Surface s is a part of the paraboloid z=4-x^2-y^2 that lies above the plane z=0.(6+7+7=20pt) a)Find the parametric equation vecr(u,v) of the surface with polar coordinates x=ucos(v), y=usin(v) and find the domain D for u and v.

Surface s is a part of the paraboloid $z=4-{x}^{2}-{y}^{2}$ that lies above the plane $z=0$.$\left(6+7+7=20pt\right)$

a) Find the parametric equation $\stackrel{\to }{r}\left(u,v\right)$ of the surface with polar coordinates $x=u\mathrm{cos}\left(v\right),y=u\mathrm{sin}\left(v\right)$ and find the domain D for u and v.

b) Find ${\stackrel{\to }{r}}_{u},{\stackrel{\to }{r}}_{v},$ and ${\stackrel{\to }{r}}_{u}\cdot {\stackrel{\to }{r}}_{v}$.

c) Find the area of the surface

You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Viktor Wiley
Given : $z=4-{x}^{2}-{y}^{2}$, $z=0$
a) $x=u\mathrm{cos}v,y=u\mathrm{sin}v,z=4-{u}^{2}{\mathrm{cos}}^{2}v-{u}^{2}\mathrm{sin}v$
$z=4-{u}^{2}$
$r\left(u,v\right)=$
The domain of $0\le u\le 2$
$0\le v\le 2\pi$
b) ${r}_{u}=<\mathrm{cos}v,\mathrm{sin}v,-2u>$
${r}_{v}=<-u\mathrm{sin}v,u\mathrm{cos}v,0>$
$ruxrv=\left[\begin{array}{ccc}i& j& k\\ \mathrm{cos}v& \mathrm{sin}v& -2u\\ -u\mathrm{sin}v& u\mathrm{cos}v& 0\end{array}\right]$
$=i\left(0+2{u}^{2}\mathrm{cos}v\right)-j\left(-2{u}^{2}\mathrm{sin}v\right)+k\left(u{\mathrm{cos}}^{2}v+u{\mathrm{sin}}^{2}v\right)=<2{u}^{2}\mathrm{cos}v,2{u}^{2}\mathrm{sin}v,u>$
c) Surface Area =${\int }_{0}^{2}\pi {\int }_{0}^{2}|{r}_{u}x{r}_{v}|dudv$
$|{r}_{u}x{r}_{v}|=\sqrt{4}{u}^{2}{\mathrm{cos}}^{2}v+4{u}^{4}{\mathrm{sin}}^{2}v+{u}^{2}=\sqrt{8}{u}^{4}+{u}^{2}$
$=u\sqrt{1}+8{u}^{2}$
A=${\int }_{0}^{2}\pi {\int }_{0}^{2}u\sqrt{1}+8{u}^{2}dudv$
$1+{84}^{2}=f$
$={\int }_{0}^{2}\pi dv\int \sqrt{f}d\frac{f}{16}$
$16udu=df$