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I'm having some trouble figuring out how to compute this answer. The problem is as follows:

The plane

$\frac{x}{\sqrt 2}$ $+$ $y + \frac {z}{\sqrt 2}$ $ = 1$

cuts a unit sphere centered at the origin into 2 pieces. Find the volume of both parts.

I tried to solve this problem using cylindrical coordinates and a triple integral, but the issue is I don't know the projection of the plane intersection on the xy plane so it is difficult to see what r goes from. My $\theta$ goes from $0$ to $2\pi$ and z is the difference of the unit sphere minus the plane equation.

Quanto
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Abhimanyu Deshpande
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1 Answers1

2

Note that distance from origin to the plane is

$$\frac{|0\cdot\frac1{\sqrt2}+0\cdot1+ 0\cdot\frac1{\sqrt2}-1|}{\sqrt{\frac12+1+\frac12}}=\frac1{\sqrt2}$$

So, equivalently, the unit sphere $x^2+y^2 + z^2 = 1$ can be viewed as being cut by the plane $z = \frac1{\sqrt2}$. The volume of the portion above the plane is then given by,

$$V_1 = 2\pi\int_0^{\pi/4}\int_{\frac1{\sqrt2\cos\theta}}^1 r^2\sin\theta \>drd\theta =\frac{2\pi}3\int_0^{\pi/4}\sin\theta \left( 1-\frac1{2\sqrt2\cos^3\theta}\right) d\theta=\frac\pi{12}(8-5\sqrt2)$$

The volume of the portion below the plane is,

$$V_2 = \frac43\pi - V_1 =\frac\pi{12}(8+5\sqrt2) $$

Quanto
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