Helping my lil brother with work

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

Alenkasestr [34]

$\mathbf f(x,y,z)=\langle z,y,x\rangle\implies\nabla\cdot\mathbf f=\dfrac{\partial z}{\partial x}+\dfrac{\partial y}{\partial y}+\dfrac{\partial x}{\partial z}=0+1+0=1$

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.

3 0

3 years ago

A recent study reported that high school students spend an average of 94 minutes per day texting. Jenna claims that the average

larisa [96]

Answer:

We have sufficient evidence to support the claim that the average for the students at Jenna's large high school is greater than 94 minutes.

Step-by-step explanation:

Jenna claims that the average time of texting at her larger high school is greater than 94 minutes per day.

From here we can see that we have to perform a hypothesis test about a sample mean. The null and alternate hypothesis will be:

Null Hypothesis: $\mu \leq 94$

Alternate Hypothesis: $\mu > 94$

Jenna collected data from a sample of 32 students. So, sample size will be:

Sample Size = n = 32

Sample Mean = x = 96.5

Sample Standard Deviation = s = 6.3

We have to perform a hypothesis test, to test Jenna's claim. Since, the value of Population Standard Deviation is unknown and the value of Sample Standard Deviation is known, we will use One Sample t-test in this case.

The formula to calculate the test statistic is:

$t=\frac{x-\mu}{\frac{s}{\sqrt{n}}}$

Using the values, we get:

$t=\frac{96.5-94}{\frac{6.3}{\sqrt{32} } }=2.245$

The degrees of freedom will be:

df = n - 1 = 32 - 1 = 31

We have to convert the t-score 2.245 with 31 degrees of freedom to its equivalent p-value. From t-table this value comes out to be:

p-value = 0.0160

The significance level is:

$\alpha =0.05$

Since, the p-value is lesser than the level of significance, we reject the Null Hypothesis.

Conclusion:

We have sufficient evidence to support the claim that the average for the students at Jenna's large high school is greater than 94 minutes.

3 0

3 years ago

I will make you the Brainliest

Stells [14]

It’s the first one bc if u add -3 to the input it gives the outputs

4 0

3 years ago

Read 2 more answers

Two coins and one six sided number cube are tossed together

Triss [41]

Answer:

1/24

Step-by-step explanation:

two coins each have a 1/2 probability of landing on heads so we multiply

1/2 x 1/2 = 1/4 probability of getting heads twice

we have a 1/6 chance of landing on a 4

so we multiply 1/6 x 1/4

which equals 1/24

7 0

3 years ago

11 / 2 plus 11 / 4 in fraction

dybincka [34]

Answer:

8and1/4

Step-by-step explanation:

8 0

3 years ago