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Be the sets: $$C:= \lbrace (x,y,0)\in\mathbb{R}^{3}: (x-1)^2+y^2=1\rbrace$$ $$C':= \lbrace (x,0,z)\in\mathbb{R}^{3}: (x+1)^2+z^2=1\rbrace $$ $$\overline{C}= \lbrace tx+(1-t)x': x\in C, x' \in C', t\in [0,1]\rbrace$$

Calculate the volume of $\overline{C}$. I drawed the sets $C$ and $C'$, but I can't see how is the set $\overline{C}$

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  • $\begingroup$ How did you start? $\endgroup$ – Code-Guru Jun 6 '13 at 22:48
  • $\begingroup$ @Code-Guru I drawed the surface $\overline{C}$. I think that $\overline{C}$ is the convex hull.. $\endgroup$ – El Peluca Sapeeee Jun 7 '13 at 2:12
  • $\begingroup$ This is challenging, indeed. Was this a homework problem a mean teacher came up with? My students think I'm mean :D $\endgroup$ – Ted Shifrin Jun 7 '13 at 3:50
  • $\begingroup$ @TedShifrin Well, the truth is that this was my 3rd test of the semester, with another two questions, and I had only 3 hours... :( $\endgroup$ – El Peluca Sapeeee Jun 7 '13 at 4:47
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Using the parametric equivalents

$$C:=\lbrace (\cos(\theta_1)+1,\sin(\theta_1),0)\rbrace$$

$$C':=\lbrace (\cos(\theta_2)-1,0,\sin(\theta_2))\rbrace$$

therefore

$$\overline{C}:=\lbrace(t\cos(\theta_1)+t+(1-t)\cos(\theta_2)-(1-t),t\sin(\theta_1),(1-t)\sin(\theta_2))\rbrace$$

Can you take it from there?

OK guys - try this:

Let

$$y=ty', y' \in [-1,1]$$

$$z=(1-t)z', z'\in [-1,1]$$

Therefore

$$x=t(\cos(\pm \arcsin(y'))+1)+(1-t)(\cos(\pm\arcsin(z'))-1)$$

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    $\begingroup$ I got here, too. Remember that when you apply the Change of Variables formula (which I assume the OP is expected to use), you need to take the absolute value of the Jacobian determinant. Doing the $\theta_1\theta_2$ integral is sufficiently tricky that Mathematica can't do it, and neither can I. :) $\endgroup$ – Ted Shifrin Jun 7 '13 at 0:45
  • $\begingroup$ @TedShifrin Surely this can be simplified by observing that for any value of $t$, $cos(\alpha)$ takes all the values from $-1\dots 1$? You only then need to consider the limiting cases. $\endgroup$ – Dale M Jun 7 '13 at 1:35
  • $\begingroup$ Try working it out. Something funny happens when $(\cos\theta_1-1/2)(\cos\theta_2+1/2)=-1/4$. There's gotta be a better way. $\endgroup$ – Ted Shifrin Jun 7 '13 at 1:46
  • $\begingroup$ Employing the usual integration in $t$, $\theta_1$, and $\theta_2$ is a bit hard, I tried a little but stuck too. Could you please show a little bit more? $\endgroup$ – Shuhao Cao Jun 7 '13 at 1:52
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    $\begingroup$ @TedShifrin I know isn't the same, is just an idea, but I really don't know how to do it. Yes! sorry, is a solid. $\endgroup$ – El Peluca Sapeeee Jun 7 '13 at 3:16

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