Find the volume of $V\subset \mathbb{R^3}$ limited by a plane and a paraboloid

Question

Let $V=\{(x,y,z)\in\mathbb{R^3}: x^2+y^2\le z, z\le x+2\}$

Then the volume of V is:

(A) Vol(V) = $\frac{75}{8}\pi$

(B) Vol(V) = $\frac{81}{32}\pi$

(C) Vol(V) = $\frac{9}{4}\pi$

(D) Vol(V) = $\frac{45}{8}\pi$

I already understand how the figure sits on 3D space, is limited by a paraboloid on the bottom and a plane on the top, which is inclined. I guessed it would be easier if I used cylindrical or spherical coordinates but I'm having a hard time with the inclined plane. I would appreciate any hint in or guided resolution in order to understand what thing I'm missing. Thank you.

Answer 1

$V=\{(x,y,z)\in\mathbb{R^3}: x^2+y^2\le z, z\le x+2\}$

You can either use cylindrical coordinates or cartesian coordinates. Spherical coordinates is not the right way to go for this when you have a choice.

Let's first look at the intersection of the paraboloid and the plane.

$z = x + 2 = x^2 + y^2 \implies x^2 - x + y^2 = 2$

You can rewrite it as $(x-\frac{1}{2})^2 + y^2 = \frac{9}{4}$

Hence we use the substitution as below -

$x = \frac{1}{2} + r \cos \theta, y = r \sin \theta$

$0 \leq r \leq \frac{3}{2}, 0 \leq \theta \leq 2\pi$

For the bound of $z$, as you said it is bound below by the paraboloid with $z = x^2 + y^2 = (\frac{1}{2} + r \cos \theta)^2 + r^2 \sin^2\theta$, and above by the plane $z = x + 2 = \frac{1}{2} + r \cos \theta + 2$

So the integral to find the desired volume becomes,

$V = \displaystyle \int_0^{2\pi} \int_0^{3/2} \int_{\frac{1}{4} + r^2 + r \cos \theta}^{\frac{5}{2} + r \cos\theta} r \ dz \ dr \ d\theta$

which is a lot more straightforward integration than if we used $x = r \cos\theta, y = r\sin\theta$.

Answer 2

Start with $x^2\leq z\leq x+2.$ So $x^2-x-2\leq 0,$ and $x\in[-1,2].$

Given $x,$ $z\in [x^2,x+2].$

Given $x$ and $z,$ $y\in [-\sqrt{z-x^2},\sqrt{z-x^2}].$

So this gives the integral:

$$\int_{-1}^2 \int_{x^2}^{x+2}\int_{-\sqrt{z-x^2}}^{\sqrt{z-x^2}}dy\,dz\,dx$$