# 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 $$\displaystyle{z}={4}-{x}^{{2}}-{y}^{{2}}$$ that lies above the plane $$z=0$$.$$(6+7+7=20pt)$$

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

b) Find $$\displaystyle\vec{{r}}_{{u}},\vec{{r}}_{{v}},$$ and $$\displaystyle\vec{{r}}_{{u}}\cdot\vec{{r}}_{{v}}$$.

c) Find the area of the surface

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Viktor Wiley
Given : $$\displaystyle{z}={4}-{x}^{{2}}-{y}^{{2}}$$, $$\displaystyle{z}={0}$$
a) $$\displaystyle{x}={u}{\cos{{v}}},{y}={u}{\sin{{v}}},{z}={4}-{u}^{{2}}{{\cos}^{{2}}{v}}-{u}^{{2}}{\sin{{v}}}$$
$$\displaystyle{z}={4}-{u}^{{2}}$$
$$\displaystyle{r}{\left({u},{v}\right)}={<}{u}{\cos{{v}}},{u}{\sin{{v}}},{4}-{u}^{{2}}{>}$$
The domain of $$\displaystyle{0}\le{u}\le{2}$$
$$\displaystyle{0}\le{v}\le{2}\pi$$
b) $$\displaystyle{r}_{{u}}={<}{\cos{{v}}},{\sin{{v}}},-{2}{u}{>}$$
$$\displaystyle{r}_{{v}}={<}-{u}{\sin{{v}}},{u}{\cos{{v}}},{0}{>}$$
$$\displaystyle{r}{u}{x}{r}{v}={\left[\begin{array}{ccc} {i}&{j}&{k}\\{\cos{{v}}}&{\sin{{v}}}&-{2}{u}\\-{u}{\sin{{v}}}&{u}{\cos{{v}}}&{0}\end{array}\right]}$$
$$\displaystyle={i}{\left({0}+{2}{u}^{{2}}{\cos{{v}}}\right)}-{j}{\left(-{2}{u}^{{2}}{\sin{{v}}}\right)}+{k}{\left({u}{{\cos}^{{2}}{v}}+{u}{{\sin}^{{2}}{v}}\right)}={<}{2}{u}^{{2}}{\cos{{v}}},{2}{u}^{{2}}{\sin{{v}}},{u}{>}$$
c) Surface Area =$$\displaystyle{\int_{{0}}^{{2}}}\pi{\int_{{0}}^{{2}}}{\left|{r}_{{u}}{x}{r}_{{v}}\right|}{d}{u}{d}{v}$$
$$\displaystyle{\left|{r}_{{u}}{x}{r}_{{v}}\right|}=\sqrt{{4}}{u}^{{2}}{{\cos}^{{2}}{v}}+{4}{u}^{{4}}{{\sin}^{{2}}{v}}+{u}^{{2}}=\sqrt{{8}}{u}^{{4}}+{u}^{{2}}$$
$$\displaystyle={u}\sqrt{{1}}+{8}{u}^{{2}}$$
A=$$\displaystyle{\int_{{0}}^{{2}}}\pi{\int_{{0}}^{{2}}}{u}\sqrt{{1}}+{8}{u}^{{2}}{d}{u}{d}{v}$$
$$\displaystyle{1}+{84}^{{2}}={f}$$
$$\displaystyle={\int_{{0}}^{{2}}}\pi{d}{v}\int\sqrt{{f}}{d}\frac{{f}}{{16}}$$
$$\displaystyle{16}{u}{d}{u}={d}{f}$$
$$\displaystyle=\frac{{1}}{{16}}\int{v}{\int_{{0}}^{{2}}}\pi{\left[\frac{{f}^{{\frac{{3}}{{2}}}}}{{\frac{{3}}{{2}}}}\right]}_{{{u}={0}}}$$
$$\displaystyle=\frac{\pi}{{8}}\cdot\frac{{2}}{{3}}{\left[\frac{{\left({1}+{8}{u}^{{2}}\right)}^{{3}}}{{2}}{\int_{{0}}^{{2}}}\right.}$$
$$\displaystyle=\frac{\pi}{{12}}{\left[\frac{{\left({33}\right)}^{{3}}}{{2}}-{1}\right]}_{{u}}$$