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How can I find the volume of the solid bounded by the cylinder $y^2+z^2=9$ and the planes $x=2y$, $x=0$, $z=0$ in the first octant?

I'm not really sure how I should even approach this, but I know I should somehow solve it using a double integral. I know how to solve double integrals, but I need help setting up a definite double integral for this problem.

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If your outer integral is over $z$ (you get to pick-the answer should come out the same) you need to figure out the range of $z$. In this case it runs from $0$ to $3$ so we have $\int_0^3 dz ($something). For the next integral, we get to consider $z$ to be fixed. If we do it over $y$, we need to figure out the range of $y$ at a given $z$. Now we have $\int_0^3 dz \int_{y_{min}}^{y_{max}} dy ($range of $x$ at this $(y,z))$. Now, considering $y$ and $z$ are fixed, you need to figure out the range of $x$, put that in the parentheses and you have your double integral.

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  • $\begingroup$ I'm not sure I'm following you; could you go a tad further, particularly with the range of x? $\endgroup$
    – GPhys
    Oct 20 '12 at 23:43
  • $\begingroup$ Also, I think I was trying something similar/equivalent (I think?), but the other way around because it worked out nicer: $$\int_{-3}^3 \int_0^{\sqrt{9-y^2}} (?)\,\mathrm{d}z\,\mathrm{d}y$$ I'm still not sure what goes inside, though (or if that's even for sure correct so far). $\endgroup$
    – GPhys
    Oct 20 '12 at 23:53
  • $\begingroup$ @Gnintendo: the lower limit should be $0,$ not $-3,$ to stay in the first octant. You are fine to do $y$ on the outside. What goes inside is $x_{max}-x_{min}$ at a given $(y,z)$. Again it will be $0$ due to the octant restriction. What is the greatest $x$ can be? $\endgroup$ Oct 21 '12 at 0:37
  • $\begingroup$ Oh, I see how the first octant restriction comes in with that now. Is the greatest $x$ can be...$2y$? $\endgroup$
    – GPhys
    Oct 21 '12 at 1:27
  • $\begingroup$ @Gnintendo: That is right. So you have $\int_0^3\int_0^{\sqrt{9-y^2}}2y\;dz\;dy$. Maybe your order was a better choice than mine: the $z$ integral is easy. $\endgroup$ Oct 21 '12 at 1:34

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