Finding the Volume of Region in Three-Dimensional Space.

Question

I am currently working on calculating the volume of the region $W$ in $\mathbb{R}^3$, defined by the constraints:

$$W=\{ (x,y,z)\in\mathbb{R}^3 ∣ z≥2,  x^2+y^2≤6−z,  y≥x \} .$$

My approach involves using triple integration, and the volume integral I have formulated is as follows:

$$V = \int_{2}^{4} \int_{x}^{\sqrt{6-x^2}} \int_{2}^{6-x^2-y^2} \, dz \, dy \, dx .$$

However, I find myself at a stage where I need assistance confirming if my approach is correct. Could you provide guidance on how to tackle this analytical development? Are there any suggestions for simplifying the integration or considerations I should be aware of?

I appreciate any help or insights you can offer. Thank you!

Answer 1

The paraboloid $x^2 + y^2 \le 6-z$ intersects the plane $z=2$

$x^2+y^2 \le 4$ in a disk of radius 2.

The image of the region in the plane $z = 2$

If we want to integrate in cartesian coordinates, we are going to have to break up this integral into two regions. That which is to the left of the line $x = -\sqrt 2$ and what is to the right of that line.

your integral should be:

$\int_{-2}^{-\sqrt 2} \int_{-\sqrt {6-x^2}}^{\sqrt {6-x^2}} \int_2^{\sqrt {6-x^2-y^2}} \ dz\ dy\ dx + \int_{-\sqrt 2}^{\sqrt 2} \int_{x}^{-\sqrt {6-x^2}} \int_2^{\sqrt {6-x^2-y^2}} \ dz\ dy\ dx$

Rather than splitting this into two integrals, I would suggest you try one of two things.

1. We could rotate the coordinates 45 degrees.

$x = \frac {\sqrt 2}{2} (u+v)\\y = \frac {\sqrt 2}{2} (v-u)$

Solving for $u,v$
$u = \frac {\sqrt 2}{2} (x+y)\\v = \frac {\sqrt 2}{2} (y-x)$

$y>x \implies v > 0$

$\int_{-2}^{2} \int_{0}^{\sqrt {6-u^2}} \int_2^{\sqrt {6-u^2-v^2}} \ dz\ dv\ du $

You could probably just make the argument that this figure can be rotated without formally going through the coordinate conversion.

2. We could convert to polar.

$x = r\cos \theta\\ y = r\sin\theta\\ dy\ dx = r\ dr\ d\theta$

$\int_{\frac {\pi}{4}}^{\frac {5\pi}{4}}\int_0^2 \int_2^{6-r^2} r\ dz \ dr\ d\theta$

or you could do both.

$x = r\cos (\theta+\frac {\pi}{4})\\ y = r\sin(\theta+\frac {\pi}{4})\\ dy\ dx = r\ dr\ d\theta$

$\int_{0}^{\pi}\int_0^2 \int_2^{6-r^2} r\ dz \ dr\ d\theta$

Answer 2

If you draw the $2$D domain you get half circle meaning that the volume described by $W$ is half a paraboloid with base at $z=2$:

$$ \begin{aligned} V_W=\dfrac{V_{\text{paraboloid}}}{2}&=\dfrac{1}{2}\int_0^{2\pi} d\theta\int_2^6\int_0^{\sqrt{6-z}}\rho d\rho dz\\ &=\pi\int_2^6\left[\dfrac{\rho^2}{2}\right|_{0}^{\sqrt{6-z}}dz\\ &=\dfrac{\pi}{2}\int_2^6(6-z)dz=\dfrac{\pi}{2}[(36-18)-(12-2)]=4\pi\ \text{u}^2. \end{aligned} $$