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Section 7.3 Week 7 Activity

Subsection 7.3.1 Parametric Surfaces and Areas

Activity 7.3.1.

Consider the portion of the plane \(3x - 4y - z = 4\) which lies above the interval \([3,6] \times [-1,4]\) in the \(xy\) plane.

  • Describe this as a parametric surface.

  • Calculate its surface area.

Solution.

I would like to use Example 7.1.2 to parametrization this. To make this the graph of a function, I can solve for \(z\) in the equation of the plane as \(z = 3x-4y-4\text{.}\) The bounds of the parameter domain are precisely the bounds on \(x\) and \(y\) given in the descriptions.

\begin{align*} \sigma(u,v) \amp = (u,v,3u - 4v - 4) \\ D \amp = (u,v) \in [3,6] \times[-1,4] \end{align*}

To calculate the surface area, I'll need the partial derivative, the normal, and the length of the normal.

\begin{align*} \sigma_u \amp = (1,0,3) \\ \sigma_v \amp = (0,1,-4) \\ \sigma_u \times \sigma_v \amp = (-3,4,1) \\ |\sigma_u \times \sigma_v| \amp = \sqrt{9+16+1} = \sqrt{26} \end{align*}

Then I can calculate the surface area by integrating the length of the normal over the parameter domain.

\begin{align*} A \amp = \int_D |\sigma_u \times \sigma_v| du dv \\ \amp = \int_{-1}^4 \int_3^6 \sqrt{26} du dv = \sqrt{26} (4-(-1))(6-3)\\ \amp= 15 \sqrt{26} \end{align*}

Activity 7.3.2.

Consider the sphere of radius \(4\) centred at \((-1,-2,3)\text{.}\)

  • Describe this as a parametric surface.

  • Calculate its surface area.

Solution.

From Example 7.1.4, I already have the parametrization of the sphere. However, now I have shifted the centre. How to I adjust the parametrization in the example? I can simply at the offset to each coordinate, which moves each coordinate to give the correct centre point.

\begin{align*} \sigma(\phi,v) \amp = (4 \sin \phi \cos v - 1, 4 \sin \phi \sin v - 2, 4 \cos \phi + 3) \\ D \amp = (\phi,\theta) \in [0,\pi] \times [0, 2\pi] \\ \sigma_\phi \amp = (4 \cos \phi \cos \theta, 4 \cos \phi \sin \theta, -4 \sin \phi)\\ \sigma_\theta \amp = (-4 \sin \phi \sin \theta, 4 \sin \phi \cos \theta, 0)\\ \sigma_\phi \times \sigma_\theta \amp = (16 \sin^2 \phi \cos \theta, 16 \sin^2 \phi \sin \theta, 16 \cos \phi \sin \phi )\\ |\sigma_\phi \times \sigma_\theta| \amp = 16 \sqrt{\sin^4 \phi \cos^2 \theta+ \sin^4 \phi \sin^2 \theta + \cos^2 \phi \sin^2\phi}\\ \amp = 16 \sqrt{\sin^4 \phi + \cos^2 \phi \sin^2 \phi } = 16 \sqrt{\sin^2 \phi} = 16 \sin \phi \end{align*}

Then I can calculate the surface area by integrating the length of the normal over the parameter domain.

\begin{align*} A = \amp = \int_0^{2\pi} \int_0^{\pi} 16 \sin \phi d\phi d\theta = (16) 2\pi (-\cos \phi) \bigg|_0^\pi = 64 \pi \\ \int_\sigma f dA \amp = \int_0^{2\pi} \int_0^\pi (4\sin \phi \cos \theta - 1)(4 \sin \phi \sin \theta - 2) (4 \cos \phi + 3)(16 \sin \phi) d\phi d\theta \end{align*}

This gets very intense. I asekd a computer algebra system for the expanded form.

\begin{align*} \amp = \int_0^{2\pi} \int_0^\pi -199 \sin^2 \phi \sin \theta + 1024 \sin^3\phi \cos \phi \sin \theta \cos \theta + 768 \sin^3 \phi \sin \theta \cos \theta \\ \amp - 512 \sin^2 \phi \cos \phi \cos \theta - 384 \sin^2 \phi \cos \theta - 256 \sin^2\phi \cos \phi \sin \theta\\ \amp + 96 \sin \phi + 128 \sin \phi \cos \phi d\phi d\theta \end{align*}

This is a bit ridiculous, but the bounds of \(\theta\) are \([0,2\pi]\text{.}\) Therefore, any integral involving \(\sin \theta\text{,}\) \(\cos \theta\text{,}\) or \(\sin \theta \cos \theta\) is zero by symmetry. That (thankfully) removes all the but last two integrals.

\begin{align*} \amp = \int_0^{2\pi} \int_0^\pi 96 \sin \phi + 128 \sin \phi \cos \phi d\phi d\theta \\ \amp = 96 \int_0^{2\pi} d\theta \int_0^\pi \sin d\phi + 128 \int_0^{2\pi} d\theta \int_0^\pi \sin \phi \cos \phi d\phi d\theta = 192 \end{align*}

Activity 7.3.3.

Consider the ellipsoid centred at the origin described by the following locus.

\begin{equation*} \frac{x^2}{9} + y^2 + \frac{z^2}{25} = 1 \end{equation*}

  • Describe this as a parametric surface.

  • Calculate its surface area. (Start the integral, but you can stop when you get to a point where the integration is impossibly by elementary functions.)

Solution.

The approach is like previous examples for ellipses and ellipsoids. I use a version of spherical coordinates where I adjust the coordinates of \(x\text{,}\) \(y\) and \(z\) by a scaling factor.

\begin{align*} \sigma(\phi,\theta) \amp = (3 \sin \phi \cos \theta, \sin \phi \sin \theta, 5 \cos \phi) \\ D \amp = (\phi,\theta) \in [0,\pi] \times [0, 2\pi] \end{align*}

To calculate the surface area, I'll need the partial deri\theta, the normal, and the length of the normal.

\begin{align*} \sigma_\phi \amp = (3 \cos \phi \cos \theta, \cos \phi \sin \theta, -5 \sin \phi)\\ \sigma_\theta \amp = (-3 \sin \phi \sin \theta, \sin \phi \cos \theta, 0) \\ \sigma_\phi \times \sigma_\theta \amp = (5\sin^2 \phi \cos \theta, 15 \sin^2 \phi \sin \theta, 3\cos \phi \sin \phi)\\ |\sigma_\phi \times \sigma_\theta| \amp = \sqrt{25 \sin^4 \phi \cos^2 \theta+ 225 \sin^4 \phi \sin^2 \theta + 9 \cos^2 \phi \sin^2 \phi} \end{align*}

I can already see a problem arising here. The scalaing factors mess with the lovely trig identities that simplify this square root in the spherical case. I can't get rid of this square root. The integral is, like other elliptic integrals, impossibly by elementary functions.

\begin{align*} A \amp = \int_0^{2\pi} \int_0^{\pi} \sqrt{25 \sin^4 \phi \cos^2 \theta + 225 \sin^4 \phi \sin^2 \theta + 9 \cos^2 \phi \sin^2 \phi} d\phi d\theta \end{align*}

This is as far as I can go with elementary functions. Notice the parallel with the circle/ellipse perimeter problem. The circumference of the circle is easy to calculate with parametric curves, but the circumference of the ellipse leads to impossible elliptic integrals. The same is true here for parametric surfaces, spheres and ellipsoids.

Activity 7.3.4.

Consider the portion of hyperboloid \(x^2 + y^2 -z^2 = 1\) which has \(z \in [-3,3]\text{.}\)

  • Describe this as a parametric surface.

  • Calculate its surface area. (Start the integral, but you can stop when you get to a point where the integration is impossibly by elementary functions.)

Solution.

There are a variety of ways to approach this. I could take \(x\) and \(y\) as parameter and solve for \(z\text{;}\) however, the square root gives multiple values, so I can't get a whole parametric description this way. I could also treat this as a surface of revolution about the \(z\) axis and use Example 7.1.3. Notice that, in the example, the axis of revolution was the \(x\) axis. I'll just adjust for rotation around the \(z\) axis by switching the variables.

However, I'm going to do something else entirely. This hyperboloid is very similar to the sphere. I'd like to parallel the coordinate system of the sphere to get a coordinate system for this, building a kind-of hyperbolic coordinate system. For rotation about the \(z\) axis, this is still like the longitude of the sphere. However, instead of the coltatitue of the sphere, I have hyperbolic cross-section for vertical slices. Therefore, I can use hyperbolic functions for the ‘colatitude’ instead of trig function. That leads to the following parametrization.

\begin{align*} \sigma(\phi,\theta) \amp = (\cosh \phi \sin \theta, \cosh \phi \cos \theta, \sinh \phi)\\ D \amp = (\phi,\theta) \in [\arcsinh (-3), \arcsinh (3)] \times [0, 2\pi] \end{align*}

To calculate the surface area, I'll need the partial derivative, the normal, and the length of the normal.

\begin{align*} \sigma_\phi \amp = (\sinh \phi \sin \theta, \sinh \phi \cos \theta, \cosh \phi \phi) \\ \sigma_\theta \amp = (\cosh \phi \cos \theta, -\cosh \phi \sin \theta, 0) \\ \sigma_\phi \times \sigma_\theta \amp = (\cosh^2 \phi \sin \theta, \cosh^2 \phi \cos \theta, -\sinh \phi \cosh \phi )\\ |\sigma_\phi \times \sigma_\theta| \amp = \sqrt{\cosh^4 \phi \sin^2 \theta + \cosh^4 \phi \cos^2 \theta + \sinh^2 \phi \cosh^2 \phi}\\ \amp = \cosh \phi \sqrt{\cosh^2 \phi + \sinh^2 \phi} \end{align*}

Here I run into a similar program as previous question. The situation is not as nasty, as some simplification does occur. But there is no identity for \(\sinh^2 \phi + \cosh^2 \phi\) which can remove the square root (unlike the case of the sphere). Therefore, like the last question, I am stuck with this square root. For the surace area integral, this will lead to integral without elementary anti-derivatives.

\begin{align*} A \amp = \int_0^{2\pi} \int_{\arcsinh (-3)}^{\arcsinh (3)} \cosh \phi \sqrt{\cosh^2 \phi + \sinh^2 \phi} d\phi d\theta \end{align*}

Activity 7.3.5.

Consider the portion of the cuspoidal horn \(z^3 = x^2 + y^2\) which has \(z \in [0,4]\text{.}\)

  • Describe this as a parametric surface.

  • Calculate its surface area. (Start the integral, but you can stop when you get to a point where the integration is impossibly by elementary functions.)

Solution.

I'd like to treat this as a surface of revolution about the \(z\) axis. I'll use the setup from Example 7.1.3, but I switch the variables to reflect the axis of revolution.

\begin{align*} \sigma(z,\theta) \amp = \left (z^{\frac{3}{2}} \cos \theta, z^{\frac{3}{2}} \sin \theta, z \right)\\ D \amp = (z,\theta) \in [0,4] \times [0,2\pi] \end{align*}

To calculate the surface area, I'll need the partial derivative, the normal, and the length of the normal.

\begin{align*} \sigma_z \amp = \left( \frac{3}{2} z^{\frac{1}{2}} \cos \theta, \frac{3}{2} z^{\frac{1}{2}} \sin \theta, 1 \right) \\ \sigma_\theta \amp = \left( -z^{\frac{3}{2}} \sin \theta, z^{\frac{3}{2}} \cos \theta, 0 \right) \\ \sigma_z \times \sigma_\theta \amp = \left( -z^{\frac{3}{2}} \cos \theta, -z^{\frac{3}{2}} \sin \theta, \frac{3}{2}z^2 \right) \\ |\sigma_z \times \sigma_\theta| \amp = \sqrt{z^3 + \frac{9z^4}{4}} \end{align*}

Again, the problems are similar. I don't have a nice technique like trigononmetric subsitution to deal with this strange square root term. It leads to non-elementary antiderivative for the surface area integarl.

\begin{align*} A = \int_D |\sigma_z \times \sigma_\theta| dz d\theta \amp = \int_0^{2\pi} \int_0^4 \sqrt{z^3 + \frac{9z^4}{4}} dz d\theta \end{align*}

Subsection 7.3.2 Conceptual Review Questions

  • What is a parametric surface?

  • How does the normal to a parametric surface determine its properties?