Question

Find the area of the portion of the paraboloid x=4-y²-z²that lies above the ring 1<=y²+z²<=4in the yz plane.

Answer 1

For any point $(x,y,z)$ on the surface $x=4-y^2-z^2$

$F=x+y^2+z^2-4$

Then,

$\triangledown F=i+yj+2zk$

$|\triangledown F|=\sqrt{1+4y^2+4z^2}$

Here, $p=i, \ |\triangledown F\cdot p|=|1|=1$

Surface of paraboloid $x=4-y^2-z^2$ that lies above the ring $1\leq y^2+z^2\leq 4$ in the y-z plane.

Take $y=rcos\theta,\ z=rsin\theta\ \ then\ \ 1\leq r^2\leq 4$

Then, $dydz=rdrd\theta,\ 1\leq r\leq 2\ and\ \ 0\leq\theta\leq 2\theta$

$\therefore \ \ Surface\ area\ \ S=\int\int_R\dfrac{|\triangledown F|}{|\triangledown F\cdot p|}dA=\int\int_R\sqrt{1+4y^2+4z^2}dydz$

$= \int_{\theta=0}^{2\pi}\int_{r=1}^2\sqrt{1+4r^2}rdrd\theta\\\ \\=\dfrac{1}{8}\int_{\theta =0}^{2\pi}\int_{r=1}^2\dfrac{d}{dr}(1+4r^2)(1+4r^2)^{\frac{1}{2}}drd\theta\\\ \\=\dfrac{1}{8}\int_{\theta =0}^{2\pi}\dfrac{(1+4r^2)^{3/2}}{3/2}|_{r=1}^{2}d\theta\\\ \\=\dfrac{1}{12}\int_{\theta=0}^{2\pi}(17\sqrt{17}-5\sqrt 5)d\theta\\\ \\=\dfrac{(17\sqrt{17}-5\sqrt 5)\pi}{6}$

$$\therefore \boxed{\text{Surface Area }=\dfrac{(17\sqrt{17}-5\sqrt 5)\pi}{6}}$$