Will someone please help me with the following problem?
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?
Best Answer
The plane and the paraboloid intersect at a circle whose projection on XY plane is given by $$x^2+y^2=2x+3y+1\implies(x-1)^2+\left(y-{3\over2}\right)^2={17\over4}$$
From the above figure it is clear that the plane(blue) lies above the paraboloid(yellow) in the region of interest.
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 $x=1+r\cos\theta,y={3\over2}+r\sin\theta$.