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Calculate the area of the sphere $x^2+y^2+z^2=2Rz$ where $R>0$ inside of the cone $x^2+y^2=z^2$ where $z \geq 0$

Attempt

First we should use the elemental area formula given by $$\int_{S}f \dot dS=\int_{D} ||T_{u} \times T_{v}|| dudv$$ where $T_{u}$ and $T_{v}$ are the tangent vectors to the surface $S$ which is the surface parametrized .

Notice that these area is the upper semi sphere, since the sphere has center $C=(0,0,R)$ and radius $R$ and it is given by $$x^2+y^2+(z-R)^2=R^2$$ we only need to parametrize the sphere and get the upper area.

Let the parametrization $$x=R\cos \theta \sin \phi$$ $$y=R\sin\theta \sin \phi$$ $$z=R \cos \phi$$ notice that the upper sphere begin in $z=R$ this is one projection of the cone $x^2+y^2=z^2$ in $XY$ plane. since $$R\cos \phi=\sqrt{(R\cos \theta \sin \phi)^2+(R\sin\theta \sin \phi)^2}$$ $$R\cos \phi=\sqrt{R^2\cos^2 \theta \sin^2 \phi+R^2\sin^2\theta \sin^2 \phi}$$ $$Rcos \phi=\sqrt{R^2sin^2\phi}$$ $$Rcos \phi=R\sin\phi$$ $$\cos \phi= \sin \phi$$ since $\phi\in [0, \pi]$ we deduce that $0 \leq \phi \leq \frac{\pi}{4}$.

Now for calculating the tangent vectors to the surface we get $$T_{\theta}=(-R\sin \theta \sin \phi)i+(R\sin \phi \cos \theta )j+0k$$ $$T_{\phi}=(R\cos \phi \cos \theta )i+(R\cos \phi \sin \theta)j+(-R \sin \phi)k$$ Now calculating the cross product $$T_{\theta} \times T_{\phi}=(-R^2\sin^2 \phi \cos \theta)i+(-R^2\sin \phi \sin \theta)j+(-R^2 \cos \phi \sin \phi)k$$ $$||T_{\theta} \times T_{\phi}||=R^2\sin \phi$$ Finally applying the formula of the elemental área we get $$\int_{0}^{2\pi} \int_{0}^{\pi/4}R^2 \sin \phi d\phi d\theta=\pi R^2(2-\sqrt{2})$$

Is my answer right or have I done a mistake with the interpretation of the problem.

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    $\begingroup$ Shouldn't the parametrization be $z=R+R\cos(\phi)$, it seems like you're using a sphere about $(0,0,0)$ instead of $(0,0,R)$. $\endgroup$ Commented Nov 12, 2020 at 17:10

1 Answer 1

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Please note this is a sphere with center at $(0, 0, R)$ and radius $R$.

So the correct parametrization is $ \, \rho = 2R \cos \phi$

$x = \rho \cos \theta \sin \phi = 2R \cos \theta \sin \phi \cos \phi = R \cos \theta \sin2\phi$

Similarly,

$y = R \sin \theta \sin2\phi$

$z = \rho \cos \phi = 2R \cos^2\phi$

$T_{\theta}=(-R\sin \theta \sin 2\phi, R\cos \theta \sin 2\phi,0)$

$T_{\phi} = (2R\cos \theta \cos 2\phi, 2R\sin \theta \cos 2\phi, -2R \sin 2\phi)$

You can now find $|T_{\theta} \times T_{\phi}|$.

You should see the integral become -

$\displaystyle \int_{0}^{2\pi} \int_{0}^{\pi/4} 2R^2 \sin 2\phi \, d\phi \, d\theta$

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  • $\begingroup$ I get a mistake parametrizing the surface, thanks so much for you lastest advice in my other post $\endgroup$
    – user795628
    Commented Nov 12, 2020 at 18:04
  • $\begingroup$ You are welcome! $\endgroup$
    – Math Lover
    Commented Nov 12, 2020 at 18:06

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