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I am given the following multivariate calculus question:

Integrate $f(x,y,z)=14xz$ over the region in the first octant $(x,y,z \ge 0)$ above the parabolic cylinder $z=y^2$ and below the paraboloid $z=8-2x^2-y^2$.

Wikipedia has multiple definitions for octant. Some of which are on circles. It would seem to me the easiest version of this question is when octant refers to the cubic region ($x\ge0, y\ge0, z\ge0$), which is also what is explicitly stated on the question.

from the 2 equations I can get:

$2y^2\le8-2x^2-y^2 \iff y \le \sqrt{4-x^2}$ (since $y\ge0$)

And that gives me a bounbdary for $y$.

but how can I find a boundary for $x$?

The only boundary I can come up with would be $0$ to infinity.

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  • $\begingroup$ The first octant refers to the portion of the xyz plane for which $x$, $y$, and $z$ are all positive. (And perhaps the last octant is the portion of the xyz plane for which $x$, $y$, and $z$ are all negative.) As far as I know, there's no standard convention for naming the octants other than the first and last. In other words, if I say "look at the second octant," one can't actually be sure of what I'm talking about. $\endgroup$ – Tiwa Aina Nov 17 '17 at 3:36
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$x$ ranges from $0$ to $2$: The octant constraint gives the $0$. As for the $2$: it should be apparent from what you've done that the region (projected to the $xy$ plane) should be a quarter circle:

enter image description here

Because of course, when $x>2$, the constraint on $y$ becomes undefined.

The final integral's bounds are:

$$\int_0^2\int_0^{\sqrt{4-x^2}}\int_{y^2}^{8-2x^2-y^2}(\cdots)dzdydx.$$

Does this make things clearer?

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  • $\begingroup$ Yes I beleive this is correct, and now I have no desire of answering this question because those bounds are... ermmm... ugly But thanks $\endgroup$ – Makogan Nov 17 '17 at 4:34
  • $\begingroup$ To actually solve the integral, it is beneficial to transform the $x$ and $y$ bounds to polar. The answer ends up being $\frac{1024}3$. $\endgroup$ – Austin Weaver Nov 17 '17 at 14:28

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