Chapter 6: Vector Analysis

${\mathbf{\int }}_{\mathbf{C}}\mathbf{ }\left(x^2-y\right)\mathbf{dx}\mathbf{+}\left(x+y^3\right)\mathbf{dy}$: where $\mathbf{C}$is the parallelogram with vertices $\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{3}\mathbf{,}\mathbf{1}\mathbf{)}$.

$\mathbf{\int }\left(y^2-x^2\right)\mathit{d}\mathit{x}\mathbf{+}\left(2xy+3\right)\mathit{d}\mathit{y}$along the x axis from (0,0) to$\left(\sqrt{5},0\right)$ and along a circular are from$\mathbf{(}\sqrt{\mathbf{5}}\mathbf{,}\mathbf{0}\mathbf{)}$ to (1,2).

The force on a charge moving with velocity $\mathbf{v}\mathbf{}\mathbf{=}\mathbf{dr}\mathbf{/}\mathbf{dt}$in a magnetic field B is$\mathbf{F}\mathbf{=}\mathbf{q}\left(v\times B\right)$we can write B as$\mathbf{B}\mathbf{=}\mathbf{\nabla }\mathbf{\times }\mathbf{A}$where A (called the vector potential) is a vector function of x,y,z,t . If the position vector$\mathbf{r}\mathbf{=}\mathbf{ix}\mathbf{+}\mathbf{jy}\mathbf{+}\mathbf{kz}$of the charge is a function of time, show that

$\frac{\mathbf{dA}}{\mathbf{dt}}\mathbf{=}\frac{\mathbf{\partial }\mathbf{A}}{\mathbf{\partial }\mathbf{t}}\mathbf{+}\mathbf{v}\mathbf{.}\mathbf{\nabla }\mathbf{A}$

Thus show that

$\mathbf{F}\mathbf{=}\mathbf{qv}\mathbf{\times }\mathbf{(}\mathbf{\nabla }\mathbf{\times }\mathbf{A}\mathbf{)}\mathbf{=}\mathbf{q}\left[\nabla (v.A)-\frac{\mathrm{dA}}{\mathrm{dt}}+\frac{\partial A}{\partial t}\right]$

Use either Stokes' theorem or the divergence theorem to evaluate each of the following integrals in the easiest possible way.

${\mathbf{\iint }}_{\mathbf{\text{surface}}\mathbf{\sigma }}\mathbf{ }\mathbf{curl}\mathbf{}\left(x^{2}i+z^{2}j-y^{2}k\right)\mathbf{-}\mathbf{nd\sigma }$, where $\mathbf{\sigma }$ is the part of the surface $\mathbf{z}\mathbf{=}\mathbf{4}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}{\mathbf{y}}^{\mathbf{2}}$above the $\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}$ plane.

Find the derivative of ${\mathbf{xy}}^{\mathbf{2}}\mathbf{+}\mathbf{yz}$ at $\left(1,1,2\right)$ in the direction of the vector $\mathbf{2}\mathbf{i}\mathbf{-}\mathbf{j}\mathbf{+}\mathbf{2}\mathbf{k}$.

$\mathit{x}\mathit{y}\mathit{d}\mathit{x}\mathbf{+}{\mathit{x}}^{\mathbf{2}}\mathit{d}\mathit{y}$where C is as selected.

Evaluate each of the integrals in Problems 3 to 8 as either a volume integral or a surface integral, whichever is easier.

$\mathbf{\iint }\mathbf{r}\mathbf{.}\mathbf{nd\sigma }$Over the whole surface of the cylinder bounded by${\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{=}\mathbf{}\mathbf{,}\mathbf{}\mathbf{z}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{and}\mathbf{}\mathbf{z}\mathbf{=}\mathbf{3}\mathbf{}\mathbf{;}\mathbf{}\mathbf{r}\mathbf{}\mathbf{means}\mathbf{}\mathbf{ix}\mathbf{+}\mathbf{jy}\mathbf{+}\mathbf{k}z$

Evaluate the line integral $\mathbf{\int }\mathbf{xydx}\mathbf{+}\mathbf{xdy}\mathbf{}\mathbf{from}\left(0,0\right)\mathbf{to}\left(1,2\right)$ along the paths shown in the sketch.

Find the total work done by forces and if the object undergoes the displacement $\mathit{C}$ . Hint: Can you add the two forces first?

Show that $\mathbf{\nabla }\mathbf{.}\left(U\times r\right)\mathbf{=}\mathbf{r}\mathbf{.}\left(\nabla \times U\right)$where U is a vector function of $\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{z}$and$\mathbf{r}\mathbf{=}\mathbf{xi}\mathbf{+}\mathbf{yj}\mathbf{+}\mathbf{zk}$ .