How to find this area?

Question

Find the area of the surface $$L: z^2 =144(x^2+y^2)$$ that is limited by the surface $$K: z=28-x^2-y^2$$

Their intersection (I have found it) $$D_1: x^2+y^2 \le 2^2$$ $$D_2: x^2+y^2\le 14^2$$

May someone give me helpful information?

Answer 1

$z=24$ for:

when $x=0$ then $y=-2$ || $y=2$ and when $y=0$ then $x=-2$ || $x=2$

Calculations:

$z=28-x^2-y^2$

$x^2+y^2+z=28$

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$z^2=144(x^2+y^2)$

$z=\sqrt{144(x^2+y^2)}=12\sqrt{x^2+y^2}$

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$x^2+y^2+12\sqrt{x^2+y^2}=28$

Let $c=x^2+y^2$

$c+12\sqrt{c}=28$

In this polynomial you can calculate Δ which is $36864$ and find out two possible values:

$c_1=196$

$c_2=4$

Since $196$ can be express as a difference of two squares ($50^2-48^2$) the other number is a solution.

$24^2=144 \cdot 4=576$

$24=28-4$