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Andru [333]
3 years ago
9

G verify that the divergence theorem is true for the vector field f on the region

Mathematics
1 answer:
Alenkasestr [34]3 years ago
3 0
\mathbf f(x,y,z)=\langle z,y,x\rangle\implies\nabla\cdot\mathbf f=\dfrac{\partial z}{\partial x}+\dfrac{\partial y}{\partial y}+\dfrac{\partial x}{\partial z}=0+1+0=1

Converting to spherical coordinates, we have

\displaystyle\iiint_E\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV=\int_{\varphi=0}^{\varphi=\pi}\int_{\theta=0}^{\theta=2\pi}\int_{\rho=0}^{\rho=6}\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=288\pi

On the other hand, we can parameterize the boundary of E by

\mathbf s(u,v)=\langle6\cos u\sin v,6\sin u\sin v,6\cos v\rangle

with 0\le u\le2\pi and 0\le v\le\pi. Now, consider the surface element

\mathrm d\mathbf S=\mathbf n\,\mathrm dS=\dfrac{\mathbf s_v\times\mathbf s_u}{\|\mathbf s_v\times\mathbf s_u\|}\|\mathbf s_v\times\mathbf s_u\|\,\mathrm du\,\mathrm dv
\mathrm d\mathbf S=\mathbf s_v\times\mathbf s_u\,\mathrm du\,\mathrm dv
\mathrm d\mathbf S=36\langle\cos u\sin^2v,\sin u\sin^2v,\sin v\cos v\rangle\,\mathrm du\,\mathrm dv

So we have the surface integral - which the divergence theorem says the above triple integral is equal to -

\displaystyle\iint_{\partial E}\mathbf f\cdot\mathrm d\mathbf S=36\int_{v=0}^{v=\pi}\int_{u=0}^{u=2\pi}\mathbf f(x(u,v),y(u,v),z(u,v))\cdot(\mathbf s_v\times\mathbf s_u)\,\mathrm du\,\mathrm dv
=\displaystyle36\int_{v=0}^{v=\pi}\int_{u=0}^{u=2\pi}(12\cos u\cos v\sin^2v+6\sin^2u\sin^3v)\,\mathrm du\,\mathrm dv=288\pi

as required.
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3.9111 km

Explanation:

The length in kilometers is equal to the meters divided by 1,000. So, if you divide 3911.1 by 1,000 that equals 3.911. You move the decimal point to the left 3 times based on the number of zeros.

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7:11

Step-by-step explanation:

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70:110

Then reduce it to the lowest term. In this case by 10.

70 ÷10=7

110÷10=11Therefore, the answe is

7:11

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The Earth completely rotates on its axis once every 24 hours.
strojnjashka [21]

Answer:

A)

62/3 = 20.67 hours

B)

60.00 hours

C)

2074.29 miles

Step-by-step explanation:

If we assume the earth is a perfect circle, then in a complete rotation the earth covers 360 degrees or 2π radians.

A)

In 24 hours the earth rotates through an angle of 360 degrees. We are required to determine the duration it takes to rotate through 310 degrees. Let x be the duration it takes the earth to rotate through 310 degrees, then the following proportions hold;

(24/360) = (x/310)

solving for x;

x = (24/360) * 310 = 62/3 = 20.67 hours

B)

In 24 hours the earth rotates through an angle of 2π radians. We are required to determine the duration it takes to rotate through 5π radians. Let x be the duration it takes the earth to rotate through 5π radians, then the following proportions hold;

(24/2π radians) = (x/5π radians)

Solving for x;

x = (24/2π radians)*5π radians = 60 hours

C)

If the diameter of the earth is 7920 miles, then in 24 hours a point on the equator will rotate through a distance equal to the circumference of the Earth. Using the formula for the circumference of a circle we have;

circumference = 2*π*R = π*D

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Therefore, the speed of the earth is approximately;

(7920π miles)/(24 hours) = 330π miles/hr

The distance covered by a point in 2 hours will thus be;

330π * 2 = 660π miles = 2074.29 miles

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