im pretty sure true because they do not intersect and are not parallel :)
![\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](https://tex.z-dn.net/?f=%5Cmathbf%20f%28x%2Cy%2Cz%29%3D%5Clangle%20z%2Cy%2Cx%5Crangle%5Cimplies%5Cnabla%5Ccdot%5Cmathbf%20f%3D%5Cdfrac%7B%5Cpartial%20z%7D%7B%5Cpartial%20x%7D%2B%5Cdfrac%7B%5Cpartial%20y%7D%7B%5Cpartial%20y%7D%2B%5Cdfrac%7B%5Cpartial%20x%7D%7B%5Cpartial%20z%7D%3D0%2B1%2B0%3D1)
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](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Ciiint_E%5Cnabla%5Ccdot%5Cmathbf%20f%28x%2Cy%2Cz%29%5C%2C%5Cmathrm%20dV%3D%5Cint_%7B%5Cvarphi%3D0%7D%5E%7B%5Cvarphi%3D%5Cpi%7D%5Cint_%7B%5Ctheta%3D0%7D%5E%7B%5Ctheta%3D2%5Cpi%7D%5Cint_%7B%5Crho%3D0%7D%5E%7B%5Crho%3D6%7D%5Crho%5E2%5Csin%5Cvarphi%5C%2C%5Cmathrm%20d%5Crho%5C%2C%5Cmathrm%20d%5Ctheta%5C%2C%5Cmathrm%20d%5Cvarphi%3D288%5Cpi)
On the other hand, we can parameterize the boundary of
![E](https://tex.z-dn.net/?f=E)
by
![\mathbf s(u,v)=\langle6\cos u\sin v,6\sin u\sin v,6\cos v\rangle](https://tex.z-dn.net/?f=%5Cmathbf%20s%28u%2Cv%29%3D%5Clangle6%5Ccos%20u%5Csin%20v%2C6%5Csin%20u%5Csin%20v%2C6%5Ccos%20v%5Crangle)
with
![0\le u\le2\pi](https://tex.z-dn.net/?f=0%5Cle%20u%5Cle2%5Cpi)
and
![0\le v\le\pi](https://tex.z-dn.net/?f=0%5Cle%20v%5Cle%5Cpi)
. 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](https://tex.z-dn.net/?f=%5Cmathrm%20d%5Cmathbf%20S%3D%5Cmathbf%20n%5C%2C%5Cmathrm%20dS%3D%5Cdfrac%7B%5Cmathbf%20s_v%5Ctimes%5Cmathbf%20s_u%7D%7B%5C%7C%5Cmathbf%20s_v%5Ctimes%5Cmathbf%20s_u%5C%7C%7D%5C%7C%5Cmathbf%20s_v%5Ctimes%5Cmathbf%20s_u%5C%7C%5C%2C%5Cmathrm%20du%5C%2C%5Cmathrm%20dv)
![\mathrm d\mathbf S=\mathbf s_v\times\mathbf s_u\,\mathrm du\,\mathrm dv](https://tex.z-dn.net/?f=%5Cmathrm%20d%5Cmathbf%20S%3D%5Cmathbf%20s_v%5Ctimes%5Cmathbf%20s_u%5C%2C%5Cmathrm%20du%5C%2C%5Cmathrm%20dv)
![\mathrm d\mathbf S=36\langle\cos u\sin^2v,\sin u\sin^2v,\sin v\cos v\rangle\,\mathrm du\,\mathrm dv](https://tex.z-dn.net/?f=%5Cmathrm%20d%5Cmathbf%20S%3D36%5Clangle%5Ccos%20u%5Csin%5E2v%2C%5Csin%20u%5Csin%5E2v%2C%5Csin%20v%5Ccos%20v%5Crangle%5C%2C%5Cmathrm%20du%5C%2C%5Cmathrm%20dv)
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](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Ciint_%7B%5Cpartial%20E%7D%5Cmathbf%20f%5Ccdot%5Cmathrm%20d%5Cmathbf%20S%3D36%5Cint_%7Bv%3D0%7D%5E%7Bv%3D%5Cpi%7D%5Cint_%7Bu%3D0%7D%5E%7Bu%3D2%5Cpi%7D%5Cmathbf%20f%28x%28u%2Cv%29%2Cy%28u%2Cv%29%2Cz%28u%2Cv%29%29%5Ccdot%28%5Cmathbf%20s_v%5Ctimes%5Cmathbf%20s_u%29%5C%2C%5Cmathrm%20du%5C%2C%5Cmathrm%20dv)
![=\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](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle36%5Cint_%7Bv%3D0%7D%5E%7Bv%3D%5Cpi%7D%5Cint_%7Bu%3D0%7D%5E%7Bu%3D2%5Cpi%7D%2812%5Ccos%20u%5Ccos%20v%5Csin%5E2v%2B6%5Csin%5E2u%5Csin%5E3v%29%5C%2C%5Cmathrm%20du%5C%2C%5Cmathrm%20dv%3D288%5Cpi)
as required.
You can scan it on brainly, It’ll help you
9514 1404 393
Answer:
D
Step-by-step explanation:
The leading term is of odd degree and negative coefficient. The odd degree tells you the end behaviors will have opposite signs. The negative coefficient tells you the left end behavior will be positive, and the right end behavior will be negative.
![\textbf{D. }\lim\limits_{x\to -\infty}f(x)=\infty\quad\lim\limits_{x\to \infty}f(x)=-\infty](https://tex.z-dn.net/?f=%5Ctextbf%7BD.%20%20%7D%5Clim%5Climits_%7Bx%5Cto%20-%5Cinfty%7Df%28x%29%3D%5Cinfty%5Cquad%5Clim%5Climits_%7Bx%5Cto%20%5Cinfty%7Df%28x%29%3D-%5Cinfty)
The answer for this question is B because - -e/-f is the same as - -e/f