# 1. The subject is surface integrals. Know how to parametrize surfaces of the following cases: ...

## Question:

1. The subject is surface integrals. Know how to parametrize surfaces of the following cases:

Case 1) Given {eq}z = f(x,y), {/eq} over a region R. Then the surface can be parametrized as {eq}r(u,v) = < u, v, f(u,v) > {/eq}

Case 2) Sphere of radius a, centered at the origin: {eq}r(u,v) = < a \ sin \ u \ cos \ v, a \ sin \ u \ sin \ v, a \ cos \ u >, {/eq} for {eq}0 \leq u \leq \Pi, \ 0\leq v \leq 2 \Pi. {/eq}.

Case 3) A plane determined by three points.

Explain each of these cases thoroughly and how to parameterize them.

## Parametric Surfaces:

When evaluating a surface integral, it is often useful to use a parametric description of the surface. In an effort to familiarize ourselves with the process, we will investigate a few special types of surfaces below and come up with a general way to parameterize them.

## Answer and Explanation:

Case 1

Parameterizing this is exactly how it looks. Here we have {eq}z = f (x,y) {/eq}. So what we really have is

{eq}\begin{align*} \vec r (x,y) &= \left< x, y, f (x,y) \right> \end{align*} {/eq}

For some reason, instructors prefer us to change the parameter, no matter how transparent they are (the surface here is already parameterized with respect to {eq}x {/eq} and {eq}y {/eq}). So all we do is assign {eq}x \to u {/eq} and {eq}y \to v {/eq}.

For example, say we have {eq}f (x,y) = xy {/eq}. Then a surface parameterization is

{eq}\begin{align*} \vec r(u,v) &= \left< u,v, uv\right> \end{align*} {/eq}

Case 2

To parameterize a sphere, we simply use spherical coordinates with constant radius {eq}a {/eq}, i.e.

{eq}\begin{align*} x &= a \cos \theta \sin \phi \\ y &= a \sin \theta \sin \phi \\ z &= a \cos \phi \end{align*} {/eq}

This is why the parameters {eq}u {/eq} and {eq}v {/eq} are bounded as they are, they are simply {eq}\theta {/eq} and {eq}\phi {/eq}, respectively. So to parameterize a sphere, we simply use spherical coordinates and assign {eq}\theta \to u {/eq} and {eq}\phi \to v {/eq}.

For example, to parameterize the sphere with radius 3, we write

{eq}\begin{align*} \vec r (u,v) &= \left< 3 \cos \theta \sin \phi, 3 \sin \theta \sin \phi, 3 \cos \phi \right> \end{align*} {/eq}

Case 3

This case is best seen through an example first. Suppose we have the points

{eq}\begin{align*} (1,1,1), (0, 1, 2), (3, 0, 4) \end{align*} {/eq}

Then our plane is parallel to the vector

{eq}\begin{align*} \left< 0, 1, 2 \right> - \left< 1, 1, 1 \right> &= \left< -1, 0, 1 \right> \end{align*} {/eq}

and

{eq}\begin{align*} \left< 3, 0, 4 \right> - \left< 1, 1, 1 \right> &= \left< 2, -1, 3 \right> \end{align*} {/eq}

The plane also passes through the point we used as our tail: {eq}(1,1,1) {/eq}. So a parameterization is

{eq}\begin{align*} \vec r (u,v) &= \left< 1,1,1 \right> + u \left< -1, 0, 1 \right> + v \left< 2, -1, 3 \right> \\ &= \left< 1-u+2v, 1-v, 1+u+3v \right> \end{align*} {/eq}

In general, we create a couple vectors that must lie in the plane using one point as the tail and the other two as the heads, then we build the plane by treating these vectors as spanning vectors and adding them to the position vector described by the point we used as the tail. We can visualize this by picturing ourselves as standing at the point we chose to be the tail, then heading directly to either point we used as heads, and sweeping out the floor created by the vectors.

#### Learn more about this topic:

Evaluating Parametric Equations: Process & Examples

from Precalculus: High School

Chapter 24 / Lesson 3
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