Calculus: Finding Volume with Triple Integrals

Question

**Problem:**A shape is bounded by the following elliptical function $4x^2 + y^2 +z = 128$ and the planes $x=0, x=4, y=0, y=4$. Find the volume of the shape.
My attempt:
$4x^2 + y^2 +z = 128 \implies z = 128-4x^2-y^2$, then $$V=\int_{0}^{4}\int_{0}^{4}128-4x^2-y^2\,dydx=\frac{4864}{3}$$ Now, I want to do this with a triple-integral. My problem is with the lower bound for $z$.My initial thought would be $z=48$, where $x=4 \land y=4$, but I got a different answer.Through trial and error, I got $$V=\int_{0}^{4}\int_{0}^{4}\int_{\color{red}{0}}^{128-4x^2-y^2}dzdydx=\frac{4864}{3}$$ Question: Why is the lower bound $z=0$ (in red)? The elliptical function $z = 128-4x^2-y^2$ is unbounded below.

Answer 1

The comment by Math Lover is correct.

In the first attempt, even though you don't explicitly set a lower bound on $z$ of $0$, you are implicitly doing so with the integrand: the function $128-4x^2-y^2$ gives the distance between the value of $z$ at the top of the region and $0$. The volume is unbounded unless we add the plane $z=0$ to the problem conditions.