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Rina8888 [55]
3 years ago
6

Which one ????????????

Mathematics
1 answer:
gregori [183]3 years ago
5 0
B is your answer because if u look at it the
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We need to see the "given system".

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Find the surface area of x^2+y^2+z^2=9 that lies above the cone z= sqrt(x^@+y^2)
Mashcka [7]
The cone equation gives

z=\sqrt{x^2+y^2}\implies z^2=x^2+y^2

which means that the intersection of the cone and sphere occurs at

x^2+y^2+(x^2+y^2)=9\implies x^2+y^2=\dfrac92

i.e. along the vertical cylinder of radius \dfrac3{\sqrt2} when z=\dfrac3{\sqrt2}.

We can parameterize the spherical cap in spherical coordinates by

\mathbf r(\theta,\varphi)=\langle3\cos\theta\sin\varphi,3\sin\theta\sin\varphi,3\cos\varphi\right\rangle

where 0\le\theta\le2\pi and 0\le\varphi\le\dfrac\pi4, which follows from the fact that the radius of the sphere is 3 and the height at which the sphere and cone intersect is \dfrac3{\sqrt2}. So the angle between the vertical line through the origin and any line through the origin normal to the sphere along the cone's surface is

\varphi=\cos^{-1}\left(\dfrac{\frac3{\sqrt2}}3\right)=\cos^{-1}\left(\dfrac1{\sqrt2}\right)=\dfrac\pi4

Now the surface area of the cap is given by the surface integral,

\displaystyle\iint_{\text{cap}}\mathrm dS=\int_{\theta=0}^{\theta=2\pi}\int_{\varphi=0}^{\varphi=\pi/4}\|\mathbf r_u\times\mathbf r_v\|\,\mathrm dv\,\mathrm du
=\displaystyle\int_{u=0}^{u=2\pi}\int_{\varphi=0}^{\varphi=\pi/4}9\sin v\,\mathrm dv\,\mathrm du
=-18\pi\cos v\bigg|_{v=0}^{v=\pi/4}
=18\pi\left(1-\dfrac1{\sqrt2}\right)
=9(2-\sqrt2)\pi
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3 years ago
What is value of x? Please help! Will reward
Andre45 [30]
X = 50
.................
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3 years ago
Help me please!!!!!!!
Cerrena [4.2K]

Answer:

I have no idea.

Step-by-step explanation:

I failed geometry.

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3 years ago
Reconocimiento de suceciones aritmeticas y geometricas
Monica [59]

Answer:

el producto es 4 o 3 por que dice hallar el producto de CUATRO termino estonses 3×4= 12

Step-by-step explanation:

si necesita ayuda o necesita responder sus preguntas en cuestión de segundos o minutos, hágamelo saber y lo ayudaré :)

7 0
3 years ago
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