22. Parametric Surfaces and Surface Integrals

a. Introduction to Parametric Surfaces

1. Position and Plots

In a plane, \(\mathbb{R}^2\), the geometrical domains on which we can integrate are curves and regions.

  1. A parametric curve is defined by an \(\mathbb{R}^2\) vector valued function of \(1\) variable: \[ \vec r(t)=\left\langle x(t),y(t)\right\rangle \]
  2. A parametrized 2D region or 2D curvilinear coordinate system is defined by an \(\mathbb{R}^2\) vector valued function of \(2\) variables: \[ \vec R(u,v)=\left\langle x(u,v),y(u,v)\right\rangle \]

In space, \(\mathbb{R}^3\), the geometrical domains on which we can integrate are curves, surfaces and regions.

  1. A parametric curve is defined by an \(\mathbb{R}^3\) vector valued function of \(1\) variable: \[ \vec r(t)=\left\langle x(t),y(t),z(t)\right\rangle \]
  2. A parametric surface is defined by an \(\mathbb{R}^3\) vector valued function of \(2\) variables: \[ \vec R(u,v)=\left\langle x(u,v),y(u,v),z(u,v)\right\rangle \]
  3. A parametrized 3D region or 3D curvilinear coordinate system is defined by an \(\mathbb{R}^3\) vector valued function of \(3\) variables: \[ \vec R(u,v,w)=\left\langle x(u,v,w),y(u,v,w),z(u,v,w)\right\rangle \]

We have studied how to integrate on all of these geometric domains except surfaces. That is the topic of this chapter. In some ways parametric surfaces behave like parametric curves but in most ways they generalize 2D curvilinear coordinate systems.

Position

A parametric surface gives the position on the surface as a function of \(2\) parameters, usually regarded as coordinates on the surface. \[ (x,y,z)=\vec R(u,v)=\left\langle x(u,v),y(u,v),z(u,v)\right\rangle \]

The parametrization can also be specified in component form: \[ x=x(u,v), \qquad y=y(u,v), \qquad z(u,v) \]

As a primary example of a parametric surface, we will study the sphere (centered at the origin) of radius \(2\), which may be parametrized as \[ (x,y,z)=\vec R(\phi,\theta) =\left\langle 2\sin\phi\cos\theta,2\sin\phi\sin\theta,2\cos\phi\right\rangle \] Notice that this is exactly the same as the spherical coordinate system, except that the radius \(\rho\) (the distance from the origin) has been replaced by \(2\).

The plot shows a sphere with two blue lines from the North pole to the South pole and an arrow pointing South. It also shows two red lines making horizontal circles on the sphere and an arrow pointing East.

The coordinates (parameters) on the surface are
(1) the polar angle, \(\phi\), measured down from the \(+z\) axis (the North pole) to the \(-z\) axis (the South pole), and
(2) the azimuthal angle, \(\theta\), measured counterclockwise (West to East) from the \(+x\) axis.
In the plot, the blue lines are lines of longitude: \(\theta\) is a constant and \(\phi\) is changing.
Similarly, the red lines are lines of latitude: \(\phi\) is a constant and \(\theta\) is changing.
The arrows point in the direction \(\phi\) or \(\theta\) is increasing.

Plot

The plot of a parametric curve is the image of the function \(\vec r(t)\), i.e. the set of points \(\left\langle x(t),y(t),z(t)\right\rangle\) for all values of \(t\). For example, the plot of the helix, \(\vec r(t)=\left\langle 4\cos t,4\sin t,3t\right\rangle\) is shown at the right.

The plot shows a helix circling counterclockwise around the z axis as it moves upward.

Similarly, the plot of a parametric surface is the image of the function \(\vec R(u,v)\), i.e. the set of points \(\left\langle x(u,v),y(u,v),z(u,v)\right\rangle\) for all values of \(u\) and \(v\). For example, the plot of the sphere, \[ \vec R(\phi,\theta) =\left\langle 2\sin\phi\cos\theta,2\sin\phi\sin\theta,2\cos\phi\right\rangle \] is shown at the right.

The plot shows a sphere of radius 2.

Parameterizing a Surface

A surface is frequently given by an equation, either as the graph of a function of \(2\) variables, \(z=f(x,y)\), or as the level set of a function of \(3\) variables, \(F(x,y,z)=C\). At times, you may want (or need) to rewrite it in parametric form, for instance to be able to integrate on the surface. Frequently a good method is to start with the position vector \(\vec R\) of an appropriate 3D coordinate system (rectangular, cylindrical, spherical, etc), and then eliminate one of the coordinates by using the equation of the surface. The resulting formula for \(\vec R\) will have only two coordinates which means that you have a parametric surface. The choice of coordinate system will depend on the shape of the surface and the shape of the boundary of the piece of the surface you are interested in. The graph of a function, \(z=f(x,y)\) or \(z=f(r,\theta)\), can always be parametrized in rectangular or cylindrical coordinates by replacing \(z\) by \(f(x,y)\) or \(f(r,\theta)\); the choice between them will depend on the domain of the function and the shape of the boundary.

Rectangular coordinates are useful when the surface is given as the graph of a function over a region in the \(xy\)-plane given in rectangular coordinates.

Parametrize the piece of the elliptic paraboloid \(z=\dfrac{x^2+y^2}{5}\) above the rectangle \([-3,3]\times[-4,4]\). Give the range of the parameters (which will be the limits of integration in future problems).

The plot shows a bowl shape whose top edge is 4 arcs bending upward to form 4 points.

If we let \(x=u\) and \(y=v\), then the equation of the surface is \(z=\dfrac{u^2+v^2}{5}\) and the parametrization is: \[ \vec R(u,v)=\left\langle u,v,\dfrac{u^2+v^2}{5}\right\rangle \] on the rectangle, \([-3,3]\times[-4,4]\). So the parameter ranges are: \[ -3 \le u \le 3 \qquad \text{and} \qquad -4 \le v \le 4 \] Actually, there was no reason to introduce the extra variables \(u\) and \(v\). We could simply give the parametrization as \[ \vec R(x,y)=\left\langle x,y,\dfrac{x^2+y^2}{5}\right\rangle \] with parameter ranges \(-3 \le x \le 3\) and \(-4 \le y \le 4\). In this form, the parametrization is simply the rectangular coordinates \((x,y,z)\) with \(z\) replaced by \(\dfrac{x^2+y^2}{5}\).

Cylindrical coordinates are useful when the surface is a piece of a cylinder or has axial symmetry or is given as the graph of a function given in cylindrical coordinates.

Parametrize the piece of the paraboloid \(z=\dfrac{x^2+y^2}{5}\) below \(z=5\). Give the range of the parameters.

The plot shows a bowl shape whose top edge is a horizontal circle.

Start with cylindrical coordinates. Express the equation of the paraboloid in cylindrical coordinates.

The parametrization is: \[ \vec R(r,\theta)=\left\langle r\cos\theta,r\sin\theta,\dfrac{r^2}{5}\right\rangle \] The bounds are: \[ 0 \le r \le 5 \qquad \text{and} \qquad 0 \le \theta \le 2\pi \]

Since the surface (\(z=\dfrac{x^2+y^2}{5}\)) and the boundary (\(x^2+y^2=25\)) are axisymmetric, it is useful to start with cylindrical coordinates: \[ \vec R(r,\theta,z)=\left\langle r\cos\theta,r\sin\theta,z\right\rangle \] In cylindrical coordinates, the equation of the surface is \(z=\dfrac{r^2}{5}\). So the parametrization of the surface becomes \[ \vec R(r,\theta)=\left\langle r\cos\theta,r\sin\theta,\dfrac{r^2}{5}\right\rangle \] Since the boundary is \(r^2=25\) and there is no restriction on \(\theta\), the parameter ranges are: \[ 0 \le r \le 5 \qquad \text{and} \qquad 0 \le \theta \le 2\pi \]

In example 2 and exercise 3, the function is the same but the limits on the region are different. The first requires rectangular coordinates. The second requires cylindrical coordinates.

Spherical coordinates are useful when the surface is a piece of a sphere or has spherical symmetry or is given in spherical coordinates.

Parametrize the apple given in spherical coordinates by \(\rho=\phi\). Give the range of the parameters.

The plot shows an apple with its dimple at the top and a point at the bottom.

We start with spherical coordinates: \[ \vec R(\rho,\phi,\theta) =\left\langle \rho\sin\phi\cos\theta,\rho\sin\phi\sin\theta,\rho\cos\phi\right\rangle \] Since \(\rho=\phi\), we replace \(\rho\) by \(\phi\). So the parametrization of the surface is: \[ \vec R(\phi,\theta) =\left\langle \phi\sin\phi\cos\theta,\phi\sin\phi\sin\theta,\phi\cos\phi\right\rangle \] On the positive \(z\) axis, \(\rho=\phi=0\). On the negative \(z\) axis, \(\rho=\phi=\pi\). There is no restriction on \(\theta\). So the parameter ranges are: \[ 0 \le \phi \le \pi \qquad \text{and} \qquad 0 \le \theta \le 2\pi \]

Now a couple which are not quite as obvious:

Parametrize the piece of the hyperbolic paraboloid \(z=\dfrac{x^2-y^2}{5}\) within the cylinder \(x^2+y^2=25\). Give the range of the parameters.
The cylinder is shown in blue, the hyperbolic paraboloid is shown in red and their intersection curve is in yellow.

The plot shows a saddle shaped surface cutting through a vertical cylinder. They intersect in a curve which starts up, goes down, then up, then down again and back up.

Start with cylindrical coordinates and express \(z\) in turns of \(r\).

The parameterization is:   \(\vec R(r,\theta) =\left\langle r\cos\theta,r\sin\theta,\dfrac{r^2}{5}(\cos^2\theta-\sin^2\theta\right\rangle\)
and the parameter ranges are:   \(0 \le r \le 5\)   and   \(0 \le \theta \le 2\pi\)

We start with cylindrical coordinates: \[ \vec R(r,\theta,z)=\left\langle r\cos\theta,r\sin\theta,z\right\rangle \] In cylindrical coordinates, the hyperbolic paraboloid is \[ z=\dfrac{x^2-y^2}{5}=\dfrac{r^2}{5}(\cos^2\theta-\sin^2\theta) \] So the parametrization of the hyperbolic paraboloid becomes \[ \vec R(r,\theta) =\left\langle r\cos\theta,r\sin\theta,\dfrac{r^2}{5}(\cos^2\theta-\sin^2\theta)\right\rangle \] The piece of the surface is within the cylinder \(x^2+y^2=25\) or \(r^2=25\) or \(r=5\). So the parameter ranges are: \[ 0 \le r \le 5 \qquad \text{and} \qquad 0 \le \theta \le 2\pi \]

Parametrize the piece of the cylinder \(x^2+z^2=4\) between \(y=0\) and \(y=10+2x+3z\). Give the range of the parameters.
The cylinder is shown in blue. The \(y=0\) plane is cyan. The plane \(y=10+2x+3z\) is red. And the intersection curves are in yellow.

The plot shows a horizontal cylinder with a vertical plane at y = 0, which intersects the cylinder in a circle. It also shows a slanted plane which intersects the cylinder in an ellipse.

Start with cylindrical coordinates along the \(y\)-axis: \[ \vec R(r,\theta,y)=\left\langle r\cos\theta,y,r\sin\theta\right\rangle \] where \(r=\sqrt{x^2+z^2}\). Then the cylinder is \(r=2\). Express the planes in these coordinates.

The parameterization is:   \(\vec R(\theta,y)=\left\langle 2\cos\theta,y,2\sin\theta\right\rangle\)
and the parameter ranges are:   \(0 \le \theta \le 2\pi\)   and   \(0 \le y \le 10+4\cos\theta+6\sin\theta\)

Since the cylinder is along the \(y\)-axis, we start with cylindrical coordinates along the \(y\)-axis: \[ \vec R(r,\theta,y)=\left\langle r\cos\theta,y,r\sin\theta\right\rangle \] where \(r=\sqrt{x^2+z^2}\). Since the equation of the cylinder is \(x^2+z^2=4\) or \(r^2=4\), we plug \(r=2\) into parametrization to get: \[ \vec R(\theta,y)=\left\langle 2\cos\theta,y,2\sin\theta\right\rangle \] The angle, \(\theta\), is unrestricted. To get the \(y\)-ranges, we plug \(x=2\cos\theta\) and \(z=2\sin\theta\) into the boundaries \(y=0\) and \(y=10+2x+3z\). So the ranges are: \[ 0 \le \theta \le 2\pi \qquad \text{and} \qquad 0 \le y \le 10+4\cos\theta+6\sin\theta \]

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