Stokes' theorem

Let $\overrightarrow {i}$, $\overrightarrow {j}$, and $\overrightarrow {k}$ be the usual unit vectors in $\mathbb {R}^{3}$. Let $\overrightarrow {F}$ denote the vector field

Using Stokes' formula, compute the line integral

where $C$ is the circle $x^{2}+y^{2}+z^{2}=a^{2}$, $x+y+z=0$, traversed counterclockwise when viewed from the positive direction of the $Ox$ axis.

Check the result by direct computation.

Let $C$ be a closed contour lying in the plane $x\cos \alpha +y\cos \beta +z\cos \gamma -p=0$ (where $\cos \alpha$, $\cos \beta$, $\cos \gamma$ are the direction cosines of the plane's normal), enclosing an area $S$; find

where the contour $C$ is traversed in the positive direction.

Using Stokes' formula, compute the integral: $\displaystyle \oint_{C} (y+z)\, dx+(z+x)\, dy+(x+y)\, dz$, where $C$ is the ellipse $x=a\sin^{2}t$, $y=2c\sin t\cos t$, $z=a\cos^{2}t$ $(0\leqslant t\leqslant 2\pi )$, traversed in the direction of increasing parameter $t$.

Using Stokes' formula, compute the integral: $\displaystyle \oint_{C} (y-z)\, dx+(z-x)\, dy+(x-y)\, dz$, where $C$ is the ellipse $x^{2}+y^{2}=a^{2}$, $\dfrac {x}{a}+\dfrac {z}{h}=1$ $(a>0,\ h>0)$, traversed counterclockwise when viewed from the positive direction of the $Ox$ axis.

Using Stokes' formula, compute the integral: $\displaystyle \oint_{C} (y^{2}+z^{2})\, dx+(x^{2}+z^{2})\, dy+(x^{2}+y^{2})\, dz$, where $C$ is the curve $x^{2}+y^{2}+z^{2}=2Rx$, $x^{2}+y^{2}=2rx$ $(0<r<R,\ z>0)$, traversed so that the smallest region it encloses on the outer surface of the sphere $x^{2}+y^{2}+z^{2}=2Rx$ stays on the left.

Using Stokes' formula, compute the integral: $\displaystyle \oint_{C} (y^{2}-z^{2})\, dx+(z^{2}-x^{2})\, dy+(x^{2}-y^{2})\, dz$, where $C$ is the trace cut on the surface of the cube $0<x<a$, $0<y<a$, $0<z<a$ by the plane $x+y+z=\dfrac {3}{2}a$, traversed counterclockwise when viewed from the positive direction of the $Ox$ axis.

Using Stokes' formula, compute the integral: $\displaystyle \oint_{C} y^{2}z^{2}\, dx+x^{2}z^{2}\, dy+x^{2}y^{2}\, dz$, where $C$ is the closed curve $x=a\cos t$, $y=a\cos 2t$, $z=a\cos 3t$, traversed in the direction of increasing parameter $t$.

Given the function

where $S$ is the area bounded by the contour $C$, $\overrightarrow {n}$ is the normal to the surface $S$, and $\overrightarrow {r}$ is the radius vector joining the point $M(x,y,z)$ in space to the moving point $A(\xi ,\eta ,\zeta )$ on the surface $S$, prove that this function is the potential of the magnetic field $\overrightarrow {H}$ produced by the current $i$ flowing through the contour $C$.

Setup of problem 4340+: by the Biot--Savart law, the current $i$ through the line element $ds$ produces, at the point $M(x,y,z)$ in space, a magnetic field whose intensity is $d\overrightarrow {H}=ki\, \dfrac {\overrightarrow {r}\times d\overrightarrow {s}}{r^{3}}$, where $\overrightarrow {r}$ is the vector joining the element $d\overrightarrow {s}$ to the point $M$, and $k$ is a proportionality constant; for a closed conductor $C$, the intensity of the magnetic field $\overrightarrow {H}$ at the point $M$ is $\overrightarrow {H}=ki\displaystyle \oint_{C} \dfrac {\overrightarrow {r}\times d\overrightarrow {s}}{r^{3}}$ (see problem 4340+).

By the Biot-Savart law, each element $d\overrightarrow {s}$ of a current of strength $I$ creates at the point $M(x,y,z)$ a magnetic field whose intensity can be expressed by the vector product $-\dfrac {kI}{r^{2}}d\overrightarrow {s}\times \overrightarrow {r}$, where $\overrightarrow {r}$ is the vector going from $d\overrightarrow {s}$ to $M$. Prove that the field intensity $\overrightarrow {H}$ created by a closed current flowing along the contour $S$ is expressed by the vector integral $\overrightarrow {H}=-kI\int_{S} \dfrac {d\overrightarrow {s}\times \overrightarrow {r}}{r^{2}}$; here $k$ is the proportionality coefficient.

Using the previous result (problem 3813: $\overrightarrow {H}=-kI\int_{S} \frac{d\overrightarrow {s}\times \overrightarrow {r}}{r^{2}}$, the intensity of the magnetic field of a closed current by the Biot-Savart law) and Stokes's formula

(the circulation of a vector around a closed contour $L$ equals the flux of the curl of the vector through the surface $\sigma$ bounded by the contour), prove the equality $\overrightarrow {H}=-kI\operatorname {grad}W$, where $W$ is the solid angle at which the area $\sigma$ with contour $S$ is seen from the point $M$, equal to the integral $\iint_{\sigma } \dfrac {\cos (\overrightarrow {r},\overrightarrow {n})}{r^{2}} \, d\sigma$.

Consider the integral $\displaystyle \int_{s_{1}} d\overrightarrow {s}_{1}\int_{s_{2}} \frac{d\overrightarrow {s}_{2}\times \overrightarrow {r}}{r^{3}}$, taken over two closed contours $s_{1}$ and $s_{2}$, where $\overrightarrow {r}$ is the vector joining variable points on the contours. Regarding $\dfrac {d\overrightarrow {s}_{1}\times d\overrightarrow {s}_{2}\cdot \overrightarrow {e}_{r}}{r^{2}}$ as a solid angle, prove that this double integral equals $0$ if one contour is not linked with the other, and equals $\pm 4\pi$ if it is linked (once).

Using Stokes' formula, calculate the integrals. $\int_{L}(x+z) d x+(x-y) d y+x d z, L$ — the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} = 1, z = c$, oriented negatively with respect to the vector $(0 ; 0 ; 1)$.

Using Stokes' formula, calculate the integrals. $\int_{L} y^{2} d x+z^{2} d y+x^{2} d z, L$ — the boundary of the triangle with vertices at the points $(a ; 0 ; 0),(0 ; a ; 0),(0 ; 0 ; a)$, oriented positively with respect to the vector $(0 ; 1 ; 0)$.

Using Stokes' formula, calculate the integrals.

Using Stokes' formula, calculate the integrals. $\int_{L}\left(y^{2}-z^{2}\right) d x+\left(z^{2}-x^{2}\right) d y+\left(x^{2}-y^{2}\right) d z, L$ — the curve of intersection of the surface of the cube $\left|x\right| \leqslant a,\left|y\right| \leqslant a,\left|z\right| \leqslant a$ with the plane $x + y+z = 3 a / 2$, oriented positively with respect to the vector $(1 ; 0 ; 0)$.

Using Stokes' formula, calculate the integrals. $\int_{L}(y-z) d x+(z-x) d y+(x-y) d z$, where:

Using Stokes' formula, calculate the integrals. $\int_{L} y d x-z d y+x d z, L$ — the curve $x^{2}+y^{2}+2 z^{2} = 2 a^{2}, y-x = 0$, oriented positively with respect to the vector $(1 ; 0 ; 0)$.

Using Stokes' formula, calculate the integrals. $\int_{L}\left(y^{2}+z^{2}\right) d x+\left(z^{2}+x^{2}\right) d y+\left(x^{2}+y^{2}\right) d z, L$ — the curve $x^{2} + y^{2}+z^{2} = 2 a x, x^{2}+y^{2} = 2 b x, z > 0,0 < b < a$, oriented positively with respect to the vector $(0 ; 0 ; 1)$.

Using Stokes' formula, calculate the integrals. $\int_{L} z^{3} d x+x^{3} d y+y^{3} d z, L$ — the curve $2 x^{2}-y^{2}+z^{2} = a^{2}, x+y = 0$, oriented positively with respect to the vector $(1 ; 0 ; 0)$.

Calculate the integrals, if the curve $L$ is oriented in the direction of increasing parameter $t$. $\int_{L} x d x+(x+y) d y+(x+y+z) d z, L$ — the curve $x = a \sin t$, $y = a \cos t, z = a(\sin t+\cos t), t \in [0 ; 2 \pi )$.

Calculate the integrals, if the curve $L$ is oriented in the direction of increasing parameter $t$. $\int_{L} y^{2} z^{2} d x+x^{2} z^{2} d y+x^{2} y^{2} d z, L$ — the curve $x = a \cos t, y = a \cos 2 t, z = a \cos 3 t, t \in [0 ; 2 \pi )$.

Calculate the integrals, if the curve $L$ is oriented in the direction of increasing parameter $t$. $\int_{L}(y+z) d x+(z+x) d y+(x+y) d z, L$ — the curve $x = a \sin^{2} t$, $y = a \sin 2 t, z = a \cos^{2} t, t \in [0 ; \pi )$.

Calculate the integrals, if the curve $L$ is oriented in the direction of increasing parameter $t$. $\int_{L}\left(x^{2}-y z\right) d x+\left(y^{2}-z x\right) d y+\left(z^{2}-x y\right) d z, L$ — the curve $x = a \cos t, y = a \sin t, z = h t /(2 \pi ), t \in [0 ; 2 \pi ]$.

Stokes' formula (19): if $\mathbf{a}$ is a continuously differentiable vector field in a region $\Omega$ and $S$ is a piecewise smooth surface in $\Omega$ oriented by the unit normal vector $\mathbf{n}$, with boundary $\partial S$ oriented consistently with the orientation of the surface, then $\oint_{\partial S} \mathbf{a} d \mathbf{r} = \iint_{S} \mathbf{n} \cdot \operatorname {rot} \mathbf{a} d S = \iint_{S} (\mathbf{n}, [\nabla , \mathbf{a}]) d S$.

Using Stokes' formula (19), find the circulation of the field $\mathbf{a}$ along the contour $\Gamma$, oriented clockwise as viewed from the origin, if:

Find the circulation of the field $\mathbf{a} = (-y \mathbf{i}+x \mathbf{j}) /\left(x^{2}+y^{2}\right)+z \mathbf{k}$ along the circle

oriented counterclockwise as viewed from points of the axis $O z$ where $z > 1$.

Under the conditions of problem 64, find the circulation of the field $\mathbf{v}$ :

Under the conditions of problem 64, take the axis of rotation as the axis $O z$, directing it along the angular velocity vector. Let $G$ be a bounded simply connected region in the plane $O x y$ with boundary $\gamma$, a piecewise smooth simple closed contour, and let $C$ be the cylinder with base $\bar{G}$ and generators parallel to the axis of rotation. Let $\Gamma$ be a closed piecewise smooth curve on the surface of the cylinder $C$ that projects bijectively onto $\gamma$. Prove that the circulation of the field $\mathbf{v}$ along $\Gamma$ equals $2 \omega \cdot \mu G$, where $\mu G$ is the area of $G$.

Find, with accuracy up to $o\left(\varepsilon^{2}\right)$, the absolute value of the circulation of the field a along the circle $(x-1)^{2}+(y-1)^{2}+(z-1)^{2} = \varepsilon^{2}, \quad x + y+z = 3$, if:

Prove the formula below. Here $\mathbf{a}$ and $u$ are continuously differentiable fields in a region $\Omega$, and $S$ is a piecewise smooth surface in $\Omega$ oriented by the unit normal vector $\mathbf{n}$, with boundary $\partial S$ oriented consistently with the orientation of the surface, so that the Stokes formula (19)

holds.

Fix $\delta , 0 < \delta < 1$. Let $D$ be the set of all $(\theta , t) \in \mathbb {R}^{2}$ such that $0 \leq \theta \leq \pi ,-\delta \leq t \leq \delta$. Let $\Phi$ be the 2 -surface in $\mathbb {R}^{3}$, with parameter domain $D$, given by

where $(x, y, z) = \Phi (\theta , t)$. Note that $\Phi (\pi , t) = \Phi (0,-t)$, and that $\Phi$ is one-to-one on the rest of $D$.

The range $M = \Phi (D)$ of $\Phi$ is known as a Möbius band. It is the simplest example of a nonorientable surface.

Prove the various assertions made in the following description: Put $\mathbf{p}_{1} = (0,-\delta ), \mathbf{p}_{2} = (\pi ,-\delta ), \mathbf{p}_{3} = (\pi , \delta ), \mathbf{p}_{4} = (0, \delta ), \mathbf{p}_{5} = \mathbf{p}_{1}$. Put $\gamma_{i} = \left[\mathbf{p}_{i}, \mathbf{p}_{i+1}\right]$, $i = 1, \ldots , 4$, and put $\Gamma_{i} = \Phi \circ \gamma_{i}$. Then

Put $\mathbf{a} = (1,0,-\delta ), \mathbf{b} = (1,0, \delta )$. Then

and $\partial \Phi$ can be described as follows. $\Gamma_{1}$ spirals up from a to $\mathbf{b}$; its projection into the $(x, y)$-plane has winding number +1 around the origin. (See Exercise 23, Chap. 8). $\Gamma_{2} = [b, a]$. $\Gamma_{3}$ spirals up from a to $\mathbf{b}$; its projection into the $(x, y)$ plane has winding number -1 around the origin. $\Gamma_{4} = [b, a]$. Thus $\partial \Phi = \Gamma_{1}+\Gamma_{3}+2 \Gamma_{2}$. If we go from $\mathbf{a}$ to $b$ along $\Gamma_{1}$ and continue along the "edge" of $M$ until we return to a, the curve traced out is

which may also be represented on the parameter interval $[0,2 \pi ]$ by the equations

It should be emphasized that $\Gamma \neq \partial \Phi$ : Let $\eta$ be the 1 -form discussed in Exercises 21 and 22. Since $d \eta = 0$, Stokes' theorem shows that

But although $\Gamma$ is the "geometric" boundary of $M$, we have

In order to avoid this possible source of confusion, Stokes' formula (Theorem 10.50) is frequently stated only for orientable surfaces $\Phi$.