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maw [93]
4 years ago
10

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

Mathematics
1 answer:
AnnyKZ [126]4 years ago
3 0

By the divergence theorem, the flux of \mathbf f(x,y,z)=4x\,\mathbf i+xy\,\mathbf j+5xz\,\mathbf k across the boundary \partial E of the cube E is equal to the integral of \nabla\cdot\mathbf f over E:

\displaystyle\iint_{\partial E}\mathbf f\cdot\mathrm d\mathbf S=\iiint_E\nabla\cdot\mathbf f\,\mathrm dV

We have divergence

\nabla\cdot\mathbf f=\dfrac{\partial(4x)}{\partial x}+\dfrac{\partial(xy)}{\partial y}+\dfrac{\partial(5xz)}{\partial z}=4+6x

and so the flux is

\displaystyle\iiint_E(4+6x)\,\mathrm dV=\int_{z=0}^{z=2}\int_{y=0}^{y=2}\int_{x=0}^{x=2}(4+6x)\,\mathrm dx\,\mathrm dy\,\mathrm dz

=4(4x+3x^2)\bigg|_{x=0}^{x=2}=80

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Answer:

The relation is not a function

The domain is {1, 2, 3}

The range is {3, 4, 5}

Step-by-step explanation:

A relation of a set of ordered pairs x and y is a function if

  • Every x has only one value of y
  • x appears once in ordered pairs

<u><em>Examples:</em></u>

  • The relation {(1, 2), (-2, 3), (4, 5)} is a function because every x has only one value of y (x = 1 has y = 2, x = -2 has y = 3, x = 4 has y = 5)
  • The relation {(1, 2), (-2, 3), (1, 5)} is not a function because one x has two values of y (x = 1 has values of y = 2 and 5)
  • The domain is the set of values of x
  • The range is the set of values of y

Let us solve the question

∵ The relation = {(1, 3), (2, 3), (3, 4), (2, 5)}

∵ x = 1 has y = 3

∵ x = 2 has y = 3

∵ x = 3 has y = 4

∵ x = 2 has y = 5

→ One x appears twice in the ordered pairs

∵ x = 2 has y = 3 and 5

∴ The relation is not a function because one x has two values of y

∵ The domain is the set of values of x

∴ The domain = {1, 2, 3}

∵ The range is the set of values of y

∴ The range = {3, 4, 5}

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By inspection, the two smallest numbers which when multiplied together yielded 850 were 25 and 34.
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