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Here is one question :

Find the volume of region bounded above by paraboloid $z = 9-x^2 -y^2$ and below by the $x -y$ plane lying outside the cylinder $ x^2+ y^2=1$

I am trying to solve this question via cylindrical coordinates:

Now I can figure out that $z$ ranges from $0$ to $9 - x^2 -y^2$ But I do not understand how can I find the range for $r$ and $\theta$, The phrase "outside the cylinder" is causing some confusion due to which I cannot find $r$ and $\theta$

Can anyone please help me here ?

Thank you.

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    $\begingroup$ HINT: For what $r$ values are you outside the cylinder (this means outside $x^2+y^2\le 1$) and inside the paraboloid above the $xy$-plane (this means $9-x^2-y^2\ge 0$)? $\endgroup$ – Ted Shifrin Oct 9 at 17:04
  • $\begingroup$ @Ted Shifrin: So, if I am outside the cylinder means $x^2 + y^2 \gt 1$ and since we are inside the paraboloid $x^2 + y^2 \le 9$ hence $r$ varies from $1$ to $3$ while $\theta$ takes values from $0$ to $2\pi$ . Is this correct ? $\endgroup$ – sat091 Oct 9 at 17:11
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    $\begingroup$ Yes, that's it. $\endgroup$ – Ted Shifrin Oct 9 at 17:26
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The condition "beeing outside the cylinder" implies $x^2 + y^2 \geq 1$, while lying above the x-y-plane means $z\geq 0$, so $x^2 + y^2 \leq 9$. In cylindrical coordinates with $r = \sqrt(x^2 + y^2)$, the conditions are therefore

$$ 1 \leq r \leq 3 $$

Integrating the function over this range gives

$$ \int_{0}^{2\pi}\int_{1}^{3} (9-r^2)\cdot r \cdot d\phi\,dr = 32\pi $$

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