Mathematics: Find the surface area of the part of the sphere x2+y2+z2=81 that lies above the cone z=x2+y2−−−−−−√

Find the surface area of the part of the sphere x2+y2+z2=81

Ответ:

kukisbae

13.03.2022

Parameterize the part of the sphere by

$\mathbf s(u,v)=(9\cos u\sin v,9\sin u\sin v,9\cos v)$

with $0\le u\le2\pi$ and $0\le v\le\dfrac\pi4$ . Then

$\mathrm dS=\|\mathbf s_u\times\mathbf s_v\|=81\sin v\,\mathrm du\,\mathrm dv$

So the area of

(the part of the sphere above the cone) is given by

$\displaystyle\iint_S\mathrm dS=81\int_{v=0}^{v=\pi/4}\int_{u=0}^{u=2\pi}\sin v\,\mathrm du\,\mathrm dv=81(2\pi)\left(1-\dfrac1{\sqrt2}\right)=(162-81\sqrt2)\pi$

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Ответ:

ronniethefun

06.06.2021

Mathematics

Volume of a cylinder = πr²h

Volume of the cylinder = π x 6² x 4

Volume of the cylinder = 452.4 in³ (nearest tenth)

Answer : Volume = 452.4 in³

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Find the surface area of the part of the sphere x2+y2+z2=81 that lies above the cone z=x2+y2−−−−−−√

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