# Calculating the volume bounded between a paraboloid and a plane

Calculate the volume bounded between $z=x^2+y^2$ and $z=2x+3y+1$.

As far as I understand, I need to switch to cylindrical coordinates: $(h,\theta, r)$.

The problem is, that I can't understand how to find the region of each new coordinate . I guess that the region for $\theta$ will be $[0,2\pi]$. But what about $h,r$?

In addition, I do not want to use symmetry . I want to calculate the entire volume , without dividing it into several smaller volumes.

Will you help me?

The plane and the paraboloid intersect at a circle whose projection on XY plane is given by $$x^2+y^2=2x+3y+1\implies(x-1)^2+\left(y-{3\over2}\right)^2={17\over4}$$
So the required volume is \begin{align} &\int_{1-{\sqrt{17}\over2}}^{1+{\sqrt{17}\over2}}\int_{{3\over2}-\sqrt{{17\over4}-(x-1)^2}}^{{3\over2}+\sqrt{{17\over4}-(x-1)^2}}\int^{2x+3y+1}_{x^2+y^2}dzdydx\\ =&\int_{1-{\sqrt{17}\over2}}^{1+{\sqrt{17}\over2}}\int_{{3\over2}-\sqrt{{17\over4}-(x-1)^2}}^{{3\over2}+\sqrt{{17\over4}-(x-1)^2}}\left[{17\over4}-(x-1)^2-\left(y-{3\over2}\right)^2\right]dydx\\ =&\int_0^{2\pi}\int_0^{\sqrt{17}\over2}\left[{17\over4}-r^2\right]rdrd\theta={289\pi\over32} \end{align}
where we've used the transformation $x=1+r\cos\theta,y={3\over2}+r\sin\theta$.
Lets see where both surfaces intersect: $$x^2+y^2=2x+3y+1\quad \Rightarrow \quad (x-1)^2+(y-\frac{3}{2})^2=\frac{17}{4}$$ which is a circle of radius $\frac{\sqrt{17}}{2}$ centered in $(1,3/2)$. Let $D$ be the surface inside this circle. Therefore, your volume equals (in cylindrical coordinates) $$V=\iint_D \int_{r^2}^{2r\cos\theta+3r\sin\theta+1} r\;dz dr d\theta %=\int_0^{2\pi}\int_0^{\sqrt{2r\cos\theta+3r\sin\theta}}\int_{r^2}^{2r\cos\theta+3r\sin\theta} r\;dz dr d\theta$$