Answer:
91.4
Step-by-step explanation:
Answer:
Yes-
Step-by-step explanation:
It repeats .121212 infinitely... I assume. You didn't say if it repeated or not.
the first one is <u>A</u>, because you divide both numbers by two, and make sure they are in the right order.
the answer to the second one is <u>2/3</u>, because 28 divided by 42 is .66666
Parameterize
in cylindrical coordinates by
![\vec s(u,v)=x(u,v)\,\vec\imath+y(u,v)\,\vec\jmath+z(u,v)\,\vec k](https://tex.z-dn.net/?f=%5Cvec%20s%28u%2Cv%29%3Dx%28u%2Cv%29%5C%2C%5Cvec%5Cimath%2By%28u%2Cv%29%5C%2C%5Cvec%5Cjmath%2Bz%28u%2Cv%29%5C%2C%5Cvec%20k)
where
![\begin{cases}x(u,v)=u\cos v\\y(u,v)=x(u,v)^2+z(u,v)^2=u^2\\z(u,v)=u\sin v\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7Dx%28u%2Cv%29%3Du%5Ccos%20v%5C%5Cy%28u%2Cv%29%3Dx%28u%2Cv%29%5E2%2Bz%28u%2Cv%29%5E2%3Du%5E2%5C%5Cz%28u%2Cv%29%3Du%5Csin%20v%5Cend%7Bcases%7D)
with
and
.
Take the normal vector to
to be
![\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}=-2u^2\cos v\,\vec\imath+u\,\vec\jmath-2u^2\sin v\,\vec k](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cpartial%5Cvec%20s%7D%7B%5Cpartial%20v%7D%5Ctimes%5Cdfrac%7B%5Cpartial%5Cvec%20s%7D%7B%5Cpartial%20u%7D%3D-2u%5E2%5Ccos%20v%5C%2C%5Cvec%5Cimath%2Bu%5C%2C%5Cvec%5Cjmath-2u%5E2%5Csin%20v%5C%2C%5Cvec%20k)
Then the flux of
across
is
![\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^2(3u^2\,\vec\imath+4\,\vec\jmath-3u^2\cos v\sin v\,\vec k)\cdot\left(\frac{\partial\vec s}{\partial v}\times\frac{\partial\vec s}{\partial u}\right)\,\mathrm du\,\mathrm dv](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Ciint_S%5Cvec%20F%5Ccdot%5Cmathrm%20d%5Cvec%20S%3D%5Cint_0%5E%7B2%5Cpi%7D%5Cint_0%5E2%283u%5E2%5C%2C%5Cvec%5Cimath%2B4%5C%2C%5Cvec%5Cjmath-3u%5E2%5Ccos%20v%5Csin%20v%5C%2C%5Cvec%20k%29%5Ccdot%5Cleft%28%5Cfrac%7B%5Cpartial%5Cvec%20s%7D%7B%5Cpartial%20v%7D%5Ctimes%5Cfrac%7B%5Cpartial%5Cvec%20s%7D%7B%5Cpartial%20u%7D%5Cright%29%5C%2C%5Cmathrm%20du%5C%2C%5Cmathrm%20dv)
![=\displaystyle\int_0^{2\pi}\int_0^2(4u-6u^4\cos^3v)\,\mathrm du\,\mathrm dv=\boxed{16\pi}](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle%5Cint_0%5E%7B2%5Cpi%7D%5Cint_0%5E2%284u-6u%5E4%5Ccos%5E3v%29%5C%2C%5Cmathrm%20du%5C%2C%5Cmathrm%20dv%3D%5Cboxed%7B16%5Cpi%7D)