Because I've gone ahead with trying to parameterize $S$ directly and learned the hard way that the resulting integral is large and annoying to work with, I'll propose a less direct approach.

Rather than compute the surface integral over $S$ straight away, let's close off the hemisphere with the disk $D$ of radius 9 centered at the origin and coincident with the plane $y=0$ . Then by the divergence theorem, since the region $S\cup D$ is closed, we have

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iiint_R(\nabla\cdot\vec F)\,\mathrm dV$

where $R$ is the interior of $S\cup D$ . $\vec F$ has divergence

$\nabla\cdot\vec F(x,y,z)=\dfrac{\partial(xz)}{\partial x}+\dfrac{\partial(x)}{\partial y}+\dfrac{\partial(y)}{\partial z}=z$

so the flux over the closed region is

$\displaystyle\iiint_Rz\,\mathrm dV=\int_0^\pi\int_0^\pi\int_0^9\rho^3\cos\varphi\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=0$

The total flux over the closed surface is equal to the flux over its component surfaces, so we have

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iint_S\vec F\cdot\mathrm d\vec S+\iint_D\vec F\cdot\mathrm d\vec S=0$

$\implies\boxed{\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=-\iint_D\vec F\cdot\mathrm d\vec S}$

Parameterize $D$ by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec k$

with $0\le u\le9$ and $0\le v\le2\pi$ . Take the normal vector to $D$ to be

$\vec s_u\times\vec s_v=-u\,\vec\jmath$

Then the flux of $\vec F$ across $S$ is

$\displaystyle\iint_D\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^9\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_u\times\vec s_v)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^{2\pi}\int_0^9(u^2\cos v\sin v\,\vec\imath+u\cos v\,\vec\jmath)\cdot(-u\,\vec\jmath)\,\mathrm du\,\mathrm dv$

$=\displaystyle-\int_0^{2\pi}\int_0^9u^2\cos v\,\mathrm du\,\mathrm dv=0$

$\implies\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\boxed{0}$

8 0

3 years ago

Which one is right ?

sergey [27]

Answer:

6/10

Step-by-step explanation:

all can be reduced to 2/5, except 6/10:

2/5

8/20=2/5 (divide out 4)

12/30=2/5 (divide out 6)

but not

6/10 = 3/5!

7 0

3 years ago

Please help asap!!!

sp2606 [1]

Answer:

$y=\frac{1}{6} (x+5)^2-4.5$

$y=\frac{-1}{20} (x-10)^2+1$

Step-by-step explanation:

focus at (-5, -3), and directrix y = -6

Directrix y=-6 so its a vertical parabola

so equation is

(x-h)^2 = 4p(y-k)

(h,k) is the center

P is the distance between focus and vertex

distance between focus and directrix = 2p

distance between -3 and y=-6 is 3

2p = 3

p = 3/2 or p = 1.5

Focus is (h, k+p)

given focus is (-5, -3) so h= -5 and k+p = -3

k+p=-3, plug in 1.5 for p

k + 1.5 = -3

subtract 1.5 on both sides

k = -4.5

(x-h)^2 = 4p(y-k)

$(x+5)^2= 4(1.5) (y+4.5)$

$(x+5)^2= 6(y+4.5)$

divide by 6 on both sides

then subtract 4.5 on both sides

$y=\frac{1}{6} (x+5)^2-4.5$

focus at (10, -4), and directrix y = 6.

Directrix y=6 so its a vertical parabola

so equation is

(x-h)^2 = 4p(y-k)

distance between focus and directrix = 2p

distance between -4 and y=6 is -4-6=-10

2p = -10

p = -5

Focus is (h, k+p)

given focus is (10, -4) so h= 10 and k+p = -4

k+p=-4, plug in 5 for p

k - 5 = -4

add 5 on both sides

k = 1

(x-h)^2 = 4p(y-k)

$(x-10)^2= 4(-5) (y-1)$

$(x-10)^2= -20(y-1)$

divide by -20 on both sides and add 1 on both sides

$y=\frac{-1}{20} (x-10)^2+1$

8 0

3 years ago

W is the midpoint of VX and Z is the midpoint of VY . If XY=p and WZ=p–30, what is WZ?

juin [17]

$\large{ \tt{❃ \: EXPLANATION}} :$

You'll have to know : Mid-point theorem states that A straight line segment joining the mid-points of any two triangle is parallel to the third side and it is equal to half of the length of the third side.