let C be the eleptic cylinder C = {$\frac{y^2}{4} + z^2\leq 1$}.
let D be the cylinder {$x^2 + z^2 \leq 1$} find the volume of intersection of C and D
this is what i did : $$V = \ \int_{z=-1}^{z=1}\int_{y=-\sqrt{4-4z^2}}^{y=\sqrt{4-4z^2}}\int_{x=-\sqrt{1-z^2}}^{x=\sqrt{1-z^2}} 1{ \ dx}{\ dy}{\ dz} $$
$$= \ \int_{z=-1}^{z=1}\int_{y=-\sqrt{4-4z^2}}^{y=\sqrt{4-4z^2}}2\sqrt{1-z^2}{\ dy}{\ dz}=$$ $$= \ \int_{z=-1}^{z=1}2\sqrt{1-z^2}*2\sqrt{4-4z^2} {\ dz}= \int_{z=-1}^{z=1}4\sqrt{4z^4-8z^2+4} {\ dz}=\int_{z=-1}^{z=1}4\sqrt{(2z^2 -2)^2} {\ dz} = \ 10\frac{2}{3}$$
why did i get negative volume ? did i do it correctly ? or am i missing something
You are correct up to the last integral. Recall that $\sqrt{z^2}=|z|,$ so $\sqrt{(z^2-1)^2}=|z^2-1|$ When $-1\le z\le1,$ we have $|z^2-1|=1-z^2,$ so you've got the sign wrong.