How do you parameterize a surface?
Jenn, Founder Calcworkshop®, 15+ Years Experience (Licensed & Certified Teacher)
This is the big question we need to answer in order to calculate surface integrals and eventually flux integrals, Stoke’s theorem, and divergence theorem in our quest for understanding vector calculus.
In fact, this lesson is all about finding parametric surfaces and their areas.
So, let’s get to it.
Parameterization
First, we need to begin with finding a parametric representation for a surface.
Why?
Because relating a surface to a subset of a plane makes things easier to calculate, and parameterization is the tool we need to accomplish this goal.
Parametric Curve Vs. Surface
Okay, so from our previous studies of parametric equations, we know that a curve is a one-dimensional object represented by three parametric equations involving one parameter.
Whereas a surface is a two-dimensional object that can be represented by three vector equations using two parameters.
\begin{equation}
\begin{array}{c|c}
\text { Parametric Curve in } \mathbb{R}^{3} & \text { Parametric Surface in } \mathbb{R}^{3} \\
\hline \text { Vector Equation: } & \text { Vector Equation: } \\
\vec{r}(t)=\langle x(t), y(t), z(t)\rangle & \vec{r}(u, v)=\langle x(u, v), y(u, v), z(u, v)\rangle \\
\text { where } a \leq t \leq b & \text { where }(u, v) \text { lie in a region D of the } u v \text { plane } \\
\text { Parametric Equations: } & \text { Parametric Equations: } \\
x=x(t) & x=x(u, v) \\
y=y(t) & y=y(u, v) \\
z=z(t) & z=z(u, v) \\
\text { One parameter, because a curve is a } & \begin{array}{c}
\text { Two parameters, because a surface is a two- } \\
\text { dimensional object }
\end{array} \\
\text { one-dimensional object } & \text { There are three component functions because } \\
\text { There are three component functions } & \text { the curve lines in 3-space }
\end{array}
\end{equation}
So, a parameterization is a function defined on a region \(D\) in the \(uv\) plane in \({\mathbb{R}^3}\) denoted as:
\begin{equation}
\vec{r}(u, v)=\langle x(u, v), y(u, v), z(u, v)\rangle
\end{equation}
where \(x=x(u, v), y=y(u, v)\), and \(z=z(u, v)\) are real valued continuous functions called parametric equations.
And a parametric surface is the image of a region \(D\) in the \(uv\) plane under a parameterization on \(D\).
Example
Great. So, let’s put this knowledge to use with a few problems.
Problem #1 – Find the parametric equations for the surface \(f\left( {x,y} \right) = 9 – {x^2} – {y^2}\).
One way to parameterize the surface is to take \(x\) and \(y\) as parameters and writing the parametric equation as \(x = x,y = y,\) and \(z = f\left( {x,y} \right)\) such that the parameterizations for this paraboloid is:
\begin{equation}
x=x, \quad y=y, \quad z=f(x, y)=9-x^{2}-y^{2} \quad,(x, y) \in \mathbb{R}^{2}
\end{equation}But parameterizations are not unique, as we can also represent this surface using polar coordinates.
If we let \(x = r\cos \theta ,y = r\sin \theta ,\) and \(z = f\left( {r,\theta } \right)\) then the paraboloid has the following parametric representation
\begin{equation}
x=r \cos \theta, \quad y=r \sin \theta, \quad z=f(x, y)=9-r^{2}
\end{equation}\begin{equation}
\text { where } r>0 \text { and } 0 \leq \theta \leq 2 \pi
\end{equation}
Problem #2 Find the parametric representation for \({x^2} + {y^2} + {z^2} = {a^2}\).
First, we must notice that we are being asked to parameterize a sphere. Therefore, we now need to ask ourselves, what coordinate system best represents a sphere?
Spherical coordinates, of course!
So, we will use the following parametric equations: \(x = \rho \sin \phi \cos \theta ,y = \rho \sin \phi \sin \theta ,\) and \(z = \rho \cos \phi \) such that the parameterizations for this sphere where \(a = \rho \) is:
\begin{equation}
x=a \sin \phi \cos \theta, \quad y=a \sin \phi \sin \theta, \quad z=a \cos \phi
\end{equation}\begin{equation}
\text { where } a>0,0 \leq \theta \leq 2 \pi, \text { and } 0 \leq \phi \leq \pi
\end{equation}
Easy, right?
And in our lesson, we will look at how to find the parametric representation for a surface, visualize and determine the surface given the parametric representation using the following coordinates:
- Cartesian
- Polar
- Cylindrical
- Spherical
Tangent Planes
Okay, so now that we know how to define parametric surfaces, it’s time to turn our attention to learning how to find the tangent plane to a parametric surface at a point.
Now, we’ve already seen how to find tangent planes to a level surface of a function using the gradient vector. But now, we will learn how to find tangent planes to parametric surfaces.
Consider a parametric surface \(S\) that is traced out by a vector function:
\begin{equation}
\vec{r}(u, v)=\langle x(u, v), y(u, v), z(u, v)\rangle
\end{equation}
\begin{equation}
\text { at a point } P_{0} \text { with position vector } \vec{r}\left(u_{0}, v_{0}\right)
\end{equation}
Well, all we have to do is find a point through the plane and the normal vector.
We already know our point, \({P_0}\), so we just need a normal vector to this plane. Thankfully, we can calculate our normal vector to the tangent plane by using the cross product of our partial derivatives of \(\vec r\) with respect to \(u\) and \(v\) at point \(P_{0}\).
\begin{equation}
\vec{n}=\vec{r}_{u}\left(u_{0}, v_{0}\right) \times \vec{r}_{v}\left(u_{0}, v_{0}\right)
\end{equation}
It is important to know that if \({\vec r_u} \times {\vec r_v} \ne \vec 0\), then the surface \(S\) is called smooth and has no corners at the point \(\vec r\left( {{u_0},{v_0}} \right)\)
Example
Let’s look at an example.
Find the tangent plane to the unit sphere \({x^2} + {y^2} + {z^2} = 1\) at \(\left( {\frac{1}{2},\frac{1}{2},\frac{1}{{\sqrt 2 }}} \right)\).
First, we need parameterize our unit sphere with:
\begin{equation}
x=\rho \sin \phi \cos \theta, y=\rho \sin \phi \sin \theta \text {, and } z=\rho \cos \phi
\end{equation}knowing that
\begin{equation}
\rho=1 \quad x=\sin \phi \cos \theta, \quad y=\sin \phi \sin \theta, \quad z=\cos \phi
\end{equation}\begin{equation}
\text { where } 0 \leq \theta \leq 2 \pi \text {, and } 0 \leq \phi \leq \pi
\end{equation}Okay, now that we have our parameterization, we need to take our first order partial derivatives with respect to \(\theta\) and \(\phi\).
\begin{equation}
\begin{aligned}
&\vec{r}_{\theta}=\langle-\sin \phi \sin \theta, \sin \phi \cos \theta, 0\rangle \\
&\vec{r}_{\phi}=\langle\cos \phi \cos \theta, \cos \phi \sin \theta,-\sin \phi\rangle
\end{aligned}
\end{equation}Now we need to find \(\theta\) and \(\phi\) from our given point \(\left( {\frac{1}{2},\frac{1}{2},\frac{1}{{\sqrt 2 }}} \right)\) by solving our parametric equations:
\begin{equation}
\vec{r}=\langle\underbrace{\sin \phi \cos \theta}_{x}, \underbrace{\sin \phi \sin \theta}_{y}, \underbrace{\cos \phi}_{z}\rangle \text { and }(\underbrace{\frac{1}{2}}_{x}, \underbrace{\frac{1}{2}}_{y}, \underbrace{\frac{1}{\sqrt{2}}}_{z})
\end{equation}\begin{equation}
\text { If } \cos \phi=\frac{1}{\sqrt{2}} \Rightarrow \phi=\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi}{4}
\end{equation}\begin{equation}
\text { Then } \sin \left(\frac{\pi}{4}\right) \cos \theta=\frac{1}{2} \Rightarrow \frac{1}{\sqrt{2}} \cos \theta=\frac{1}{2} \Rightarrow \theta=\cos ^{-1}\left(\frac{\sqrt{2}}{2}\right)=\frac{\pi}{4}
\end{equation}Now, we’re ready to evaluate our partials knowing that \(\theta = \frac{\pi }{4}\) and \(\phi = \frac{\pi }{4}\).
\begin{equation}
\vec{r}_{\theta}\left(\frac{\pi}{4}, \frac{\pi}{4}\right)
\end{equation}\begin{equation}
=\left\langle-\sin \left(\frac{\pi}{4}\right) \sin \left(\frac{\pi}{4}\right), \sin \left(\frac{\pi}{4}\right) \cos \left(\frac{\pi}{4}\right), 0\right\rangle
\end{equation}\begin{equation}
=\left\langle-\frac{1}{2}, \frac{1}{2}, 0\right\rangle
\end{equation}\begin{equation}
\vec{r}_{\phi}\left(\frac{\pi}{4}, \frac{\pi}{4}\right)
\end{equation}\begin{equation}
=\left\langle\cos \left(\frac{\pi}{4}\right) \cos \left(\frac{\pi}{4}\right), \cos \left(\frac{\pi}{4}\right) \sin \left(\frac{\pi}{4}\right),-\sin \left(\frac{\pi}{4}\right)\right\rangle
\end{equation}\begin{equation}
=\left\langle\frac{1}{2}, \frac{1}{2},-\frac{\sqrt{2}}{2}\right\rangle
\end{equation}Next, we will calculate the normal vector to the tangent plane.
\begin{equation}
\vec{n}=\vec{r}_{\theta} \times \vec{r}_{\phi}
\end{equation}\begin{equation}
=\left\langle-\frac{1}{2}, \frac{1}{2}, 0\right\rangle \times\left\langle\frac{1}{2}, \frac{1}{2},-\frac{\sqrt{2}}{2}\right\rangle
\end{equation}\begin{equation}
\vec{n}=\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
-\frac{1}{2} & \frac{1}{2} & 0 \\
\frac{1}{2} & \frac{1}{2} & -\frac{\sqrt{2}}{2}
\end{array}\right|
\end{equation}\begin{equation}
=\left\langle-\frac{\sqrt{2}}{4},-\frac{\sqrt{2}}{4},-\frac{1}{2}\right\rangle
\end{equation}Now, recall the formula for writing an equation of the tangent plane with point \(\left( {{x_0},{y_{_0}},{z_0}} \right)\) and normal vector \(\vec n = \left\langle {a,b,c} \right\rangle \) is \(a\left( {x – {x_0}} \right) + b\left( {x – {x_0}} \right) + b\left( {x – {x_0}} \right) = 0\).
Therefore, an equation of the tangent plane at \(\left( {\frac{1}{2},\frac{1}{2},\frac{1}{{\sqrt 2 }}} \right)\) is:
\begin{equation}
-\frac{\sqrt{2}}{4}\left(x-\frac{1}{2}\right)-\frac{\sqrt{2}}{4}\left(x-\frac{1}{2}\right)-\frac{1}{2}\left(x-\frac{1}{\sqrt{2}}\right)=0
\end{equation}
Pretty straightforward, right?
Surface Area Of Parametric Surfaces
But there’s more!
We can extend our parameterization skills to find the surface area of parametric surfaces using double integrals.
If a smooth parametric surface \(S\) is given by the equation:
\begin{equation}
\vec{r}(u, v)=\langle x(u, v), y(u, v), z(u, v)\rangle
\end{equation}
where \(\left( {u,v} \right) \in D\) and \(S\) is covered just once as \(\left( {u,v} \right)\) ranges throughout the parameter domain D, then the surface area of \(S\) is:
\begin{equation}
A(S)=\iint_{D}|| \vec{r}_{u} \times \vec{r}_{v}|| d A
\end{equation}
Example
Let’s prove why this formula works.
Well, if we take our function \(z = f\left( {x,y} \right)\) and parameterize it so that \(\vec r\left( {x,y,z} \right) = \left\langle {x,y,f\left( {x,y} \right)} \right\rangle \) by letting \(x = x,\)\(y = y,\)and \(z = f\left( {x,y} \right)\)
Then,
\begin{equation}
\vec{r}_{x}=\left\langle 1,0, f_{x}\right\rangle \text { and } \vec{r}_{y}=\left\langle 0,1, f_{y}\right\rangle
\end{equation}And
\begin{equation}
\vec{r}_{x} \times \vec{r}_{y}=\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
1 & 0 & f_{x} \\
0 & 1 & f_{y}
\end{array}\right|=\left\langle-f_{x},-f_{y}, 1\right\rangle
\end{equation}Thus,
\begin{equation}
\left\|\vec{r}_{x} \times \vec{r}_{y}\right\|=\sqrt{\left(-f_{x}\right)^{2}+\left(-f_{y}\right)^{2}+1^{2}}
\end{equation}Which is exactly the same formula we learned about in our previous lesson on surface area of the graph of a function \(z = f\left( {x,y} \right)\):
\begin{equation}
A(S)=\iint_{D} \sqrt{1+\left(\frac{\partial z}{\partial x}\right)^{2}+\left(\frac{\partial z}{\partial x}\right)^{2}} d A
\end{equation}
Super cool!
Together we will learn how to identify and visualize the surface of a given vector equation, find a parametric representation for a surface, find equations of tangent planes to a parametric surface at a point and even find the surface area.
Gosh, there is a lot of information to cover and discuss in this lesson so let’s get after it!
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