Chapter 6: Vector Analysis

$\mathbf{\iint }\mathbf{V}\mathbf{-}\mathbf{nd\sigma }$over the curved part of the hemisphere in Problem $\mathbf{24}$, if role="math" localid="1657355269158" $\mathbf{V}\mathbf{=}\mathbf{curl}\mathbf{(}\mathbf{yi}\mathbf{-}\mathbf{xj}\mathbf{)}$.

$\mathbf{\iint }\mathbf{V}\mathbf{\times }\mathbf{nd\sigma }$over the entire surface of the cube in the first octant with three faces in the three coordinate planes and the other three faces intersecting at $\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{2}\mathbf{)}$, where $\mathbf{V}\mathbf{=}\mathbf{(}\mathbf{2}\mathbf{-}\mathbf{y}\mathbf{)}\mathbf{i}\mathbf{+}\mathbf{xzj}\mathbf{+}\mathbf{xyzk}$.

Problem$\mathbf{26}\mathbf{,}\mathbf{}$but integrate over the open surface obtained by leaving out the face of the cube in the$\mathbf{}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{}\mathbf{y}\mathbf{)}$ plane.

$\mathbf{\oint }\mathit{F}\mathbf{.}\mathit{d}\mathit{r}$around the circle over the curved part of the hemisphere in Problem 24, if ${\mathit{x}}^{\mathbf{2}}\mathbf{+}{\mathit{y}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathit{x}\mathbf{=}\mathbf{0}$, where $\mathit{F}\mathbf{=}\mathit{y}\mathit{i}\mathbf{-}\mathit{x}\mathit{j}$.

$\mathbf{\text{∮}}\mathbf{V}\mathbf{\times }\mathbf{dr}$around the boundary of the square with vertices$\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{-}\mathbf{1}\mathbf{)}$ , if$\mathbf{V}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\mathbf{i}\mathbf{+}\mathbf{5}\mathbf{xj}$

If A and B are the diagonals of a parallelogram, find a vector formula for the area of the parallelogram.

Given the vector.$\mathbf{A}\mathbf{=}\left(x^2-y^2\right)\mathbf{i}\mathbf{+}\mathbf{2}\mathbf{xyj}$

(a) Find $\mathbf{\nabla }\mathbf{\times }\mathbf{A}$.

(b) Evaluate$\mathbf{\iint }\mathbf{(}\mathbf{\nabla }\mathbf{\times }\mathbf{A}\mathbf{)}\mathbf{\times }\mathbf{d\sigma }$ over a rectangle in the$\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}$ plane bounded by the lines $\mathbf{x}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{x}\mathbf{=}\mathbf{a}\mathbf{,}\mathbf{}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{y}\mathbf{=}\mathbf{b}$.

(c) Evaluate around the boundary of the rectangle and thus verify Stokes' theorem for this case.

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

Question:Evaluate the line integral $\mathbf{\oint }\left(x+2y\right)\mathit{d}\mathit{x}\mathbf{-}\mathbf{2}\mathit{x}\mathit{d}\mathit{y}$along each of the following closed paths taken counterclockwise:

(a) The circle ${\mathit{x}}^{\mathbf{2}}\mathbf{+}{\mathit{y}}^{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{;}$.

(b) The square with corners at $\left(1,1\right)\mathbf{,}\mathbf{}\left(-1,1\right)\mathbf{,}\mathbf{}\left(-1,-1\right)\mathbf{,}\mathbf{}\left(1,-1\right)\mathbf{;}$

(c) The square with corners$\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{,}\mathbf{}\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{,}\mathbf{}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{;}$

Given, integrate $\mathit{V}\mathbf{}\mathbf{-}\mathit{n}\mathit{d}\mathbf{}\mathit{\sigma }$over the whole surface of the cube of side 1 with four of its vertices at $\left(0,0,0\right)\mathbf{,}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}$Evaluate the same integral by means of the divergence theorem.

$\mathbf{\oint }\mathbf{2}\mathit{x}\mathit{d}\mathit{y}\mathbf{-}\mathbf{3}\mathit{y}\mathit{d}\mathit{x}$around the square with vertices$\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{2}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{-}\mathbf{2}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{-}\mathbf{2}\mathbf{)}\mathbf{.}\mathbf{}$