Stokes' Formula

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.

Compute the integral

taken from the point $A(a,0,0)$ to the point $B(a,0,h)$ along the helix

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+).