Surface s is a part of the paraboloid

a) Find the parametric equation

b) Find

c) Find the area of the surface

OlmekinjP
2021-09-12
Answered

Surface s is a part of the paraboloid

a) Find the parametric equation

b) Find

c) Find the area of the surface

You can still ask an expert for help

Viktor Wiley

Answered 2021-09-13
Author has **84** answers

Given : $z=4-{x}^{2}-{y}^{2}$ , $z=0$

a)$x=u\mathrm{cos}v,y=u\mathrm{sin}v,z=4-{u}^{2}{\mathrm{cos}}^{2}v-{u}^{2}\mathrm{sin}v$

$z=4-{u}^{2}$

$r(u,v)=<u\mathrm{cos}v,u\mathrm{sin}v,4-{u}^{2}>$

The domain of$0\le u\le 2$

$0\le v\le 2\pi$

b)${r}_{u}=<\mathrm{cos}v,\mathrm{sin}v,-2u>$

${r}_{v}=<-u\mathrm{sin}v,u\mathrm{cos}v,0>$

$ruxrv=\left[\begin{array}{ccc}i& j& k\\ \mathrm{cos}v& \mathrm{sin}v& -2u\\ -u\mathrm{sin}v& u\mathrm{cos}v& 0\end{array}\right]$

$=i(0+2{u}^{2}\mathrm{cos}v)-j(-2{u}^{2}\mathrm{sin}v)+k(u{\mathrm{cos}}^{2}v+u{\mathrm{sin}}^{2}v)=<2{u}^{2}\mathrm{cos}v,2{u}^{2}\mathrm{sin}v,u>$

c) Surface Area =${\int}_{0}^{2}\pi {\int}_{0}^{2}\left|{r}_{u}x{r}_{v}\right|dudv$

$\left|{r}_{u}x{r}_{v}\right|=\sqrt{4}{u}^{2}{\mathrm{cos}}^{2}v+4{u}^{4}{\mathrm{sin}}^{2}v+{u}^{2}=\sqrt{8}{u}^{4}+{u}^{2}$

$=u\sqrt{1}+8{u}^{2}$

A=${\int}_{0}^{2}\pi {\int}_{0}^{2}u\sqrt{1}+8{u}^{2}dudv$

$1+{84}^{2}=f$

$={\int}_{0}^{2}\pi dv\int \sqrt{f}d\frac{f}{16}$

$16udu=df$

a)

The domain of

b)

c) Surface Area =

A=

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