Let
![R](https://tex.z-dn.net/?f=R)
be the solid. Then the volume is
![\displaystyle\iiint_R\mathrm dV=\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=\sqrt8}\int_{\zeta=r^2}^{\zeta=\sqrt{72-r^2}}r\,\mathrm d\zeta\,\mathrm dr\,\mathrm d\theta](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Ciiint_R%5Cmathrm%20dV%3D%5Cint_%7B%5Ctheta%3D0%7D%5E%7B%5Ctheta%3D2%5Cpi%7D%5Cint_%7Br%3D0%7D%5E%7Br%3D%5Csqrt8%7D%5Cint_%7B%5Czeta%3Dr%5E2%7D%5E%7B%5Czeta%3D%5Csqrt%7B72-r%5E2%7D%7Dr%5C%2C%5Cmathrm%20d%5Czeta%5C%2C%5Cmathrm%20dr%5C%2C%5Cmathrm%20d%5Ctheta)
which follows from the facts that
![\begin{cases}x=r\cos\theta\\y=r\sin\theta\\z=\zeta\end{cases}\implies\mathrm dx\,\mathrm dy\,\mathrm dz=r\,\mathrm dr\,\mathrm d\theta\,\mathrm d\zeta](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7Dx%3Dr%5Ccos%5Ctheta%5C%5Cy%3Dr%5Csin%5Ctheta%5C%5Cz%3D%5Czeta%5Cend%7Bcases%7D%5Cimplies%5Cmathrm%20dx%5C%2C%5Cmathrm%20dy%5C%2C%5Cmathrm%20dz%3Dr%5C%2C%5Cmathrm%20dr%5C%2C%5Cmathrm%20d%5Ctheta%5C%2C%5Cmathrm%20d%5Czeta)
(by computing the Jacobian)
and
![z=x^2+y^2=r^2\implies z+z^2=72\implies z=-9\text{ or }z=8](https://tex.z-dn.net/?f=z%3Dx%5E2%2By%5E2%3Dr%5E2%5Cimplies%20z%2Bz%5E2%3D72%5Cimplies%20z%3D-9%5Ctext%7B%20or%20%7Dz%3D8)
(we take the positive solution, since it's clear that
![R](https://tex.z-dn.net/?f=R)
lies above the
![x](https://tex.z-dn.net/?f=x)
-
![y](https://tex.z-dn.net/?f=y)
plane)
![r^2+z^2=72\implies z=\pm\sqrt{72-r^2}](https://tex.z-dn.net/?f=r%5E2%2Bz%5E2%3D72%5Cimplies%20z%3D%5Cpm%5Csqrt%7B72-r%5E2%7D)
(again, taking the positive root for the same reason)
![z=r^2\implies 8=r^2\implies r=\sqrt8](https://tex.z-dn.net/?f=z%3Dr%5E2%5Cimplies%208%3Dr%5E2%5Cimplies%20r%3D%5Csqrt8)
![\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=\sqrt8}\int_{\zeta=r^2}^{\zeta=\sqrt{72-r^2}}r\,\mathrm d\zeta\,\mathrm dr\,\mathrm d\theta](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_%7B%5Ctheta%3D0%7D%5E%7B%5Ctheta%3D2%5Cpi%7D%5Cint_%7Br%3D0%7D%5E%7Br%3D%5Csqrt8%7D%5Cint_%7B%5Czeta%3Dr%5E2%7D%5E%7B%5Czeta%3D%5Csqrt%7B72-r%5E2%7D%7Dr%5C%2C%5Cmathrm%20d%5Czeta%5C%2C%5Cmathrm%20dr%5C%2C%5Cmathrm%20d%5Ctheta)
![=\displaystyle2\pi\int_{r=0}^{r=\sqrt8}\int_{\zeta=r^2}^{\zeta=\sqrt{72-r^2}}r\,\mathrm d\zeta\,\mathrm dr](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle2%5Cpi%5Cint_%7Br%3D0%7D%5E%7Br%3D%5Csqrt8%7D%5Cint_%7B%5Czeta%3Dr%5E2%7D%5E%7B%5Czeta%3D%5Csqrt%7B72-r%5E2%7D%7Dr%5C%2C%5Cmathrm%20d%5Czeta%5C%2C%5Cmathrm%20dr)
![=\displaystyle2\pi\int_{r=0}^{r=\sqrt8}r(\sqrt{72-r^2}-r^2)\,\mathrm dr](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle2%5Cpi%5Cint_%7Br%3D0%7D%5E%7Br%3D%5Csqrt8%7Dr%28%5Csqrt%7B72-r%5E2%7D-r%5E2%29%5C%2C%5Cmathrm%20dr)
![=\displaystyle2\pi\int_{r=0}^{r=\sqrt8}(r\sqrt{72-r^2}-r^3)\,\mathrm dr](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle2%5Cpi%5Cint_%7Br%3D0%7D%5E%7Br%3D%5Csqrt8%7D%28r%5Csqrt%7B72-r%5E2%7D-r%5E3%29%5C%2C%5Cmathrm%20dr)
According to the passage, the correct answer has become obsolete. The correct option is C.
In Given, a passage that is shown in the attached photos.
Option C is the best answer. "obsolete" states clearly and concisely: the idea that skills can become obsolete. The meaning of obsolete is no longer in use as it has been replaced by something new and better or more trendy.
Options other than C are A, B and D are incorrect because they are unclear or because they convey an inappropriate tone for passing.
The correct choice from the given passage is therefore obsolete.
Learn more about passage from here brainly.com/question/12555695
#SPJ4
Im sorry i dont see a question so Whats the question.
The answer is y=- 7/6x+8
Plz mark me Brainly
Answer: second Congress
Step-by-step explanation:
Short and sweet