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