coexpennan
2021-01-31
Answered

Flux integrals Compute the outward flux of the following vector fields across the given surfaces S. You should decide which integral of the Divergence Theorem to use.

$F=\u27e8x\mathrm{sin}y,-\mathrm{cos}y,z\mathrm{sin}y\u27e9$ , S is the boundary of the region bounded by the planes x = 1, y = 0, $y=\frac{\pi}{2},z=0$ , and z = x.

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odgovoreh

Answered 2021-02-01
Author has **107** answers

Step 1

Given:

$F=\u27e8x\mathrm{sin}y,-\mathrm{cos}y,zz\mathrm{sin}y\u27e9$ , the surface S is the boundary of the region bounded by the planes x = 1, $y=\frac{\pi}{2},z=0$ , and z = x.

Step 2

Using Divergence theorem

$\int {\int}_{D}\int \mathrm{\nabla}\cdot Fdv=\int {\int}_{S}F\cdot nds$

Compute The divergence of the field

$\mathrm{\nabla}\cdot F=\frac{\partial}{\partial x}\left(x\mathrm{sin}y\right)+\frac{\partial}{\partial y}(-\mathrm{cos}y)+\frac{\partial}{\partial z}\left(z\mathrm{sin}y\right)$

$=\mathrm{sin}y\frac{\partial}{\partial x}\left(x\right)+\frac{\partial}{\partial y}(-\mathrm{cos}y)+\mathrm{sin}y\frac{\partial}{\partial z}\left(z\right)$

$=\mathrm{sin}y+\mathrm{sin}y+\mathrm{sin}y$

$=3\mathrm{sin}y$

Step 3

Therefore, the outward flux is

$\int {\int}_{D}\int \mathrm{\nabla}\cdot Fdv={\int}_{0}^{\frac{\pi}{2}}{\int}_{0}^{1}{\int}_{0}^{x}3\mathrm{sin}ydzdxdy$

$=3{\int}_{0}^{\frac{\pi}{2}}{\int}_{0}^{1}\mathrm{sin}y{\left[z\right]}_{0}^{x}dxdy$

$=3{\int}_{0}^{\frac{\pi}{2}}{\int}_{0}^{1}x\mathrm{sin}ydxdy$

$=3{\int}_{0}^{\frac{\pi}{2}}{\left[\frac{{x}^{2}}{2}\right]}_{0}^{1}\mathrm{sin}ydy$

$=\frac{3}{2}{\int}_{0}^{\frac{\pi}{2}}[\mathrm{sin}ydy$

$=\frac{3}{2}{\int}_{0}^{\frac{\pi}{2}}[\mathrm{sin}ydy$

$=\frac{3}{2}{[-\mathrm{cos}y]}_{0}^{\frac{\pi}{2}}$

$=\frac{3}{2}(-\mathrm{cos}\left(\frac{\pi}{2}\right)-(-\mathrm{cos}\left(0\right)))$

$=\frac{3}{2}(0-(-div>$###### Not exactly what you’re looking for?

Given:

Step 2

Using Divergence theorem

Compute The divergence of the field

Step 3

Therefore, the outward flux is

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It may just be I am searching using the wrong terminology, if so direction of where to look would be greatly appreciated too.EDIT: To add some more information, the initial problem involved four equations.

${y}_{1}={K}_{1}/{x}_{1}$

${y}_{2}={K}_{2}/{x}_{2}$

${y}_{3}={K}_{3}/{x}_{3}$

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Where K and X values are given constants, and x and y values are unknown. I am trying to find a solution that maximises the sum of the y values.

It is from this problem that I simplified it to

$\frac{{K}_{1}}{{y}_{1}}+\frac{{K}_{2}}{{y}_{2}}+\frac{{K}_{3}}{{y}_{3}}=X$

With the first equation in the initial question just being arbitrary values.

$\frac{10}{{y}_{1}}+\frac{12}{{y}_{2}}+\frac{15}{{y}_{3}}=50$

It may just be I am searching using the wrong terminology, if so direction of where to look would be greatly appreciated too.EDIT: To add some more information, the initial problem involved four equations.

${y}_{1}={K}_{1}/{x}_{1}$

${y}_{2}={K}_{2}/{x}_{2}$

${y}_{3}={K}_{3}/{x}_{3}$

${x}_{1}+{x}_{2}+{x}_{3}=X$

Where K and X values are given constants, and x and y values are unknown. I am trying to find a solution that maximises the sum of the y values.

It is from this problem that I simplified it to

$\frac{{K}_{1}}{{y}_{1}}+\frac{{K}_{2}}{{y}_{2}}+\frac{{K}_{3}}{{y}_{3}}=X$

With the first equation in the initial question just being arbitrary values.

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Use Stokes' Theorem to find the work done.

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c) What is the probability that the number of granite specimens selected for analysis is within 1 standard deviation of its mean value? (Round your answer to four decimal places.)

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