3ebfa1…: “Left of the arid parallelogram”

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Let $B$ be the sphere of radius $3$ centered at the point $(0,0,7)$ , i.e., $B = \{ (x,y,z) \in \R ^3 : x^2 + y^2 + (z-7)^2 = 3^2 \}$ and let $H$ be the northern hemisphere meaning $H = \{ (x,y,z) \in B : z \geq 7 \}.$ In both cases, arrange the orientation so that the positive-flux direction at the north pole points in the direction of the positive $z$ -axis.

Define a vector field $\vec {F} : \R ^3 \to \R ^3$ by the rule $\vec {F}(x,y,z) = \vector {z,x+y+z,y+2z}.$

$\displaystyle \oiint _B (\curl \vec {F})\dotp \uvec {n}\d S = \answer {0}$ .

This equals, by Stokes’ theorem, $\displaystyle \oint _{\partial B} \vec {F}\dotp \d \vec {p}$ .

But $\partial B$ is empty, so the integral vanishes.

$\curl \vec {F} = \vector {\answer {0},\answer {1},\answer {1}}$ .

$\displaystyle \iint _H (\curl \vec {F})\dotp \uvec {n}\d S = \answer {9\pi }$

Let $E$ be the equator meaning $E = \{ (x,y,z) \in B : z = 7 \},$ so that $E$ is a circle of radius $3$ .

By Stokes’ theorem, this integral is $\oint _E \vec {F} \dotp \d \vec {p}$ .

Let $D$ be the equatorial disk $D = \{ (x,y,z) \in \R ^3 : x^2 + y^2 \leq 3^2 \mbox { and } z = 7 \}.$

By Stokes’ theorem, these integrals are also equal to $\displaystyle \iint _D (\curl \vec {F})\dotp \uvec {n}\d S$ provided $D$ is oriented appropriately—in this case, with $\uvec {n} = \vector {0,0,1}$ .

Since $(\curl \vec {F})\dotp \uvec {n} = 1$ , these integrals are the area of $D$ , which is $9\pi$ .

$\displaystyle \oiint _B \vec {F}\dotp \uvec {n}\d S = \answer {108\pi }$ .

By the divergence theorem, the given integral is the same as the integral of $\divergence \vec {F}$ over the enclosed volume.

But $\divergence \vec {F} = 3$ and the sphere of radius $3$ has volume $36\pi$ .

So the integral is $3 \cdot 36\pi$ .

$\displaystyle \iint _H \vec {F}\dotp \uvec {n} \d S = \answer {180\pi }$

Let $D$ be the equatorial disk $D = \{ (x,y,z) \in \R ^3 : x^2 + y^2 \leq 3^2 \mbox { and } z = 7\}.$

Then $H$ and $D$ (suitably oriented) together bound an solid hemisphere.

By the divergence theorem, the difference between the flux through $H$ and $D$ equals $\divergence \vec {F}$ integrated over that solid hemisphere.

But $\divergence \vec {F} = 3$ and the hemisphere of radius $3$ has volume $18\pi$ .

So the difference between the flux through $H$ and $D$ is $54\pi$ .

The flux through $D$ can be computed by hand: in that case, $\uvec {n} = \vector {0,0,-1}$ but then on $D$ the quantity $\vec {F}\dotp \uvec {n} = -y-14$ .

By symmetry, $\iint _D -y \d A = 0$ and $\iint _D -14 \d A = -126\pi$ .

So the desired integral is $54\pi + 126\pi$ .