魅影直播

Calculate the volume integral of the function $\mathit{T}\mathbf{=}{\mathit{z}}^{\mathbf{2}}$over the tetrahedron with comers at (0,0,0), (1,0,0), (0,1,0), and (0,0,1).

Test Stokes' theorem for the function $\mathit{v}\mathbf{=}\left(xy\right)\mathit{i}\mathbf{鈥�}\mathbf{+}\left(2yz\right)\mathbf{鈥�}\mathit{j}\mathbf{+}\left(3zx\right)\mathbf{鈥�}\mathit{k}$, using the triangular shaded area of Fig. 1.34.

Express the unit vectors $\mathit{r}\mathbf{,}\mathbf{鈥�}\mathit{胃}\mathbf{,}\mathbf{鈥�}\mathit{蠒}$in terms of x, y, z (that is, deriveEq. 1.64). Check your answers several ways ( $\mathit{r}\mathbf{路}\mathit{r}\stackrel{\mathbf{?}}{\mathbf{=}}\mathbf{1}$, $\mathit{胃}\mathbf{路}\mathit{蠒}\stackrel{\mathbf{?}}{\mathbf{=}\mathbf{0}}$, $\mathit{r}\mathbf{脳}\mathit{胃}\stackrel{\mathbf{?}}{\mathbf{=}\mathbf{蠒}}$).Also work out the inverse formulas, giving x, y, z in terms of $\mathit{r}\mathbf{,}\mathbf{鈥�}\mathit{胃}\mathbf{,}\mathbf{鈥�}\mathit{蠒}$ (and $\mathbf{鈥�}\mathit{胃}\mathbf{,}\mathbf{鈥�}\mathit{蠒}$).

Let ${}^{\mathbf{r}}$be the separation vector from a fixed point ${}^{\mathbf{(}\mathbf{x}\mathbf{\text{'}}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{,}\mathbf{}\mathbf{z}\mathbf{\text{'}}\mathbf{)}}$to the point localid="1654317524404" $\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{z}\mathbf{)}$, and let r be its length. Show that

(a)localid="1654317730952" ${}^{\mathbf{鈭�}\mathbf{}\mathbf{\left(}{\mathbf{r}}^{\mathbf{2}}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{r}}$

(b) ${}^{\mathbf{鈭�}\mathbf{}\mathbf{(}\mathbf{1}\mathbf{/}\mathbf{r}\mathbf{)}\mathbf{=}\mathbf{-}\mathbf{r}\mathbf{/}{\mathbf{r}}^{\mathbf{2}}}$

(c) What is the general formula for localid="1654317981268" ${}^{\mathbf{鈭�}\mathbf{}\mathbf{(}\mathbf{}{\mathbf{r}}^{\mathbf{n}}\mathbf{}\mathbf{)}}$

Find the angle between the body diagonals of a cube.

Suppose that f is a function of two variables (y and z) only. Show that the gradient ${}^{\mathbf{鈭�}\mathbf{f}\mathbf{}\mathbf{=}\mathbf{(}\mathbf{鈭�}\mathbf{f}\mathbf{/}\mathbf{鈭�}\mathbf{y}\mathbf{)}\widehat{\mathbf{}\mathbf{y}}\mathbf{}\mathbf{(}\mathbf{鈭�}\mathbf{f}\mathbf{/}\mathbf{鈭�}\mathbf{z}\mathbf{)}\mathbf{}\hat{\mathbf{z}}}$transforms as a vector under rotations, Eq 1.29. [Hint: ${}^{\mathbf{(}\mathbf{鈭�}\mathbf{f}\mathbf{/}\mathbf{鈭�}\stackrel{\mathbf{炉}}{\mathbf{y}}\mathbf{)}\mathbf{=}\mathbf{(}\mathbf{鈭�}\mathbf{f}\mathbf{/}\mathbf{鈭�}\mathbf{y}\mathbf{)}\mathbf{(}\mathbf{鈭�}\mathbf{f}\mathbf{/}\mathbf{鈭�}\stackrel{\mathbf{炉}}{\mathbf{y}}\mathbf{)}\mathbf{+}\mathbf{(}\mathbf{鈭�}\mathbf{f}\mathbf{/}\mathbf{鈭�}\mathbf{z}\mathbf{)}\mathbf{(}\mathbf{鈭�}\mathbf{z}\mathbf{/}\mathbf{鈭�}\stackrel{\mathbf{炉}}{\mathbf{y}}\mathbf{)}\mathbf{,}}$and the analogous formula for ${}^{\mathbf{鈭�}\mathbf{f}\mathbf{/}\mathbf{鈭�}\stackrel{\mathbf{炉}}{\mathbf{z}}}$. We know that localid="1654595255202" ${}^{\stackrel{\mathbf{炉}}{\mathbf{y}}\mathbf{}\mathbf{=}\mathbf{测肠辞蝉蠒}\mathbf{}\mathbf{+}\mathbf{锄蝉颈苍蠒}}$and ${}^{\overline{\mathbf{z}}\mathbf{}\mathbf{=}\mathbf{-}\mathbf{ycos}\mathbf{鈭�}\mathbf{}\mathbf{+}\mathbf{}\mathbf{zcos}\mathbf{鈭�}\mathbf{;}}$鈥漵olve鈥� these equations for y and z (as functions of localid="1654325243865" ${}^{\stackrel{\mathbf{炉}}{\mathbf{y}}}$and ${}^{\overline{\mathbf{z}}}$(as functions of ${}^{\overline{\mathbf{y}}}$and ${}^{\overline{\mathbf{z}}}$), and compute the needed derivatives ${}^{\mathbf{鈭�}\mathbf{f}\mathbf{/}\mathbf{鈭�}{\overline{\mathbf{y}}}_{\mathbf{,}}\mathbf{}\mathbf{鈭�}\mathbf{z}\mathbf{/}\mathbf{鈭�}\overline{\mathbf{y}}\mathbf{}}$, etc]

Use the cross product to find the components of the unit vector$\hat{\mathit{n}}$ perpendicular to the shaded plane in Fig. 1.11.

Calculate the divergence of the following vector functions:

$$\begin{gathered} \mathbf{\left(}\mathit{a}\mathbf{\right)}\mathbf{}{\mathit{v}}_{\mathbf{a}}\mathbf{=}{\mathit{x}}^{\mathbf{2}}\hat{\mathbf{x}}\mathbf{+}\mathbf{3}\mathit{x}{\mathit{z}}^{\mathbf{2}}\hat{\mathbf{y}}\mathbf{}\mathbf{-}\mathbf{2}\mathit{x}\mathit{z}\hat{\mathbf{z}} \\ \mathbf{\left(}\mathit{b}\mathbf{\right)}\mathbf{}{\mathit{v}}_{\mathbf{b}}\mathbf{=}\mathit{x}\mathit{y}\hat{\mathbf{x}}\mathbf{+}\mathbf{2}\mathit{y}\mathit{z}\hat{\mathbf{y}}\mathbf{+}\mathbf{3}\mathit{z}\mathit{x}\hat{\mathbf{z}} \\ \mathbf{\left(}\mathit{c}\mathbf{\right)}\mathbf{}{\mathit{v}}_{\mathbf{c}}\mathbf{=}{\mathit{y}}^{\mathbf{2}}\hat{\mathbf{x}}\mathbf{+}\mathbf{(}\mathbf{2}\mathit{x}\mathit{y}\mathbf{+}{\mathit{z}}^{\mathbf{2}}\mathbf{)}\mathbf{+}\mathbf{2}\mathit{y}\mathit{z}\hat{\mathbf{z}} \end{gathered}$$

Prove the BAC-CAB rule by writing out both sides in component form.

Sketch the vector function

$\mathbf{v}\mathbf{=}\frac{\hat{\mathbf{r}}}{{\mathbf{r}}^{\mathbf{2}}}$

and compute its divergence. The answer may surprise you ... can you explain it?