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irga5000 [103]
3 years ago
6

Use spherical coordinates to find the volume of the region that lies outside the cone z = p x 2 + y 2 but inside the sphere x 2

+ y 2 + z 2 = 2. Write the answer as an exact answer, which should involve π and √ 2. Do not round or use a calculator.

Mathematics
1 answer:
Harman [31]3 years ago
4 0

I assume the cone has equation z=\sqrt{x^2+y^2} (i.e. the upper half of the infinite cone given by z^2=x^2+y^2). Take

\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\varphi\end{cases}\implies\mathrm dx\,\mathrm dy\,\mathrm dz=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi

The volume of the described region (call it R) is

\displaystyle\iiint_R\mathrm dx\,\mathrm dy\,\mathrm dz=\int_0^{2\pi}\int_0^{\sqrt2}\int_{\pi/4}^\pi\rho^2\sin\varphi\,\mathrm d\varphi\,\mathrm d\rho\,\mathrm d\theta

The limits on \theta and \rho should be obvious. The lower limit on \varphi is obtained by first determining the intersection of the cone and sphere lies in the cylinder x^2+y^2=1. The distance between the central axis of the cone and this intersection is 1. The sphere has radius \sqrt2. Then \varphi satisfies

\sin\varphi=\dfrac1{\sqrt2}\implies\varphi=\dfrac\pi4

(I've added a picture to better demonstrate this)

Computing the integral is trivial. We have

\displaystyle2\pi\left(\int_0^{\sqrt2}\rho^2\,\mathrm d\rho\right)\left(\int_{\pi/4}^\pi\sin\varphi\,\mathrm d\varphi\right)=\boxed{\frac43(1+\sqrt2)\pi}

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