Find the missing side.<br> 09<br> 9.5<br> 10<br> C<br> 7 cm<br> 10.6<br> 8 cm<br> HELP PLEASE

Use Stokes' Theorem to evaluate C F · dr F(x, y, z) = xyi + yzj + zxk, C is the boundary of the part of the paraboloid z = 1 − x

Serggg [28]

I assume $C$ has counterclockwise orientation when viewed from above.

By Stokes' theorem,

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

so we first compute the curl:

$\vec F(x,y,z)=xy\,\vec\imath+yz\,\vec\jmath+xz\,\vec k$

$\implies\nabla\times\vec F(x,y,z)=-y\,\vec\imath-z\,\vec\jmath-x\,\vec k$

Then parameterize $S$ by

$\vec r(u,v)=\cos u\sin v\,\vec\imath+\sin u\sin v\,\vec\jmath+\cos^2v\,\vec k$

where the $z$ -component is obtained from

$1-(\cos u\sin v)^2-(\sin u\sin v)^2=1-\sin^2v=\cos^2v$

with $0\le u\le\dfrac\pi2$ and $0\le v\le\dfrac\pi2$ .

Take the normal vector to $S$ to be

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

Then the line integral is equal in value to the surface integral,

$\displaystyle\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

$=\displaystyle\int_0^{\pi/2}\int_0^{\pi/2}(-\sin u\sin v\,\vec\imath-\cos^2v\,\vec\jmath-\cos u\sin v\,\vec k)\cdot(\vec r_v\times\vec r_u)\,\mathrm du\,\mathrm dv$

$=\displaystyle-\int_0^{\pi/2}\int_0^{\pi/2}\cos v\sin^2v(\cos u+2\cos^2v\sin u+\sin(2u)\sin v)\,\mathrm du\,\mathrm dv=\boxed{-\frac{17}{20}}$

6 0

2 years ago

Give the domain and range. domain: {2, 0, 2}, range: {1, 0, 1}

Leona [35]

<h2>Domain: {0, 2}</h2><h2>Range: {0, 1, 1}</h2><h2 /><h3>When dealing with the domain, remember that no numbers can repeat in the domain so if there is more than one of the same number, we only list it one.</h3><h3 /><h3>When dealing with the range, it's ok if numbers repeat because it's the y-value not the x-value.</h3><h3 /><h3 />

5 0

3 years ago

A new catalyst is being investigated for use in the production of a plastic chemical. Ten batches of the chemical are produced.

zzz [600]

Answer:

$CI=(66.54,78.46)$

Step-by-step explanation:

Given : A new catalyst is being investigated for use in the production of a plastic chemical. Ten batches of the chemical are produced. The mean yield of the 10 batches is 72.5% and the standard deviation is 5.8%. Assume the yields are independent and approximately normally distributed.

To find : A 99% confidence interval for the mean yield when the new catalyst is used ?

Solution :

Let X be the yield of the batches.

We have given, n=10 , $\bar{X}=72.5\%$ , s=5.8%