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Consider the surface delimited by $z=9-y^2$, $y=2x$ and $x=6$ on the first octant. How to find its volume using triple integrals?

Seems like $x$ goes from $0$ to $6$, $y$ goes from $0$ to $2x$, and then $z$ goes from $0$ to $9-y^2$. In such case the integral would be

$$\int_0^6 \int_0^{2x} \int_0^{9-y^2} dz\ dy\ dx$$ Which can be easily calculated. Are those limits correct? attaching a picture enter image description here

Valent
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    The limits are most definitely incorrect. For starters, note that $z\ge 0$ tells you that $0\le y\le 3$, whereas you're allowed $y$ to range up to $12$. I suggest putting $dy$ on the outside. The picture shows what you can do with technology, but what is the volume we're interested in, in your picture? Technology doesn't replace sketching and thinking. – Ted Shifrin Oct 08 '22 at 22:31
  • @TedShifrin I see, but then $z$ goes from $0$ to $9-y^2$ still, even when $y$ is bounded right? Otherwise if I say from $0$ to $9$ for instance, that would be a cube which is wrong. – Valent Oct 08 '22 at 22:41
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    Yes, $z$ definitely goes from $0$ to $9-y^2$. Agreed. :) And $x$ can't be any greater than $6$. – Ted Shifrin Oct 08 '22 at 22:45

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I think the solid is $$\frac{y}{2}\leq x \leq 6,$$ $$0\leq z \leq 9-y^2,$$ $$0\leq y \leq 3.$$

Hence, $V=\int_0^3\int_0^{9-y^2}\int_{\frac{y}{2}}^6dxdzdy=\int_0^3\int_0^{9-y^2}(6-\frac{y}{2})dzdy=\int_0^3(6-\frac{y}{2})(9-y^2)dy=\frac{783}{8}.$

Bob Dobbs
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  • I wish you would have let the OP and me finish our little conversation in the comments before posting the final solution. I realize that lots of people don't like that use of this site, but ... – Ted Shifrin Oct 08 '22 at 22:55
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    That's unexpected, the integral I have computed $$\int_{0}^6\int_0^{3}\int_{0}^{9-y^2}dz\ dy\ dx = 108$$ – Valent Oct 08 '22 at 22:55
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    Well, now that Bob has shown you the correct answer, can you figure out why you're wrong? What happened to poor $y=2x$? – Ted Shifrin Oct 08 '22 at 22:56
  • @TedShifrin yes, I see now since $x$ depends on $y$, I have to adjust its variation. I cannot use simply $x$ from $0$ to $6$ as it moves together with $y$. – Valent Oct 08 '22 at 23:03
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    So, as you practice more and more of these questions, you need to think about the order of integration and how the shape of the region might affect whether you do $dz,dy,dx$ or $dy,dz,dx$, etc. – Ted Shifrin Oct 08 '22 at 23:08