## 17Calculus - Describing Surfaces Using Parametric Equations

Building parametric equations of surfaces can appear to be confusing. There is an art to it but the basic techniques to get started are fairly straightforward. When you first learned parametrics, you probably used t as your parametric variable. When describing surfaces with parametric equations, we need to use two variables. We will use u and v, which is common in vector calculus.

Techniques

Something to keep in mind when building parametric representations is that there are infinite ways to describe any surface. Here are a couple of common ways to set up the equations.

1. If the description of the surface is in the form $$z=g(x,y)$$, the easiest parametric equations are
$$u=x$$, $$v=y$$ and $$z=g(u,v)$$. In vector form, $$\vec{r}(u,v)=u\hat{i}+v\hat{j}+g(u,v)\hat{k}$$.

2. A second basic technique is to describe your surface in cylindrical or spherical coordinates and then set u and v to the appropriate variable. For example, let's say we have a spherical shell of radius 3 that we want to describe in parametric form.
In general, conversion from rectangular to spherical coordinates, here are the equations.

rectangular → spherical coordinates

$$x=\rho \sin \phi \cos \theta$$

$$y=\rho \sin \phi \sin \theta$$

$$z=\rho \cos \phi$$

$$0 \leq \phi \leq \pi$$       $$0 \leq \theta \leq 2\pi$$

Since our example has a radius of 3, $$\rho = 3$$ in the above equations. We can set $$u=\phi$$ and $$v=\theta$$ giving us these parametric equations.

a parametric description of a spherical shell of radius 3

$$x=3 \sin(u) \cos(v)$$

$$y=3 \sin(u) \sin(v)$$

$$z=3 \cos(u)$$

$$0 \leq u \leq \pi$$       $$0 \leq v \leq 2\pi$$

$$\vec{r}(u,v)=3\sin(u)\cos(v)\hat{i} + 3\sin(u)\sin(v)\hat{j} + 3\cos(u)\hat{k}$$

It is important to use the variables u and v for the parametric equations, not because u and v have any significance themselves but because they need to be completely different than the spherical coordinates variables. When you work with surface integrals, you will see why.

Okay, time for a video. Here is a gentle introduction to the idea of parameterizing curves in space.

### Dr Chris Tisdell - Parametrised surfaces [32min-48secs]

video by Dr Chris Tisdell

Here are a couple of video clips discussing how to set up parametric surfaces. He shows some nice special cases which give you more of a feel for how to set them up.

### Evans Lawrence - Lecture 31 - Parametric Surfaces, Surface Integrals [22min-10secs]

video by Evans Lawrence

### Evans Lawrence - Lecture 32 - More on Parametric Surfaces, Surface Integrals [8min-36secs]

video by Evans Lawrence

Now you are ready to learn how to set up and evaluate surface integrals. However, try your hand at some practice problems first, to make sure you really understand and are able to set up various types of parameterized surfaces.

Practice

Unless otherwise instructed, parameterize these surfaces. Remember that there are multiple ways to parameterize a surface, so your answer may not be the same as given in the solution.

Basic

$$z^2=x^2+y^2$$

Problem Statement

Find a set of parametric equations for the surface of the cone $$z^2=x^2+y^2$$.

$$\vec{\Phi}(u,v)=\langle u\cos v, u\sin v, u \rangle$$, $$u \in \mathbb{R}, ~ 0 \leq v \leq 2\pi$$

Problem Statement

Find a set of parametric equations for the surface of the cone $$z^2=x^2+y^2$$.

Solution

### 1877 video

video by Dr Chris Tisdell

$$\vec{\Phi}(u,v)=\langle u\cos v, u\sin v, u \rangle$$, $$u \in \mathbb{R}, ~ 0 \leq v \leq 2\pi$$

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$$x^2+y^2=r^2$$

Problem Statement

Find a set of parametric equations for the surface of the cylinder $$x^2+y^2=r^2$$.

$$\vec{\Phi}(u,v)=\langle r\cos u, r\sin u, v \rangle$$, $$0 \leq u \leq 2\pi, ~ v \in \mathbb{R}$$

Problem Statement

Find a set of parametric equations for the surface of the cylinder $$x^2+y^2=r^2$$.

Solution

### 1878 video

video by Dr Chris Tisdell

$$\vec{\Phi}(u,v)=\langle r\cos u, r\sin u, v \rangle$$, $$0 \leq u \leq 2\pi, ~ v \in \mathbb{R}$$

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$$x^2+y^2+z^2=r^2$$

Problem Statement

Parameterize the sphere of fixed radius r, $$x^2+y^2+z^2=r^2$$. Remember that there are multiple ways to parameterize a surface, so your answer may not be the same as given in the solution.

$$\vec{\Phi}(u,v)=\langle r \sin(u)\cos(v), r\sin(u)\sin(v), r\cos(u) \rangle$$, $$0 \leq u \leq \pi, 0 \leq v \leq 2\pi$$

Problem Statement

Parameterize the sphere of fixed radius r, $$x^2+y^2+z^2=r^2$$. Remember that there are multiple ways to parameterize a surface, so your answer may not be the same as given in the solution.

Solution

### 1879 video

video by Dr Chris Tisdell

$$\vec{\Phi}(u,v)=\langle r \sin(u)\cos(v), r\sin(u)\sin(v), r\cos(u) \rangle$$, $$0 \leq u \leq \pi, 0 \leq v \leq 2\pi$$

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$$2x-4y+3z=16$$

Problem Statement

Parameterize the plane $$2x-4y+3z=16$$. Remember that there are multiple ways to parameterize a surface, so your answer may not be the same as given in the solution.

Hint

To get the same answer as we have, let $$u=x$$ and $$v=y$$.

Problem Statement

Parameterize the plane $$2x-4y+3z=16$$. Remember that there are multiple ways to parameterize a surface, so your answer may not be the same as given in the solution.

$$\vec{r}(u,v)=\langle u,v, (16-2u+4v)/3 \rangle$$, $$u \in \mathbb{R}$$, $$v \in \mathbb{R}$$

Problem Statement

Parameterize the plane $$2x-4y+3z=16$$. Remember that there are multiple ways to parameterize a surface, so your answer may not be the same as given in the solution.

Hint

To get the same answer as we have, let $$u=x$$ and $$v=y$$.

Solution

For this one, it is easiest to let $$u=x$$ and $$v=y$$. Then we can substitute into the equation of the plane and solve for z to get the last component. Also, since there is no restriction on the plane. Therefore, x and y can take on all values of the real numbers. We can either specify $$x \in \mathbb{R}$$ and $$y \in \mathbb{R}$$ or just not say anything about the range of values. In most cases, it is probably better to specify the values so that we communicate that we didn't just forget about the ranges. However, check with your instructor to see what they require.

$$\vec{r}(u,v)=\langle u,v, (16-2u+4v)/3 \rangle$$, $$u \in \mathbb{R}$$, $$v \in \mathbb{R}$$

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Intermediate

Find a parametric representation of the part of the sphere $$x^2+y^2+z^2=4$$ that lies above the cone $$z=\sqrt{x^2+y^2}$$.

Problem Statement

Find a parametric representation of the part of the sphere $$x^2+y^2+z^2=4$$ that lies above the cone $$z=\sqrt{x^2+y^2}$$.

$$x=2\sin(u)\cos(v), y=2\sin(u)\sin(v), z=2\cos(u)$$
$$0 \leq u \leq \pi/4, 0 \leq v \leq 2\pi$$
in vector form: $$\vec{r}(u,v)=2\sin(u)\cos(v)\hat{i}+2\sin(u)\sin(v)\hat{j}+2\cos(u)\hat{k}$$

Problem Statement

Find a parametric representation of the part of the sphere $$x^2+y^2+z^2=4$$ that lies above the cone $$z=\sqrt{x^2+y^2}$$.

Solution

Although her final answer is correct in this video, it would be better to use the variables u and v instead of $$\phi$$ and $$\theta$$ in the final form of the parameterized surface, especially if you are going to be doing a surface integral using this parametric surface. So setting $$u=\phi$$ and $$v=\theta$$ we get the final answer.

### 1876 video

video by Krista King Math

$$x=2\sin(u)\cos(v), y=2\sin(u)\sin(v), z=2\cos(u)$$
$$0 \leq u \leq \pi/4, 0 \leq v \leq 2\pi$$
in vector form: $$\vec{r}(u,v)=2\sin(u)\cos(v)\hat{i}+2\sin(u)\sin(v)\hat{j}+2\cos(u)\hat{k}$$

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### parametric surfaces 17calculus youtube playlist

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 precalculus parametrics

### Trig Formulas

The Unit Circle

The Unit Circle [wikipedia] Basic Trig Identities

Set 1 - basic identities

$$\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }$$

$$\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }$$

$$\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }$$

$$\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }$$

Set 2 - squared identities

$$\sin^2t + \cos^2t = 1$$

$$1 + \tan^2t = \sec^2t$$

$$1 + \cot^2t = \csc^2t$$

Set 3 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle formulas

$$\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }$$

$$\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }$$

Trig Derivatives

 $$\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }$$ $$\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }$$ $$\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }$$ $$\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }$$ $$\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }$$ $$\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }$$

Inverse Trig Derivatives

 $$\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }$$ $$\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }$$

Trig Integrals

 $$\int{\sin(x)~dx} = -\cos(x)+C$$ $$\int{\cos(x)~dx} = \sin(x)+C$$ $$\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C$$ $$\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C$$ $$\int{\sec(x)~dx} =$$ $$\ln\abs{\sec(x)+\tan(x)}+C$$ $$\int{\csc(x)~dx} =$$ $$-\ln\abs{\csc(x)+\cot(x)}+C$$

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