Define the set $ S:=\{(x,y,z)\in \mathbb R^{3}:-3<z<1,x^{2}+y^{2}=36\} $ Find the integral $$ \iiint_{S} xyz\; dx dy dz. $$
I am struggling with the equality $x^{2}+y^{2}=36$; I have only encountered inequalities until now: $x^{2}+y^{2}\leq36$ would be easy, as
$$ \int_{-3}^{1}\int_{0}^{6}\int_{-\sqrt{36-y^2}}^{\sqrt{36-y^2}}(xyz)\;dxdydz $$
(I hope this is correct for the inequality case) But how can I treat equality $x^{2}+y^{2}=36$?