Question

1. find the volume v of the set of points (x, y, z) satisfying 2z + 3 ≤ x² + y² + z² ≤ 2x + 15.

Explanation:

Step1: Rewrite inequalities as sphere equations

First, complete the square for both inequalities:
For $x^2+y^2+z^2 \geq 2z+3$:
$$x^2+y^2+(z^2-2z) \geq 3$$
$$x^2+y^2+(z-1)^2 -1 \geq 3$$
$$x^2+y^2+(z-1)^2 \geq 4$$
This is the exterior of a sphere with center $(0,0,1)$ and radius $r_1=2$.

For $x^2+y^2+z^2 \leq 2x+15$:
$$(x^2-2x)+y^2+z^2 \leq 15$$
$$(x-1)^2 -1 + y^2+z^2 \leq 15$$
$$(x-1)^2+y^2+z^2 \leq 16$$
This is the interior of a sphere with center $(1,0,0)$ and radius $r_2=4$.

Step2: Calculate distance between centers

Let $C_1=(0,0,1)$, $C_2=(1,0,0)$. The distance $d$ between them is:
$$d = \sqrt{(1-0)^2+(0-0)^2+(0-1)^2} = \sqrt{2} \approx 1.414$$

Step3: Check sphere relationship

Since $r_2 - r_1 = 4-2=2$, and $d=\sqrt{2}<2$, the smaller sphere is entirely inside the larger sphere.

Step4: Compute volume difference

Volume of a sphere is $V=\frac{4}{3}\pi r^3$. The desired volume is the volume of the larger sphere minus the volume of the smaller sphere:
$$V = \frac{4}{3}\pi r_2^3 - \frac{4}{3}\pi r_1^3$$
$$V = \frac{4}{3}\pi (4^3 - 2^3)$$
$$V = \frac{4}{3}\pi (64 - 8)$$
$$V = \frac{4}{3}\pi \times 56$$

Answer:

$\frac{224}{3}\pi$