# Math2111: Chapter 5: Integral theorems. Section 2: Stokes’ theorem

In this blog entry you can find lecture notes for Math2111, several variable calculus. See also the table of contents for this course. This blog entry printed to pdf is available here.

We now generalise Green’s theorem to vector fields in $\mathbb{R}^3.$ The following result is by Sir George Gabriel Stokes.

Stokes’ theorem
Let $\mathcal{C}$ be a simple closed positively oriented curve in the plane $\mathbb{R}^2$ parameterised by a continuously differentiable function $\boldsymbol{c}:[a,b]\to\mathbb{R}^2$ and let ${}D$ be the region enclosed by $\mathcal{C}.$ Let $\boldsymbol{\Phi}:D\to\mathbb{R}^3,$ $\boldsymbol{\Phi}=\boldsymbol{\Phi}(u,v),$ be a twice continuously differentiable and one-to-one parameterisation of a surface $S=\mbox{Image}(\boldsymbol{\Phi})$. The boundary $\partial S$ of ${}S$ is parameterised by the positively oriented curve $\boldsymbol{\Phi}(\boldsymbol{c}):[a,b]\to\mathbb{R}^3.$

Then

$\iint_D (\nabla \times \boldsymbol{F})(\boldsymbol{\Phi}(u,v)) \cdot \left(\frac{\partial \boldsymbol{\Phi}}{\partial u} \times \frac{\partial \boldsymbol{\Phi}}{\partial v}\right)(u,v) \, \mathrm{d} u \, \mathrm{d} v = \int_{\partial S} \boldsymbol{F}\cdot \mathrm{d} \boldsymbol{s}.$

Notice that the orientation of the surface ${}S$ and boundary curve $\partial S$ is specified in the theorem above. This orientation of the surface and curve can also be described in the following way: If you walk in the positive direction along the boundary curve $\partial S$ with your head pointing in the direction of the normal vector $\boldsymbol{n}$, then the surface ${}S$ will always be on your left.

Exercise
Use Stokes’ theorem to evaluate the integral $\iint_S \mbox{curl } \boldsymbol{F} \cdot \mathrm{d} \boldsymbol{\mathcal{S}},$ where $\boldsymbol{F}(x,y,z)= yz\widehat{\boldsymbol{i}} + xz \widehat{\boldsymbol{j}} + xy \widehat{\boldsymbol{k}}$ and ${}S$ is the part of the sphere $x^2+y^2+z^2=4$ that lies inside the cylinder $x^2+y^2=1$ and above the ${}xy$-plane. $\Box$

Exercise
Prove that if $\mbox{curl } \boldsymbol{F}=\boldsymbol{0}$ everywhere on $\mathbb{R}^3$, then $\boldsymbol{F}$ is conservative.
$\Box$

Exercise
We have previously seen that Green’s theorem does not apply to the case where the curve is the circle centered at $(0,0)$ with radius ${}1$ and where $\boldsymbol{F}=\nabla (\arctan (y/x)) = (-y/(x^2+y^2), x/(x^2+y^2)),$ since $\boldsymbol{F}$ is not defined at $(0,0).$

Consider now the vector field

$\displaystyle \boldsymbol{G}= \left(\frac{-y}{x^2+y^2+z^2}, \frac{x}{x^2+y^2+z^2}, 0\right).$

Notice that for $z=0$ the first two components of $\boldsymbol{G}$ are the same as the first two components of $\boldsymbol{F}.$ Hence $\int_{\mathcal{C}} \boldsymbol{F}\cdot \mathrm{d} \boldsymbol{s} = \int_{\mathcal{C}} \boldsymbol{G}\cdot\mathrm{d} \boldsymbol{s},$ where $\mathcal{C}$ is the circle centered at $(0,0)$ with radius ${}1$ in the $xy$-plane. Let the surface be the upper hemisphere, i.e. $z = \sqrt{1-x^2-y^2}$ and $x^2+y^2\le 1.$ Verify Stokes’ theorem for this case.
$\Box$

$\rhd$ Exercise
Let the curve $\mathcal{C}$ be the boundary curve of the Möbius strip parameterised by

$\displaystyle \boldsymbol{c}(u) = \frac{1}{2} \sin \frac{u}{2} \, \widehat{\boldsymbol{i}} + \left(1+\frac{1}{2}\cos\frac{u}{2}\right)\sin u \, \widehat{\boldsymbol{j}} + \left(1+\frac{1}{2}\cos\frac{u}{2}\right)\cos u \, \widehat{\boldsymbol{k}},$

where $0 \le u\le 2\pi.$ Calculate the line integral

$\displaystyle \int_{\mathcal{C}} (y-z \mathrm{e}^{\cos xz} \sin xz) \,\mathrm{d} x + x \,\mathrm{d} y - x \mathrm{e}^{\cos xz} \sin xz \,\mathrm{d} z.$

Note that the integrand is conservative. Further recall that the Möbius strip is not orientable and hence the surface integral over the Möbius strip is not defined.$\Box$

$\rhd$ Proof of Stokes’ theorem
Let the vector field $\boldsymbol{F} =F_1 \widehat{\boldsymbol{i}} + F_2\widehat{\boldsymbol{j}}+F_3 \widehat{\boldsymbol{k}}.$ Let the parameterisation of the surface ${}S$ be given by

$\displaystyle \boldsymbol{\Phi}(u,v)= X(u,v) \widehat{\boldsymbol{i}} + Y(u,v) \widehat{\boldsymbol{j}} + Z(u,v) \widehat{\boldsymbol{k}}$

and let

$\displaystyle \boldsymbol{c}(t)= x(t) \widehat{\boldsymbol{i}} + y(t) \widehat{\boldsymbol{j}}.$

Then the boundary $\partial S$ is parameterised by

$\displaystyle \boldsymbol{\Phi}(\boldsymbol{c}(t)) = X(x(t),y(t)) \widehat{\boldsymbol{i}} + Y(x(t),y(t)) \widehat{\boldsymbol{j}} + Z(x(t),y(t)) \widehat{\boldsymbol{k}}.$

We can now write the line integral as

$\displaystyle \begin{array}{ll} & \int_{\partial S} \boldsymbol{F} \cdot \mathrm{d} \boldsymbol{s} =\int_a^b \boldsymbol{F}(\boldsymbol{\Phi}(\boldsymbol{c}(t))) \cdot \frac{\mathrm{d} \boldsymbol{\Phi}(\boldsymbol{c}(t))}{\mathrm{d} t} \,\mathrm{d} t \\&\\ &=\int_a^b \left(F_1(\boldsymbol{\Phi}(\boldsymbol{c}(t))) \frac{\partial X}{\partial t}(\boldsymbol{c}(t)) + F_2(\boldsymbol{\Phi}(\boldsymbol{c}(t))) \frac{\partial Y}{\partial t}(\boldsymbol{c}(t)) + F_3(\boldsymbol{\Phi}(\boldsymbol{c}(t))) \frac{\partial Z}{\partial t}(\boldsymbol{c}(t)) \right)\,\mathrm{d} t \end{array}$

Using the chain rule we have

$\displaystyle \begin{array}{rcl} \frac{\partial X}{\partial t}&=&X_u x_t + X_v y_t, \\&&\\\frac{\partial Y}{\partial t}&=& Y_u x_t+ Y_v y_t,\\&&\\\frac{\partial Z}{\partial t}&=& Z_u x_t + Z_v y_t, \end{array}$

To shorten the expressions we assume in the following that $F_i=F_i(\boldsymbol{\Phi}(\boldsymbol{c}(t))$ and that $X_u=X_u(\boldsymbol{c}(t))$ and similarly for $X_v,Y_u,Y_v,Z_u,Z_v.$ Thus

$\displaystyle \begin{array}{ll} \int_{\partial S} \boldsymbol{F} \cdot \mathrm{d} \boldsymbol{s} = & \int_a^b F_1 (X_u x_t + X_v y_t) \,\mathrm{d}t \\&\\& +\int_a^b F_2 (Y_u x_t +Y_v y_t)\,\mathrm{d}t \\ &\\& +\int_a^b F_3 (Z_u x_t+Z_v y_t) \,\mathrm{d} t \\ &\\& = \int_{\partial D} F_1 X_u \mathrm{d} u + F_1 X_v \,\mathrm{d}v \\&\\& +\int_{\partial D} F_2 Y_u \,\mathrm{d} u + F_2 Y_v \,\mathrm{d} v \\ &\\& +\int_{\partial D} F_3 Z_u \, \mathrm{d} u+ F_3 Z_v \,\mathrm{d} v, \end{array}$

where the line integral is over the boundary curve of ${}D$ in positive orientation.

Since the last 3 integrals are over a region in $\mathbb{R}^2$ and the boundary curve $\partial D$ of ${}D$ is positively oriented, we can use Green’s theorem to obtain

$\displaystyle \begin{array}{ll} \int_{\partial S} \boldsymbol{F} \cdot \mathrm{d} \boldsymbol{s} = & \iint_D \left(\frac{\partial (F_1 X_v)}{\partial u} - \frac{\partial (F_1 X_u)}{\partial v}\right) \,\mathrm{d} u\,\mathrm{d}v \\&\\ &+ \iint_D \left(\frac{\partial (F_2 Y_v)}{\partial u} - \frac{\partial (F_2 Y_u)}{\partial v}\right) \,\mathrm{d} u\,\mathrm{d} v \\&\\ &+ \iint_D \left(\frac{\partial (F_3 Z_v)}{\partial u} - \frac{\partial (F_3 Z_u)}{\partial v}\right) \,\mathrm{d} u\,\mathrm{d}v. \end{array}$

We now apply the chain rule to the partial derivatives appearing above to obtain

$\displaystyle \begin{array}{rcl} \frac{\partial (F_1 X_v)}{\partial u} - \frac{\partial (F_1 X_u)}{\partial v} &=& \frac{\partial F_1}{\partial x} X_uX_v + \frac{\partial F_1}{\partial y} Y_uX_v + \frac{\partial F_1}{\partial z} Z_u X_v + F_1 X_{u,v} \\&&\\&&- \frac{\partial F_1}{\partial x} X_vX_u -\frac{\partial F_1}{\partial y} Y_u X_v - \frac{\partial F_1}{\partial z} Z_u X_v - F_1 X_{v,u} \\ &&\\ &=& \frac{\partial F_1}{\partial y} (Y_uX_v-Y_vX_u) +\frac{\partial F_1}{\partial z} (Z_u X_v-Z_vX_u). \end{array}$

By simplifying also the other expressions in the same manner we obtain

$\displaystyle \begin{array}{ll} \int_{\partial S} \boldsymbol{F}\cdot \mathrm{d} \boldsymbol{s} =&\iint_D \left(\frac{\partial F_3}{\partial y}-\frac{\partial F_2}{\partial z} \right) \left(Y_uZ_v-Y_vZ_u \right)\,\mathrm{d} u\,\mathrm{d} v \\&\\ & + \iint_D \left(\frac{\partial F_1}{\partial z}-\frac{\partial F_3}{\partial x} \right) \left(Z_uX_v-Z_vX_u\right)\,\mathrm{d} u\,\mathrm{d} v \\&\\ &+ \iint_D \left(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y} \right) \left(X_uY_v-X_vY_u\right)\,\mathrm{d} u\,\mathrm{d} v \\&\\& = \iint_D (\nabla \times \boldsymbol{F})(\boldsymbol{\Phi}(u,v)) \cdot \left(\frac{\partial \boldsymbol{\Phi}}{\partial u} \times \frac{\partial \boldsymbol{\Phi}}{\partial v}\right)(u,v) \, \mathrm{d} u \,\mathrm{d} v. \end{array}$

$\Box$

### 3 responses to “Math2111: Chapter 5: Integral theorems. Section 2: Stokes’ theorem”

1. Peter

One of the sections listed above has the title example, but it seems more like a question? (J. Changed it to Exercise.)

2. Chen

I’m not quit understand the equation of stocks’ theorem above. Is that(Tu*Tv) a unit normal ? why it is dS not dA? thanks (J. Updated it now, thanks.)

3. cnikki

Plz help me how to work out this problem?
Q.
Let W = { (x, y, z) є R3 : 1 ≤ x2 + y2 + z2 ≤ 4 } and F : W → R3 be defined by

F(x, y, z) = (x, y, z) / [x2 + y2 + z2 ]3/2 for (x, y, z) є W .if ∂W denotes the boundary of W oriented by the outward normal n to W ,then ∫∫ F. n d S is equal to
a) 0 b) 4pi c) 8pi d) 12 pi