# How to find the volume between the paraboloid lying outside the cylinder.

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 $$\theta$$

Thank you.

• 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 $9-x^2-y^2\ge 0$)? – Ted Shifrin Oct 9 '19 at 17:04
• @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 ? – sat091 Oct 9 '19 at 17:11
• Yes, that's it. – Ted Shifrin Oct 9 '19 at 17:26

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