Question

Find the area of the surface cut from the bottom of the paraboloid x²+y²-z=0 by the plane z = 4

Answer 1

First rewrite the equation $x²+y²-z=0$ to get

$z=x²+y²$

We can find the area using the following formula;

$A(S)=\int_0^{2\pi}\int_0^{2}\sqrt{1+4r^2}rdrd\theta$

$A(S)=2\pi\int_0^2\sqrt{1+4r^2}rdr$ , $u=1+4r^2$ , $du=8rdr$

$A(S)={2\pi \over 8}\int_1^{17} u^{1\over 2}du={2\pi \over 8}{2\over 3}(u^{3\over 2}|_1^{17})$

$\therefore$ The Area of the surface is

$A(S)={\pi \over 6}[(17)^{3\over 2}-1]$