# 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 $z \geq 0$

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.

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

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 -

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

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