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I want the volume of the sphere $x^2 + y^2 + z^2 = 4$ from $x = 0$ to $x = 4-y^2$. The integral that gives this volume is

$$\int\limits_{-2}^2 \int\limits_0^{4-y^2} \int\limits_{-\sqrt{4-x^2-y^2}}^{\sqrt{4-x^2-y^2}}\ 1\ dz\ dx\ dy$$

I don't find $T$ such that $T\bar{u} = \bar{x}$ for every $\bar{x}\in D$ where $D$ is our desired volume.

I mean, I'm trying to find $D^*$ such that for every $\bar{u} \in D^*$, $T$ is a change of variables for our problem.

So, I'm trying the find the volume between the regions $x^2 + y^2 + z^2 = 4$ and $x = 4-y^2$ but I can neither find that scalar.

Any ideas please.

David G. Stork
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Blue Tomato
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1 Answers1

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Nobody commented on my scalar in the comments section so far.

The projection of the solid on the $xy$-plane is determined by curves, $x^2+y^2=4$, $x=4-y^2$ and $x=0$. The common solution of the circle and the parabola gives $(1,\pm \sqrt{3})$. Hence the quarter volume integral due to symmetries w.r.t $xy$-plane and $x$-axis, is $$\frac{1}{4}V=\int_0^{\sqrt{3}}\int_0^{\sqrt{4-y^2}}\int_0^{\sqrt{4-x^2-y^2}}dzdxdy+\int_{\sqrt{3}}^2\int_0^{4-y^2}\int_0^{\sqrt{4-x^2-y^2}}dzdxdy$$ and $$\frac{1}{4}V=\int_0^{\sqrt{3}}\int_0^{\sqrt{4-y^2}}\sqrt{4-x^2-y^2}dxdy+\int_{\sqrt{3}}^2\int_0^{4-y^2}\sqrt{4-x^2-y^2}dxdy$$ and $$\frac{1}{4}V=\int_0^{\sqrt{3}}\frac{\pi}{4}(4-y^2)dy+\int_{\sqrt{3}}^2\left(\frac{x\sqrt{4-x^2-y^2}}{2}+\frac{(4-y^2)\arcsin(\frac{x}{\sqrt{4-y^2}})}{2}\right)\vert_0^{4-y^2}dy$$ and hence $V$ is $$V=3\sqrt{3}\pi+2\int_{\sqrt{3}}^{2}\sqrt{y^2-3}\sqrt{4-y^2}+(4-y^2)\arcsin(\sqrt{4-y^2})dy$$ By WolframAlpha, $V\approx 16.324+0.482=16.806$. The second integral which gives the tiny volume is not nice.

Bob Dobbs
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    Sorry about that. The tricky step in the original problem is the integral $$\int_{-2}^{2} \int_{0}^{4-y^2} 2\sqrt{4-x^2-y^2} dx dy$$ A formal and clearly way to solve that last integral is not trivial, at least not for me neither for WolframAlpha, so it is necessary to commit a change of variables transformation. I parameterized one little region with cylindrical coordinates. – Blue Tomato Nov 19 '22 at 15:22
  • @BlueTomato This integral?: https://www.wolframalpha.com/input?key=&i=int+%284-y%5E2%29arcsin%28sqrt%284-y%5E2%29%29 – Bob Dobbs Nov 19 '22 at 15:30
  • https://www.wolframalpha.com/input?key=&i2d=true&i=Integrate%5B2Sqrt%5B4-Power%5Bx%2C2%5D-Power%5By%2C2%5D%5D%2C%7Bx%2C0%2C4-Power%5By%2C2%5D%7D%5D

    https://www.wolframalpha.com/input?key=&i2d=true&i=Integrate%5B2Sqrt%5B4-Power%5Bx%2C2%5D-Power%5By%2C2%5D%5D%2Cx%5D

    The complete proof in the second link, even by Wolfram, is not short nor elegant as desired as with the change of variables transformation. Maybe I'm wrong. Let me know if so.

     – Blue Tomato Nov 19 '22 at 15:55
  • Second link... You can do this by hand. Problem starts after that. – Bob Dobbs Nov 19 '22 at 16:03