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I want to compute the volume given by: $$ D = \{ (x,y,z) \in \mathbb{R}^{3} : x^2 + y^2 + z^2 \leq 1, y^2 + z^2 \leq x^2 \} $$

Which acording to Geogebra looks like: Plot of D (Sorry guys, I tried to add some fancy plot of $ D $, but it looks like I am not allowed to show images).

In other words, I want to compute the 3D integral: $$ \int_{D} 1 dA $$

So, I used spherical coordinates and setup the integral in the following way:

$$ 2 \int_{0}^{1} \int_{0}^{\pi/2} \int_{0}^{\pi/2} r^2 \sin\theta \, d\varphi \, d\theta \, dr \tag{*} $$

Where $ 0 \leq \varphi \leq 2\pi $ and $ 0 \leq \theta \leq \pi $.

My explanation:

First, I reduce the problem to compute only one volume and duplicate it, to get the whole volume of $ D $. Then: the integral depeding of $ r $ is clear because of the sphere. The integral depending of $ \theta $ and the integral depending of $ \varphi $ have to have the same interval because the conic surface is simetric in $ y $ and $ z $ coordinates. At first, I tried to write it down just like: $$ 2 \int_{0}^{1} \int_{-\pi/4}^{\pi/4} \int_{-\pi/4}^{\pi/4} r^2 \sin\theta \, d\varphi \, d\theta \, dr $$ but then I remembered that both angles have to be positive. So I think that the volume of the function $ f(x,y,z) = 1 $ in $ [-\pi/4,\pi/4] $ is the same that in $ [0,\pi/2] $. Is it right? $ \tag{**} $

My questions:

  1. Is the integral $ (*) $ right? Does it really compute the volume of $ D $?

  2. Is my explanation correct?

  3. Is $ (**) $ true?

At the moment, I just want to know if everything is fine, so I can compute the value of the integral. But my dificulties are the ones I presented before, so I appreciate if you can answer my questions.

D'oh
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  • My initial thought: since $\phi$ in spherical coordinates is angle against the $z$-axis, it might prove pruductive to rotate the figure so that it is rotationally symmetric about the $z$-axis rather than the $x$-axis. – Arthur Jan 25 '16 at 11:49
  • @Arthur Let's think that I'm changing $ x $ to $ z $ and vice versa, seeing the cone just like $ x^2 + y^2 \leq z^2 $. If I understand what you are saying, it doesn't really affect the problem because the the function is $ f(x,y,z) = 1 $ and the variable changing doesn't modify anything at all in the function. I might be wrong. – D'oh Jan 25 '16 at 11:53
  • The thing is, the limits of your integration $(\text *)$ are not correct (it would be quite hard to set up the correct limits, actually, with lots of trigonometry involved). Once the figure is symmetric about the $z$-axis, it becomes much easier. – Arthur Jan 25 '16 at 11:55
  • @Arthur Oh, then I don't know how to solve this problem. How can I do that axis change that you mentioned? – D'oh Jan 25 '16 at 11:57
  • The axis change is done just the way you said, sorry. I was just telling you why $(\text *)$ was wrong. – Arthur Jan 25 '16 at 12:00
  • @Arthur But thinking $ x $ as $ z $ and vice versa doesn't make it simetric and all the stuff so I can work just the way I would if it really was $x^2 + y^2 \leq z^2$? How would you setup the integration limits of this problem? – D'oh Jan 25 '16 at 12:08
  • The limits of spherical integration for the whole of ${(x, y, z) \in \Bbb R^3 \mid x^2 + y^2 + z^2 \leq 1, x^2 + y^2 \leq z^2}$ is $$ \int_0^1 dr\int_0^{2\pi}d\theta \int_{\pi/4}^{3\pi/4}d\varphi $$ – Arthur Jan 25 '16 at 12:12

1 Answers1

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May what i did it isn't such a complex solutions as you are looking for, but i hope it helps. Let's analyze what is our domain $ D = \{ (x,y,z) \in \mathbb{R}^{3} : x^2 + y^2 + z^2 \leq 1, y^2 + z^2 \leq x^2 \}$, which gives us two restrictions,

1.-That's the sphere or ratio r=1 $$ x^2 + y^2 + z^2 \leq 1$$

And

2.-The cone $$ y^2 + z^2 \leq x^2 $$

On the other hand, we can compute a solid's volume by Solid of revolution th, $$ V=\pi\int_a^b f(x)^2 dx$$. The volume of the cone can be compute using the formula below taking $f(x)=x, a=-1, b=1$ .Furthermore we can compute the volumen of the sphere taking $g(x)=\sqrt{1-x^2}, a=-1, b=1$ (I can´t add photos to this anwer, but please try to draw this two functions). Now, let's calculate where these to functions take the same value, ie $$f(x)=g(x), x=\sqrt{1-x^2}$$ after some computations we obtain $$x=+1\sqrt{2}, x=-x=+1\sqrt{2}$$, Finally, considering the symmetry of the function, it's easy to see that que volume we are looking for is $$V=2(\pi\int_{0}^{1/\sqrt{2}} f(x)^2dx+\pi\int_{1/\sqrt{2}}^{1} g(x)^2dx);$$

$$$V=2(\pi\int_{0}^{1/\sqrt{2}}x^2dx+\pi\int_{1/\sqrt{2}}^{1} ( \sqrt{1-x^2} )^2dx)=2[\pi\frac{(1/\sqrt{2})^3}{3}+ \pi (1-\frac{1}{3})- \pi(\frac{1}{ \sqrt{2}}-\frac{(\frac{1}{\sqrt{2}})^3}{3})]=2[\frac{2\pi}{3}-\frac{\pi}{\sqrt{2}}]$$

Lucia
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