The question in my book is:
Find the volume of the solid enclosed by the surface z = $x\sec^2y$ and the planes $z = 0, \, x = 0, \, x = 2, \, y = 0, \, y = \frac{\pi}{4}$.
I understand how to solve the problem: $\int _0^2\int _0^{\pi/4}x\sec^2y\:dy\, dx$ which comes out to $2$.
My question is, what was the point in telling us that the solid was enclosed by the the plane $z = 0$. Why did it include that? What was I supposed to deduce from that?
Your integral is actually
$$\int _0^2\int _0^{\frac{\pi }{4}}xsec^2y\color{red}{-(0)}\:dydx\:\:\:$$
Suppose it is bound by $z=-1$ instead, it would become
$$\int _0^2\int _0^{\frac{\pi }{4}}xsec^2y\color{red}{-(-1)}\:dydx\:\:\:$$
If it is not specified, the interested region is not bounded.