Parametric Surfaces

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Objectives:

1.: Know what it means to parametrize a surface.
2.: Understand why surface parametrizations always have two variables.
3.: Be able to parametrize various surfaces.
4.: Understand how to find the tangent plane to a parametrized surface.
5.: Know how to calculate the surface area of a parametrized surface.

Recap Video

Here is a video highlights the main points of the section.

To summarize:

A parametrization of a surface $S$ consists of a description of the points of the surface as a function of two variables, say $\textbf {r}(u,v)$ , where $(u,v)$ live in some domain $D$ .
If a surface $S$ is parametrized as $\textbf {r}(u,v)$ , then the partials $\textbf {r}_u$ and $\textbf {r}_v$ are computed componentwise and give tangent vectors to $S$ , and their cross product gives the normal vector to the tangent plane.
The surface area of a surface $S$ given by $\textbf {r}(u,v)$ , where $(u,v)$ are in some domain $D$ , is given by $\iint _D \| \textbf {r}_u \times \textbf {r}_v \| du dv.$

Problems

Parametrizing Surfaces

The function $\textbf {r}(u,v) = \left \langle u+v,3u-v,2u+1 \right \rangle$ parametrizes a plane. The equation of the plane (in the form $ax+by+c=d$ ) is $\answer {x+y-2z}=-2$ .

Find a relation among $x$ , $y$ , and $z$ that eliminates $u$ and $v$ .

Notice that $x+y-2z$ is a constant.

Use the spherical coordinate method demonstrated in class to parametrize the portion of the sphere $x^2+y^2+z^2=9$ which lies in the first octant. (INPUT NOTE: In your answer, write p instead of phi. For example, write $\sin (p)$ instead of $\sin (\phi )$ .)

The parametrization is $\textbf {r}(\theta ,\phi ) = \left \langle \answer {3\cos (\theta )\sin (p)},\answer {3\sin (\theta )\sin (p)},\answer {3\cos (p)} \right \rangle$ and the bounds are $0\leq \theta \leq \answer {\pi /2},\quad 0 \leq \phi \leq \answer {\pi /2}.$

The spherical coordinate parametrization of a sphere of radius $R$ centered at the origin is $(R\cos (\theta )\sin (\phi ),R\sin (\theta )\sin (\phi ),R\cos (\phi )).$

The bounds are $0$ to $\pi /2$ for both because $\theta$ controls rotation around the $z$ -axis and $\phi$ controls movement from the north pole to the south pole.

The surface with parametrization $G(x,\theta ) = \left \langle x,\sin (x)\cos (\theta ),\sin (x)\sin (\theta ) \right \rangle$ for $0 \leq x \leq \pi$ and $0 \leq \theta \leq 2\pi$ can be gotten by revolving the graph of $y = \answer {\sin (x)}$ for $0 \leq x \leq \pi$ around the xyz -axis.

The $x$ in the first coordinate says that the $x$ -axis is the axis of rotation, and the $\sin (x)$ in the second and third components tell us that that is the graph.

Let $S$ be the portion of the graph of $z = 16-x^2-y^2$ where $z \geq 0$ .

A parametrization for $S$ in Cartesian coordinates could be $\textbf {r}(x,y) = \left \langle x,y,\answer {16-x^2-y^2} \right \rangle$ for $x^2+y^2 \leq 16$ .
A parametrization for $S$ using polar coordinates in $x$ and $y$ would be $\textbf {r}(r,\theta ) = \left \langle r\cos (\theta ),r\sin (\theta ),\answer {16-r^2} \right \rangle$ for $0 \leq r \leq 4$ and $0 \leq \theta \leq 2\pi$ .

Matching game: match each of the following equations to their corresponding graphs.

$\textbf {r}(u,v) = (u\cos (v),u\sin (v),u^2)$
$\textbf {r}(u,v) = (\sin (u),\cos (v),u)$
$\textbf {r}(u,v) = (\cos (u)\cos (v),\cos (u)\sin (v),u)$
$\textbf {r}(u,v) = (\sin (u),\cos (v),\cos (u))$

Which of the following parametrizes the portion of $z = \sqrt {x^2+y^2}$ with $0 \leq z \leq 4$ . Select all that apply.

$\textbf {r}(x,y) = \left \langle x,y,\sqrt {x^2+y^2} \right \rangle$ with $x^2+y^2 \leq 16$ . $\textbf {r}(x,y) = \left \langle x,2y,\sqrt {x^2+2y^2} \right \rangle$ with $x^2+y^2 \leq 16$ . $\textbf {r}(r,\theta ) = \left \langle r\cos (\theta ),r\sin (\theta ),r \right \rangle$ with $0\leq r \leq 4$ , $0 \leq \theta \leq 2\pi$ .

Tangent Planes

Consider a surface given by $\textbf {r}(u,v) = \left \langle u,v,uv \right \rangle$ . Find the tangent plane to the surface at the point $(u,v) = (1,2)$ .

The partials are $\textbf {r}_u = \left \langle \answer {1},\answer {0},\answer {v} \right \rangle ,$ $\textbf {r}_v = \left \langle \answer {0},\answer {1},\answer {u} \right \rangle .$ At the point $(u,v) = (1,2)$ , we have $\textbf {r}_u(1,2) = \left \langle \answer {1},\answer {0},\answer {2} \right \rangle ,$ $\textbf {r}_v(1,2) = \left \langle \answer {0},\answer {1},\answer {1} \right \rangle .$ The cross product is $\textbf {r}_u(1,2) \times \textbf {r}_v(1,2) = \left \langle \answer {-2},\answer {-1},\answer {1} \right \rangle .$ The point $(x,y,z)$ corresponding to $u=1$ and $v=2$ is $(x,y,z) = (\answer {1},\answer {2},\answer {2}).$ Therefore, the equation of the plane in the form $ax+by+cz=d$ is $-2x+\answer {-1}y+\answer {1}z = \answer {-2}.$

Consider the surface $S$ given by $\textbf {r}(\theta ,z) = \left \langle 2\cos (\theta ),2\sin (\theta ),z \right \rangle .$ The equation of the tangent plane to $S$ at the point $\textbf {r}(\pi /4,5)$ in the form $ax+by+cz=d$ is $\answer {x+y} = 2\sqrt {2}.$

Surface Area

Consider the surface $S$ gotten by revolving the portion of $y = x^3$ where $0 \leq x \leq 2$ around the $x$ -axis. This surface can be parametrized as $\textbf {r}(x,\theta ) = \left \langle x,x^3\cos (\theta ),x^3\sin (\theta ) \right \rangle .$ Find the surface area of $S$ .

We compute the partials: $\textbf {r}_x(x,\theta ) = \left \langle \answer {1},\answer {3x^2\cos (\theta )},\answer {3x^2\sin (\theta )} \right \rangle$ $\textbf {r}_{\theta }(x,\theta ) = \left \langle \answer {0},\answer {-x^3\sin (\theta )},\answer {x^3\cos (\theta )} \right \rangle .$ The cross product is $\textbf {r}_x\times \textbf {r}_{\theta } = \left \langle \answer {3x^5},\answer {-x^3\cos (\theta )},\answer {-x^3\sin (\theta )} \right \rangle$ which has magnitude $\textbf {r}_x \times \textbf {r}_{\theta } = \sqrt {\answer {9x^{10}+x^6}}.$ The bounds for $x$ and $\theta$ are $\answer {0} \leq x \leq \answer {2},\quad 0 \leq \theta \leq \answer {2\pi }.$ The surface area is therefore $\int _0^{2\pi } \int _0^2 \answer {\sqrt {9x^{10}+x^6}}dx d\theta = \int _0^{2\pi } \int _0^2 x^3\sqrt {9x^4+1}dxd\theta .$ The integral evaluates to $\answer {\frac {\pi }{27}(145\sqrt {145}-1)}$ .

The surface area of the portion of $y = 9-z^2$ where $0 \leq x \leq z \leq 3$ is equal to $\answer {\frac {1}{12}(37\sqrt {37}-1)}$ .

Parametrize the surface as $\textbf {r}(x,z) =(x,9-z^2,z)$ . The partials are $\textbf {r}_x = \left \langle 1,0,0 \right \rangle ,\quad \textbf {r}_z = \left \langle 0,-2z,1 \right \rangle .$ The cross product is $\left \langle 0,-1,-2z \right \rangle$ which has magnitude $\sqrt {1+4z^2}$ .