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**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.

AtKin
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    The question should state the lower bound of $z$ otherwise $z$ is not bound below and the volume is not finite. – Math Lover Apr 03 '22 at 02:48

1 Answers1

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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.

inavda
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    That's what I thought. However, I'm having trouble seeing how $128-4x^2-y^2$ gives the distance between $z$ and 0? – AtKin Apr 03 '22 at 03:13
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    If $z$ takes the value $128-4x^2-y^2$, then $z-0 = 128-4x^2-y^2-0 = 128-4x^2-y^2$. I should say the distance between the value of $z$ at the top of the region and $0$. – inavda Apr 03 '22 at 03:16
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    Can I argue that since $0\le x \le 4$ and $0\le y\le 4 \therefore 48\le z \le 128?$ – AtKin Apr 03 '22 at 03:38
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    That is correct for the surface - $z$ has its maximum value $128$ at $(0,0)$ and its minimum value $48$ on the quarter circle $x^2+y^2=4$. But the actual region is unbounded below since the problem never specified a bottom plane as a boundary. The value of $z$ can go to $-\infty$ in the region, but on the surface itself it can only go as low as $48$. – inavda Apr 03 '22 at 03:45
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    Now, I understand. Thank you for the quick responses. – AtKin Apr 03 '22 at 03:47
  • no problem - happy to help :) – inavda Apr 03 '22 at 03:50