Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\overline{C}$

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

1 Answer 1

Using the parametric equivalents

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

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



Can you take it from there?

OK guys - try this:


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

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


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

share|cite|improve this answer
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. :) – Ted Shifrin Jun 7 '13 at 0:45
@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. – Dale M Jun 7 '13 at 1:35
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. – Ted Shifrin Jun 7 '13 at 1:46
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? – Shuhao Cao Jun 7 '13 at 1:52
I think that the surface $\overline{C}$ is similar to but I'm not sure.. – Sebastián Molina Jun 7 '13 at 2:20

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.