \frac{{\partial \mathbf{r}}}{{\partial u}} \times \frac{{\partial \mathbf{r}}}{{\partial v}} = \left| {\begin{array}{*{20}{c}} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ 0&0&1\\ { - \sin v}&{\cos v}&0 \end{array}} \right| = - \cos v \cdot \mathbf{i} - \sin v \cdot \mathbf{j}.\]

Hence, the vector area element is

\[d\mathbf{S} = \left[ {\frac{{\partial \mathbf{r}}}{{\partial u}} \times \frac{{\partial \mathbf{r}}}{{\partial v}}} \right]dudv = \left( { - \cos v, - \sin v,0} \right)dudv.\]

As \(x = \cos v,\) \(y = \sin v\) and \(z = v,\) the vector field \(\mathbf{F}\) can be represented in the following form:

\[\mathbf{F}\left( {r,u,v} \right) = \left( {\sin v,\cos v,u} \right).\]

Then the original surface integral across \(S\) is

\[\iint\limits_S {\mathbf{F} \cdot d\mathbf{S}} = \iint\limits_{D\left( {u,v} \right)} {\left[ {\sin v \cdot \left( { - \cos v} \right) + \cos v \cdot \left( { - \sin v} \right) + 0} \right]dudv} = \iint\limits_{D\left( {u,v} \right)} {\left( { - 2\sin v\cos v} \right)dudv} = - \int\limits_0^2 {du} \int\limits_{\frac{\pi }{2}}^\pi {\sin 2vdv} = - 2 \cdot \left[ {\left. {\left( { - \frac{{\cos 2v}}{2}} \right)} \right|_{\frac{\pi }{2}}^\pi } \right] = \cos 2\pi - \cos \pi = 2.\]

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