Herman has $18. He earns $8 for 1 hour (h) of tutoring. He wants to buy a skateboard for $90.

Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of

tresset_1 [31]

Because I've gone ahead with trying to parameterize $S$ directly and learned the hard way that the resulting integral is large and annoying to work with, I'll propose a less direct approach.

Rather than compute the surface integral over $S$ straight away, let's close off the hemisphere with the disk $D$ of radius 9 centered at the origin and coincident with the plane $y=0$ . Then by the divergence theorem, since the region $S\cup D$ is closed, we have

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iiint_R(\nabla\cdot\vec F)\,\mathrm dV$

where $R$ is the interior of $S\cup D$ . $\vec F$ has divergence

$\nabla\cdot\vec F(x,y,z)=\dfrac{\partial(xz)}{\partial x}+\dfrac{\partial(x)}{\partial y}+\dfrac{\partial(y)}{\partial z}=z$

so the flux over the closed region is

$\displaystyle\iiint_Rz\,\mathrm dV=\int_0^\pi\int_0^\pi\int_0^9\rho^3\cos\varphi\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=0$

The total flux over the closed surface is equal to the flux over its component surfaces, so we have

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iint_S\vec F\cdot\mathrm d\vec S+\iint_D\vec F\cdot\mathrm d\vec S=0$

$\implies\boxed{\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=-\iint_D\vec F\cdot\mathrm d\vec S}$

Parameterize $D$ by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec k$

with $0\le u\le9$ and $0\le v\le2\pi$ . Take the normal vector to $D$ to be

$\vec s_u\times\vec s_v=-u\,\vec\jmath$

Then the flux of $\vec F$ across $S$ is

$\displaystyle\iint_D\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^9\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_u\times\vec s_v)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^{2\pi}\int_0^9(u^2\cos v\sin v\,\vec\imath+u\cos v\,\vec\jmath)\cdot(-u\,\vec\jmath)\,\mathrm du\,\mathrm dv$

$=\displaystyle-\int_0^{2\pi}\int_0^9u^2\cos v\,\mathrm du\,\mathrm dv=0$

$\implies\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\boxed{0}$

8 0

3 years ago

Suppose h(t) = -4t^2+ 11t + 3 is the height of a diver above the water in

nadezda [96]

Answer:

11/4 seconds

Step-by-step explanation:

We can see from the given equation that the height at the diving board is h = 3.

This is because the h(t) equation has that +3 at the end, which denotes the initial height of the diver, the height when he is standing on the board before jumping.

To find where h(t) = 3 is true, we need to set h(t) equal to 3 and solve for t.

h(t) = 3

h(t) = -4t^2 + 11t + 3 = 3

-4t^2 + 11t + 3 - 3 = 0

-4t^2 + 11t = 0

t*(-4t + 11) = 0

so h(t) = 3 when t = 0 and when t = 11/4 sec

we already know that at t = 0 the height is 3, it is the initial height given from the equation, so we want to use the other solution for t.

the diver is back at the height of the diving board at t = 11/4 sec

6 0

2 years ago

Read 2 more answers

Q3) Explain the Error: When Rashed found the difference -11 - (-4), he got-

lina2011 [118]

He maybe accidently added both of thm .
But if we difference we get 7.
Because - and - =+
So
-11+4=7.

6 0

3 years ago

Find the roots of the following equations. a) 4x+9=34 b) (x-4)(x+2)=0 c) 2x^2-6x+4=0

Masteriza [31]

A. 4x + 9 = 34
 - 9 - 9
 4x = 25
 4 4
 x = 6.25

B. (x - 4)(x + 2) = 0
 x - 4 = 0 U x + 2 = 0
 + 4 + 4 - 2 - 2
 x = 4 x = -2

C. 2x² - 6x + 4 = 0
 2(x²) - 2(3x) + 2(2) = 0
2(x² - 3x + 2) = 0
 2 2
 x² - 3x + 2 = 0
 x = -(-3) +/- √((-3)² - 4(1)(2))
 2(1)
 x = 3 +/- √(9 - 8)
 2
 x = 3 +/- √(1)
 2
 x = 3 +/- 1
 2
 x = 3 + 1 U x = 3 - 1
 2 2
 x = 4 x = 2
 2 2
 x = 2 x = 1

7 0

3 years ago

How many fractions are equivalent to 4/5 explain

Reptile [31]

There are alot it depeneds of what is the number :)

5 0

3 years ago