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Use stoke's theorem to evaluate∬m(∇×f)⋅ds where m is the hemisphere x^2+y^2+z^2=9, x≥0, with the normal in the direction of the

ludmilkaskok [199]

By Stokes' theorem,

$\displaystyle\int_{\partial\mathcal M}\mathbf f\cdot\mathrm d\mathbf r=\iint_{\mathcal M}\nabla\times\mathbf f\cdot\mathrm d\mathbf S$

where $\mathcal C$ is the circular boundary of the hemisphere $\mathcal M$ in the $y$ - $z$ plane. We can parameterize the boundary via the "standard" choice of polar coordinates, setting

$\mathbf r(t)=\langle 0,3\cos t,3\sin t\rangle$

where $0\le t\le2\pi$ . Then the line integral is

$\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=\int_{t=0}^{t=2\pi}\mathbf f(x(t),y(t),z(t))\cdot\dfrac{\mathrm d}{\mathrm dt}\langle x(t),y(t),z(t)\rangle\,\mathrm dt$
$=\displaystyle\int_0^{2\pi}\langle0,0,3\cos t\rangle\cdot\langle0,-3\sin t,3\cos t\rangle\,\mathrm dt=9\int_0^{2\pi}\cos^2t\,\mathrm dt=9\pi$

We can check this result by evaluating the equivalent surface integral. We have

$\nabla\times\mathbf f=\langle1,0,0\rangle$

and we can parameterize $\mathcal M$ by

$\mathbf s(u,v)=\langle3\cos v,3\cos u\sin v,3\sin u\sin v\rangle$

so that

$\mathrm d\mathbf S=(\mathbf s_v\times\mathbf s_u)\,\mathrm du\,\mathrm dv=\langle9\cos v\sin v,9\cos u\sin^2v,9\sin u\sin^2v\rangle\,\mathrm du\,\mathrm dv$

where $0\le v\le\dfrac\pi2$ and $0\le u\le2\pi$ . Then,

$\displaystyle\iint_{\mathcal M}\nabla\times\mathbf f\cdot\mathrm d\mathbf S=\int_{v=0}^{v=\pi/2}\int_{u=0}^{u=2\pi}9\cos v\sin v\,\mathrm du\,\mathrm dv=9\pi$

as expected.

7 0

3 years ago

A riverboat travels 54 km downstream in 2 hours. It travels 57 km upstream in 3 hours. Find the speed of the boat and the speed

irakobra [83]

Here we must write and solve a system of equations to find the speed of the boat and the speed of the stream. We will find that the boat's speed is 23km/h and the river's speed is 4km/h.

First, remember the relation:

Distance = Speed*Time.

Now let's define the variables we will be using:

B = boat's speed
R = river's speed.

When the boat travels downstream, the total speed of the boat is the speed of the boat plus the speed of the river, and we know that in that case it travels 54km in 2 hours, then:

54km = (B + R)*2h

When the boat travels upstream, we must subtract the speed of the river. In that case, we know that the boat travels 57km in 3 hours, then we have:

57km = (B - R)*3h

Then our system of equations is:

54km = (B + R)*2h

57km = (B - R)*3h

To solve this, first, we need to isolate one of the variables in one of the equations, let's isolate B in the second one:

57km = (B - R)*3h

57km/3h + R = B

19 km/h + R = B

Now we can replace that in the other equation:

54km = (B + R)*2h

54km/2h = B + R

27 km/h = B + R = (19 km/h + R) + R

Now we can solve this for R:

27 km/h = 19km/h + 2*R

27 km/h - 19km/h = 2*R

8 km/h = 2*R

(8km/h)/2 = 4km/h = R

Now that we know the value of R, we can use:

B = 19 km/h + R = 19 km/h + 4km/h = 23 km/h

So the boat's speed is 23km/h and the river's speed is 4km/h.

If you want to learn more, you can read:

brainly.com/question/12895249

5 0

3 years ago

At the beginning of this month, the balance of Nelson's checking account was $523.12. So far this month, he has received a paych

Andrews [41]

Her check was 1,582.15

6 0

3 years ago

Hi can someone please help i only have a few hours to do this please help

Sonja [21]

(4,1) because it is the ordered pair that passes through the line

5 0

3 years ago

Read 2 more answers

If you do 12 minutes of a 45 minute workout how many calories is it? (the 45 minutes is 1000 calories in total)

velikii [3]

Answer:

266 $\frac{2}{3}$

Step-by-step explanation:

if a 45 minute workout is 1000 calories, at the same rate, a 12 minute workout would be 266 calories and 2 thirds. $\frac{12}{45}$ x 1000 = 266 $\frac{2}{3}$

7 0

2 years ago