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the professor of our integration class gave us the following exercise as an assignment:

Determine the measure in $\mathbb{R}^3$ of $A=\{(x,y,z) \in \mathbb{R}^3 ,x^2+y^2+z^2-2x+2z\leq0\leq x^2-y^2-z^2 \}$

That is, the volumen resulting of the solid bounded by the sphere of center $(1,0,-1)$ and radius $\sqrt{2}$ and the cone $x^2=y^2+z^2$. Due to the lack of spherical/cylindrical symmetry (at least at a first glance), I have tried to come up with a solution using cartesian coordinates.

First of all, plotting the section for $y=0$, it can be geometrically argued that $x \in [0,1+\sqrt{2}]$. The following idea is to study the transversal sections for each value of $x \in [0,1+\sqrt{2}]$. Fix $x \in [0,1+\sqrt{2}]$, then the resulting section is the surface bounded by the equations $y^2+(z+1)^2=2-(z-1)^2$ and $y^2+z^2=x^2$. Solving this system of equations on $(y,z)$ we deduce that the solution exists for $0\leq x \leq 2$ and for each such $x$ there are exactly two solutions: $(-\sqrt{2x^3-x^4},x-x^2),(\sqrt{2x^3-x^4},x-x^2)$.

Therefore, for a fixed $x \in [0,2]$ we have that $y \in [-\sqrt{2x^3-x^4},\sqrt{2x^3-x^4}]$ as the intersetion points of the circunferences are the extreme possible values for the $y$. Then, be must set the integration limits for $z$: it's easy to see that $z \in [-\sqrt{x^2-y^2},-1+\sqrt{2-(x-1)^2-y^2}]$ as it's bounded below by the cone and bounded above by the sphere.

On the other hand, for $x \in [2,1+\sqrt{2}]$ we have no intersections in the transversal sections. Therefore, the circle corresponding to a cut of the sphere is completely inside the cut corresponding the cone. Therefore, in this case we have to integrate for each $x$ a circle of radius $\sqrt{2-(x-1)^2}$. Summing it all up, we conclude that

$$ m_3(A) = \int_0^{2} dx \int_{-\sqrt{2x^3-x^4}}^{\sqrt{2x^3-x^4}}dy \int_{-\sqrt{x^2-y^2}}^{-1+\sqrt{2-(x-1)^2-y^2}} dz + \int_2^{1+\sqrt{2}} dx \pi (2-(x-1)^2) $$

I can't find any incongruence with my reasoning and my (limited) geometric intuition makes me think I'm correct. However, as the resulting integral seems to be hard to compute, makes me think that either there's an error in my procedure or that there's a simpler/straightforward way of handling the problem. Any suggestion or objection is always welcome.

Javier Herrero
  • 725

2 Answers2

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Sphere: $(x-1)^2 + y^2 + (z+1)^2 = 2$
Cone: $y^2 + z^2 = x^2$

Using spherical coordinates, $x = \rho \cos\phi, y = \rho \cos\theta \sin \phi, z = \rho \sin\theta \sin\phi$

Sphere: $~\rho = 2 (\cos\phi - \sin\theta \sin\phi)$
Cone: $ \phi = \frac{\pi}{4}$

You can see that the center of the sphere $(1, 0, -1)$ is on the cone and the plane $x = z~$ is tangent to the sphere at $(0, 0, 0)$. Also the line $x = z, y = 0~$ in the plane is on the cone.

So for all values of $~0 \leq \theta \leq 2 \pi~$ and $~ 0 \leq \phi \leq \pi/4$, $~\rho~$ is bound above by the sphere and the integral to find volume is,

$ \displaystyle \int_0^{\pi/4} \int_0^{2\pi} \int_0^{2 (\cos\phi - \sin \phi \sin \theta)} \rho^2 \sin \phi ~ d\rho ~ d\theta ~ d\phi = \frac{3 \pi}{2}$

Math Lover
  • 52,200
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My answer is more of an outline of a solution than a full solution.

The way I did it was to take cross sectional slices along the $x$-axis on two intervals. On $x \in [0,2]$, the cross section is the intersection of two circles, one from the sphere and one from the cone, and whose differential volume is expressible as $$dV_1 = \left( x^2 \cos^{-1} (x-1) + (1 + 2x - x^2) \cos^{-1} \frac{1 + x - x^2}{\sqrt{1 + 2x - x^2}} - x \sqrt{2x - x^2} \right) dx.$$ Then on the interval $x \in [2, 1+\sqrt{2}]$, the cross section is just the circle corresponding to the sphere, thus its differential volume is $$dV_2 = \pi (1 + 2x - x^2) \, dx.$$ Consequently, the total volume is $$V = \int_{x=0}^2 dV_1 + \int_{x=2}^{1+\sqrt{2}} dV_2 = \frac{8\sqrt{2} - 19}{6} \pi + \frac{4 \sqrt{2} - 5}{3} \pi = \frac{3\pi}{2}.$$

The result suggests that perhaps there is a more elegant method of solution than what is described here.

heropup
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