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RUDIKE [14]
2 years ago
6

PLEASE HELP!

Mathematics
1 answer:
AnnZ [28]2 years ago
3 0

Answer:

A.30x3-3x2-24x+9

Step-by-step explanation

1. Separate the terms in g(x). (6x) (-3)

2. Distribute by multiplying each term in f(x) by (6x).

   (6x)(5x2+2x-3)=30x3+12x2-18x

3. Distribute the other term in g(x) to each term in f(x).

   (-3)(5x2+2x-3)= -15x2-6x+9

4. Combine the like terms of both answers.

   30x3+12x2-18x-15x2-6x+9= <em><u>30x3-3x2-24x+9</u></em>

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Given that you mention the divergence theorem, and that part (b) is asking you to find the downward flux through the disk x^2+y^2\le3, I think it's same to assume that the hemisphere referred to in part (a) is the upper half of the sphere x^2+y^2+z^2=3.

a. Let C denote the hemispherical <u>c</u>ap z=\sqrt{3-x^2-y^2}, parameterized by

\vec r(u,v)=\sqrt3\cos u\sin v\,\vec\imath+\sqrt3\sin u\sin v\,\vec\jmath+\sqrt3\cos v\,\vec k

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\vec r_v\times\vec r_u=3\cos u\sin^2v\,\vec\imath+3\sin u\sin^2v\,\vec\jmath+3\sin v\cos v\,\vec k

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b. Let D be the disk that closes off the hemisphere C, parameterized by

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\displaystyle\int_0^{2\pi}\int_0^{\sqrt3}(2\,\vec k)\cdot(\vec s_v\times\vec s_u)\,\mathrm du\,\mathrm dv=-2\int_0^{2\pi}\int_0^{\sqrt3}u\,\mathrm du\,\mathrm dv

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c. The net flux is then \boxed{4\sqrt3\pi}.

d. By the divergence theorem, the flux of \vec F across the closed hemisphere H with boundary C\cup D is equal to the integral of \mathrm{div}\vec F over its interior:

\displaystyle\iint_{C\cup D}\vec F\cdot\mathrm d\vec S=\iiint_H\mathrm{div}\vec F\,\mathrm dV

We have

\mathrm{div}\vec F=\dfrac{\partial(2z+2)}{\partial z}=2

so the volume integral is

2\displaystyle\iiint_H\mathrm dV

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\displaystyle2\int_0^{\pi/2}\int_0^{2\pi}\int_0^{\sqrt3}\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=4\sqrt3\pi

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