Week 11 Activity

Section 11.3 Week 11 Activity

Subsection 11.3.1 Parametric Surfaces and Areas

Activity 11.3.1.

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

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Describe this as a parametric surface.
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Calculate its surface area.
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Solution.

I would like to use Example 11.1.2 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.

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\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.

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\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.

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\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*}

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Activity 11.3.2.

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

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Describe this as a parametric surface.
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Calculate its surface area.
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Solution.

From Example 11.1.4, 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.

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\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.

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\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.

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\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.

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\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*}

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Activity 11.3.3.

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

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

Describe this as a parametric surface.
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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.)
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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.

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\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.

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\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.

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\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.

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Activity 11.3.4.

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

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Describe this as a parametric surface.
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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.)
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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 11.1.3. 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.

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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.

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\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.

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\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.

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\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*}

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Activity 11.3.5.

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

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Describe this as a parametric surface.
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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.)
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Solution.

I’d like to treat this as a surface of revolution about the \(z\) axis. I’ll use the setup from Example 11.1.3, but I switch the variables to reflect the axis of revolution.

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\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.

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\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.

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\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*}

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Subsection 11.3.2 Activity

Activity 11.3.6.

Consider this vector field.

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\begin{equation*} F(x,y,z) = (4z - x, 3y + 3z, x + z) \end{equation*}

Calculate the flux of this vector field over the following surface.

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The rectangle with vertices \((0,0)\text{,}\) \((0,4)\text{,}\) \((3,0)\) and \((3,4)\) in the \(xy\) plane.
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The unit sphere centred at the origin.
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The open cylinder about the \(z\) axis with radius \(3\) and \(z \in [0,2]\text{.}\)
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The same cylinder as the previous point, but now also including the top and bottom discs to make it a closed cylinder.
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Solution.

For each part, I need to parametrize the surface, calculate the normal, then use the normal and the field to calculate the flux integral.

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The rectangle can be parametrized by the variables \(x\) and \(y\) with \(z\) set to constant zero.
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\begin{align*} \sigma(x,y) \amp = (x,y,0) \amp \amp (x,y) \in [0,3] \times [0,4] \\ \sigma_x (x,y) \amp = (1,0,0) \amp \amp \\ \sigma_y (x,y) \amp = (0,1,0) \amp \amp \\ \sigma_x \times \sigma_y (x,y) \amp = (0,0,1) \end{align*}

Then I calculate the flux integral.
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\begin{align*} \int_\sigma F \cdot dA \amp = \int_D F(\sigma(x,y)) \cdot (\sigma_x \times \sigma_y) dx dy\\ \amp = \int_0^4 \int_0^3 (-u, 3v, u) \cdot (0,0,1) dx dy \\ \amp = \int_0^4 \int_0^3 x dx dy = 4 \frac{9}{2} = 18 \end{align*}

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The sphere can be parametrized by following Example 11.1.4. Here, the radius is one.
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\begin{align*} \sigma(\theta,\phi) \amp = (\sin \phi \cos \theta, \sin \phi \sin \theta, \cos \phi) \\ \amp (\theta,\phi) \in [0,2\pi] \times [0,\pi] \\ \sigma_\theta (\theta,\phi) \amp = (-\sin \phi \sin \theta, \sin \phi \sin \theta, 0) \\ \sigma_\phi (\theta,\phi) \amp = (\cos \phi \cos \theta, \cos \phi \sin \theta, - \sin \phi)\\ \sigma_\theta \times \sigma_\phi (u,\phi) \amp = -\sin \phi (\sin \phi \cos \theta, \sin \phi \sin \theta, \cos \phi) \end{align*}

Then I calculate the flux integral.
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\begin{align*} \amp \int_\sigma F \cdot dA = \int_D F(\sigma(\theta,\phi)) \cdot (\sigma_\theta \times \sigma_\phi) d\theta d\phi\\ \amp = \int_0^{2\pi} \int_0^{\pi} (4 \cos \phi - \sin \phi \cos \theta, 3 \sin \phi \sin \theta + 3 \cos \phi, \sin \phi \cos \theta + \cos \phi) \\ \amp \cdot -\sin \phi (\sin \phi \cos \theta, \sin \phi \sin \theta, \cos \phi) d \phi d \theta \\ \amp = \int_0^{2\pi} \int_0^{\pi} 4 \cos \phi \sin^2 \phi \cos \theta- \sin^3 \phi \cos^2 \phi + 3 \sin^3 \phi \sin^2 \theta \\ \amp + 3 \cos \phi \sin^2 \phi \sin \theta + \sin^2 \phi \cos \phi \cos \theta + \sin \phi \cos^2 \theta d \phi d \theta \end{align*}

There are many terms in this integral. I’m going to use some symmetry to simplify before I calculate. Any integral that involves an odd power of \(\sin \theta\text{,}\) \(\cos \theta\text{,}\) or \(\sin \theta \cos \theta\) is the integral of a trig function over a whole period, which must vanish by symmetry. Since the \(\phi\) and \(\theta\) integrals of each piece of the sum are individually seperably, if the \(\theta\) integral is zero, the integral of the whole term is zero. Therefore, I’ll remove any term that satisfies my criterion. What’s left over is this integral, which I split up by linearity, seperate, and solve. I’ll ask a computer for the values of each single-variable definite integral.
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\begin{align*} \amp = \int_0^{2\pi} \int_0^\pi -\sin^3 \phi \cos^2 \theta + 3 \sin^3 \phi \sin^2 \theta + \sin \phi \cos^2 \phi d \phi d \theta \\ \amp = \int_0^{2\pi} \cos^2 \theta d \theta \int_0^\pi -\sin^3 \phi d \phi + \int_0^{2\pi} \sin^2 \theta d \theta \int_0^\pi 3 \sin^3 \phi d \phi \\ \amp + \int_0^{2\pi} d\theta \int_0^\pi \sin \phi \cos^2 \phi d \phi \\ \amp = (\pi) \left( \frac{-4}{3} \right)+ (\pi)(4) + (2\pi) \left( \frac{2}{3} \right) = 4\pi \end{align*}

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The cylinder can be parametrized following Example 11.1.5. The radius here is \(3\text{.}\) The range for \(z\) is \([0,2]\) as given in the question.
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\begin{align*} \sigma(\theta,z) \amp = (3 \cos \theta, 3 \sin \theta, z) \amp \amp (\theta,z) \in [0,2\pi] \times [0,2] \\ \sigma_\theta (\theta,z) \amp = (-3 \sin \theta, 3 \cos \theta, 0) \amp \amp \\ \sigma_z (\theta,z) \amp = (0,0,1) \amp \amp \\ \sigma_\theta \times \sigma_z (\theta,z) \amp = (3 \cos \theta, 3 \sin \theta, 0) \end{align*}

Then I calculate the flux integral.
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\begin{align*} \amp \int_\sigma F \cdot dA = \int_D F(\sigma(\theta,z)) \cdot (\sigma_\theta \times \sigma_z) d\theta dz\\ \amp = \int_0^2 \int_0^{2\pi} (4z - 3 \cos \theta, 9 \sin \theta + 4 z, 3 \cos \theta + z) \cdot (3 \cos \theta, 3 \sin \theta, 0) \\ \amp = \int_0^2 \int_0^{2\pi} 12 z \cos \theta - 9\cos^2 \theta + 27 \sin^2 \theta + 12 z \sin \theta d \theta dz \end{align*}

The integrals of the first and last term are integrals of sine or cosine over a whole period, therefore must be zero. What remains is the two middle integrals. Both are separable and the \(z\) integral is trivial.
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\begin{align*} \amp = -9 \int_0^2 \int_0^{2\pi} \cos^2 \theta d \theta dz + 27 \int_0^2 \int_0^{2\pi} \sin^2 \theta d \theta dz \\ \amp = -18 \int_0^{2\pi} \cos^2 \theta d \theta + 54 \int_0^{2\pi} \sin^2 \theta d \theta = -18\pi + 54\pi = 26\pi \end{align*}

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There are three pieces to the surface. The first piece is the cylinder calcualted in the previous part of this question. I don’t need to recalculate this flux integral. The other two parts are the top disc (with an upward, outward normal) and the bottom disc (again with a downward, outward normal). The discs can be parametrized using something like polar coordinates with a constant \(z\) term. I must make sure I order the variables to give the correct direction of the normal. I’ll start with the top disc.
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\begin{align*} \sigma(r,\theta) \amp = (r \cos \theta, r \sin \theta, 2) \amp \amp (r,\theta) \in [0,3] \times [0,2\pi] \\ \sigma_r (r,\theta) \amp = (\cos \theta, \sin \theta, 0) \amp \amp \\ \sigma_\theta (r,\theta) \amp = (-r \sin \theta, r \cos \theta, 0) \amp \amp \\ \sigma_r \times \sigma_\theta (r,\theta) \amp = (0,0,r) \end{align*}

Then I calculate the flux integral.
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\begin{align*} \int_\sigma F \cdot dA \amp = \int_D F(\sigma(r,\theta)) \cdot (\sigma_r \times \sigma_\theta) dr d\theta\\ \amp = \int_0^{2\pi} \int_0^3 (8 - r \cos \theta, 3 r \sin \theta + 9, r \cos \theta + 3) \cdot (0,0,r) dr d\theta\\ \amp = \int_0^{2\pi} \int_0^3 r^2 \cos \theta + 3r dr d\theta = 0 + 2\pi \frac{27}{2} = 27 \pi \end{align*}

Then I do the bottom disc. I reverse the variable order to get a downward pointing normal.
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\begin{align*} \sigma(\theta,r) \amp = (r \cos \theta, r \sin \theta, 0) \amp \amp (r,\theta) \in [0,3] \times [0,2\pi] \\ \sigma_\theta (\theta,r) \amp = (-r \sin \theta, r \cos \theta, 0) \amp \amp \\ \sigma_r (\theta,r) \amp = (\cos \theta, \sin \theta, 0) \amp \amp \\ \sigma_\theta \times \sigma_r (\theta,r) \amp = (0,0,-r) \end{align*}

Then I calculate the flux integral.
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\begin{align*} \int_\sigma F \cdot dA \amp = \int_D F(\sigma(r,\theta)) \cdot (\sigma_r \times \sigma_\theta) dr d\theta\\ \amp = \int_0^{2\pi} \int_0^3 (-r\cos \theta, 3v \sin \theta, r \cos \theta) \cdot (0,0,-r) dr d\theta \\ \amp = \int_0^{2\pi} \int_0^3 -r^2 \cos \theta dr d\theta = 0 \end{align*}

Flux over the entire closed cylinder is the sum of the flux other the three pieces.
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\begin{equation*} \int_C F \cdot da = 27 \pi + 26 \pi + 0 = 53 \pi \end{equation*}

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Activity 11.3.7.

Consider this vector field.

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\begin{equation*} F(x,y,z) = (2y, -2x, 2x + 2y) \end{equation*}

Calculate the flux of this vector field over the following surface.

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The rectangle with vertices \((0,0)\text{,}\) \((0,4)\text{,}\) \((3,0)\) and \((3,4)\) in the \(xy\) plane.
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The unit sphere centred at the origin.
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The open cylinder about the \(z\) axis with radius \(3\) and \(z \in [0,2]\text{.}\)
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The same cylinder as the previous point, but now also including the top and bottom discs to make it a closed cylinder.
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Solution.

For each part, I need to parametrize the surface, calculate the normal, then use the normal and the field to calculate the flux integral.

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The rectangle can be parametrized by the variables \(x\) and \(y\) with \(z\) set to constant zero.
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\begin{align*} \sigma(x,y) \amp = (x,y,0) \amp \amp (x,y) \in [0,3] \times [0,4] \\ \sigma_x (x,y) \amp = (1,0,0) \amp \amp \\ \sigma_y (x,y) \amp = (0,1,0) \amp \amp \\ \sigma_x \times \sigma_y (x,y) \amp = (0,0,1) \end{align*}

Then I calculate the flux integral.
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\begin{align*} \int_\sigma F \cdot dA \amp = \int_D F(\sigma(x,y)) \cdot (\sigma_x \times \sigma_y) dx dy\\ \amp = \int_0^4 \int_0^3 (2y, -2x, 2x + 2y) \cdot (0,0,1) dx dy \\ \amp = \int_0^4 \int_0^3 2x + 2y dx dy = \int_0^4 (x^2 + 2xy) \Bigg|_0^3 dy \\ \amp = \int_0^4 9 + 6y dy = 36 + 48 = 84 \end{align*}

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The sphere can be parametrized by following Example 11.1.4. Here, the radius is one.
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\begin{align*} \sigma(\theta,\phi) \amp = (\sin \phi \cos \theta, \sin \phi \sin \theta, \cos \phi) \\ \amp (\theta,\phi) \in [0,2\pi] \times [0,\pi] \\ \sigma_\theta (\theta,\phi) \amp = (-\sin \phi \sin \theta, \sin \phi \sin \theta, 0) \\ \sigma_\phi (\theta,\phi) \amp = (\cos \phi \cos \theta, \cos \phi \sin \theta, - \sin \phi \\ \sigma_\theta \times \sigma_\phi (u,\phi) \amp = -\sin \phi (\sin \phi \cos \theta, \sin \phi \sin \theta, \cos \phi) \end{align*}

Then I calculate the flux integral.
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\begin{align*} \amp \int_\sigma F \cdot dA = \int_D F(\sigma(\theta,\phi)) \cdot (\sigma_\theta \times \sigma_\phi) d\theta d\phi\\ \amp = \int_0^{2\pi} \int_0^\pi (2 \sin \phi \sin \theta, -2 \sin \phi \cos \theta, 2 \sin \phi \cos \theta + 2 \sin \phi cos \theta) \\ \amp \cdot -\sin \phi( \sin \phi \cos \theta, \sin \phi \sin \theta, \cos \phi) d \phi d \theta \\ \amp = \int_0^{2\pi} \int_0^\pi 2 \sin^3 \phi \sin \theta \cos \theta - 2 \sin^2 \phi \sin \theta \cos \theta \\ \amp + 2 \sin^2 \phi \cos \phi \cos \theta + 2 \sin^2 \phi \cos \phi \sin \theta \end{align*}

By the same symmetry arguments at the previous problem, I can remove any terms in the integral which have \(\sin \theta\text{,}\) \(\cos \theta\) and \(\sin \theta \cos \theta\text{.}\) However, all of the terms have one of these expressions in \(\theta\text{,}\) so the entire integral is zero.
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The cylinder can be parametrized following Example 11.1.5. The radius here is \(3\text{.}\) The range for \(z\) is \([0,2]\) as given in the question.
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\begin{align*} \sigma(\theta,z) \amp = (3 \cos \theta, 3 \sin \theta, z) \amp \amp (\theta,z) \in [0,2\pi] \times [0,2] \\ \sigma_\theta (\theta,z) \amp = (-3 \sin \theta, 3 \cos \theta, 0) \amp \amp \\ \sigma_z (\theta,z) \amp = (0,0,1) \amp \amp \\ \sigma_\theta \times \sigma_z (\theta,z) \amp = (R \cos \theta, R \sin \theta, 0) \end{align*}

Then I calculate the flux integral.
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\begin{align*} \int_\sigma F \cdot dA \amp = \int_D F(\sigma(\theta,z)) \cdot (\sigma_\theta \times \sigma_z) d\theta dz\\ \amp = \int_0^2 \int_0^{2\pi} (2 z \sin \theta, -2 z \cos \theta, 2 z \cos \theta + 2 z \sin \theta) \cdot (0,0,z) d \theta dz \\ \amp = \int_0^2 \int_0^{2\pi} 2 z^2 \cos \theta + 2 z^2 \sin \theta d \theta d z \end{align*}

Again, by symmetry of integrating sine or cosine over a full period, both terms in this integral vanish and the integral is zero.
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There are three pieces to the surface. The first piece is the cylinder calcualted in the previous part of this question. I don’t need to recalculate this flux integral. The other two parts are the top disc (with an upward, outward normal) and the bottom disc (again with a downward, outward normal). The discs can be parametrized using something like polar coordinates with a constant \(z\) term. I must make sure I order the variables to give the correct direction of the normal. I’ll start with the top disc.
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\begin{align*} \sigma(r,\theta) \amp = (r \cos \theta, r \sin \theta, 2) \amp \amp (r,\theta) \in [0,3] \times [0,2\pi] \\ \sigma_r (r,\theta) \amp = (\cos \theta, \sin \theta, 0) \amp \amp \\ \sigma_\theta (r,\theta) \amp = (-r \sin \theta, r \cos \theta, 0) \amp \amp \\ \sigma_r \times \sigma_\theta (r,\theta) \amp = (0,0,r) \end{align*}

Then I calculate the flux integral.
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\begin{align*} \int_\sigma F \cdot dA \amp = \int_D F(\sigma(r,\theta)) \cdot (\sigma_r \times \sigma_\theta) dr d\theta\\ \amp = \int_0^3 \int_0^{2\pi} (2 r \sin \theta, -2 r \cos \theta, 2r \cos \theta + 2r \sin \theta) d\theta dr \\ \amp = \int_0^3 \int_0^{2\pi} 2 r^2 \cos \theta + 2 r^2 \sin \theta d\theta dr \end{align*}

Again, by symmetry of integrating over a full period, both terms of the integral vanish.
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Then I do the bottom disc. I reverse the variable order to get a downward pointing normal.
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\begin{align*} \sigma(\theta,r) \amp = (r \cos \theta, r \sin \theta, 0) \amp \amp (r,\theta) \in [0,3] \times [0,2\pi] \\ \sigma_\theta (\theta,r) \amp = (-r \sin \theta, r \cos \theta, 0) \amp \amp \\ \sigma_r (\theta,r) \amp = (\cos \theta, \sin \theta, 0) \amp \amp \\ \sigma_\theta \times \sigma_r (\theta,r) \amp = (0,0,-r) \end{align*}

Then I calculate the flux integral.
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\begin{align*} \int_\sigma F \cdot dA \amp = \int_D F(\sigma(r,\theta)) \cdot (\sigma_r \times \sigma_\theta) dr d\theta\\ \amp = \int_0^3 \int_0^{2\pi} (2 r \sin \theta, -2 r \cos \theta, 2 r \cos \theta + 2 r \sin \theta) d\theta dr \\ \amp = \int_0^3 \int_0^{2\pi} 2 r^2 \cos \theta + 2r^2 \sin \theta d\theta dr = 0 \end{align*}

Flux over the entire closed cylinder is the sum of the flux other the three pieces.
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\begin{equation*} \int_C F \cdot da = 0 + 0 + 0 = 0 \end{equation*}

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Activity 11.3.8.

Consider this vector field.

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\begin{equation*} F(x,y,z) = (x^2, y^2, z^2) \end{equation*}

Calculate the flux of this vector field over the following surface.

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The rectangle with vertices \((0,0)\text{,}\) \((0,4)\text{,}\) \((3,0)\) and \((3,4)\) in the \(xy\) plane.
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The unit sphere centred at the origin.
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The open cylinder about the \(z\) axis with radius \(3\) and \(z \in [0,2]\text{.}\)
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The same cylinder as the previous point, but now also including the top and bottom discs to make it a closed cylinder.
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Solution.

For each part, I need to parametrize the surface, calculate the normal, then use the normal and the field to calculate the flux integral.

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The rectangle can be parametrized by the variables \(x\) and \(y\) with \(z\) set to constant zero.
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\begin{align*} \sigma(x,y) \amp = (x,y,0) \amp \amp (x,y) \in [0,3] \times [0,4] \\ \sigma_x (x,y) \amp = (1,0,0) \amp \amp \\ \sigma_y (x,y) \amp = (0,1,0) \amp \amp \\ \sigma_x \times \sigma_y (x,y) \amp = (0,0,1) \end{align*}

Then I calculate the flux integral.
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\begin{align*} \int_\sigma F \cdot dA \amp = \int_D F(\sigma(x,y)) \cdot (\sigma_x \times \sigma_y) dx dy\\ \amp = \int_0^4 \int_0^3 (x^2, y^2, 0) \cdot (0,0,1) dy dz = \int_0^4 \int_0^3 0 dy dx = 0 \end{align*}

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The sphere can be parametrized by following Example 11.1.4. Here, the radius is one.
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\begin{align*} \sigma(\theta,\phi) \amp = (\sin \phi \cos \theta, \sin \phi \sin \theta, \cos \phi) \\ \amp (\theta,\phi) \in [0,2\pi] \times [0,\pi] \\ \sigma_\theta (\theta,\phi) \amp = (-\sin \phi \sin \theta, \sin \phi \sin \theta, 0) \\ \sigma_\phi (\theta,\phi) \amp = (\cos \phi \cos \theta, \cos \phi \sin \theta, - \sin \phi \\ \sigma_\theta \times \sigma_\phi (u,\phi) \amp = -\sin \phi (\sin \phi \cos \theta, \sin \phi \sin \theta, \cos \phi) \end{align*}

Then I calculate the flux integral.
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\begin{align*} \int_\sigma F \cdot dA \amp = \int_D F(\sigma(\theta,\phi)) \cdot (\sigma_\theta \times \sigma_\phi) d\theta d\phi\\ \amp = \int_0^{2\pi} \int_0^\pi (\sin^2 \phi \cos^2 \theta, \sin^2 \phi \sin^2 \theta, \cos^2 \phi) \\ \amp \cdot -\sin \phi (\sin \phi \cos \theta, \sin \phi \sin \theta, \cos \phi) d\phi d \theta \\ \amp = \int_0^{2\pi} \int_0^\pi \sin^4 \phi \cos^3 \theta + \sin^4 \phi \sin^3 \theta + \cos^3 \phi \sin \phi d\phi d \theta \end{align*}

The first two of the three terms all evaluate to zero by the now very familiar symmetry arguments for the \(\theta\) integrals. All that remains is the third term.
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\begin{align*} \amp = \int_0^{2\pi} \int_0^\pi \cos^3 \phi \sin \phi d \phi d \theta = 2\pi \int_0^\pi \cos^3 \phi \sin \phi d \phi = 2\pi (0) = 0 \end{align*}

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The cylinder can be parametrized following Example 11.1.5. The radius here is \(3\text{.}\) The range for \(z\) is \([0,2]\) as given in the question.
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\begin{align*} \sigma(\theta,z) \amp = (3 \cos \theta, 3 \sin \theta, z) \amp \amp (\theta,z) \in [0,2\pi] \times [0,2] \\ \sigma_\theta (\theta,z) \amp = (-3 \sin \theta, 3 \cos \theta, 0) \amp \amp \\ \sigma_z (\theta,z) \amp = (0,0,1) \amp \amp \\ \sigma_\theta \times \sigma_z (\theta,z) \amp = (R \cos \theta, R \sin \theta, 0) \end{align*}

Then I calculate the flux integral.
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\begin{align*} \int_\sigma F \cdot dA \amp = \int_D F(\sigma(\theta,z)) \cdot (\sigma_\theta \times \sigma_z) d\theta dz\\ \amp = \int_0^2 \int_0^{2\pi} (9 \cos^2 \theta, 9 \sin^2 \theta, z^2) \cdot (3 \cos \theta, 3 \sin \theta, 0) d\theta dz \\ \amp = \int_0^2 \int_0^{2\pi} 27 \cos^3 \theta + 27 \sin^3 \theta d \theta \end{align*}

By symmetry, both integrals are zero.
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There are three pieces to the surface. The first piece is the cylinder calcualted in the previous part of this question. I don’t need to recalculate this flux integral. The other two parts are the top disc (with an upward, outward normal) and the bottom disc (again with a downward, outward normal). The discs can be parametrized using something like polar coordinates with a constant \(z\) term. I must make sure I order the variables to give the correct direction of the normal. I’ll start with the top disc.
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\begin{align*} \sigma(r,\theta) \amp = (r \cos \theta, r \sin \theta, 2) \amp \amp (r,\theta) \in [0,3] \times [0,2\pi] \\ \sigma_r (r,\theta) \amp = (\cos \theta, \sin \theta, 0) \amp \amp \\ \sigma_\theta (r,\theta) \amp = (-r \sin \theta, r \cos \theta, 0) \amp \amp \\ \sigma_r \times \sigma_\theta (r,\theta) \amp = (0,0,r) \end{align*}

Then I calculate the flux integral.
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\begin{align*} \int_\sigma F \cdot dA \amp = \int_D F(\sigma(r,\theta)) \cdot (\sigma_r \times \sigma_\theta) dr d\theta\\ \amp = \int_0^3 \int_0^{2\pi} (r^2 \cos^2 \theta, r ^2 \sin^2 \theta, 4) \cdot (0,0,r) d\theta dr \\ \amp = \int_0^3 \int_0^{2\pi} 4r dr d\theta = 2\pi (18) = 36\pi \end{align*}

Then I do the bottom disc. I reverse the variable order to get a downward pointing normal.
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\begin{align*} \sigma(\theta,r) \amp = (r \cos \theta, r \sin \theta, 0) \amp \amp (r,\theta) \in [0,3] \times [0,2\pi] \\ \sigma_\theta (\theta,r) \amp = (-r \sin \theta, r \cos \theta, 0) \amp \amp \\ \sigma_r (\theta,r) \amp = (\cos \theta, \sin \theta, 0) \amp \amp \\ \sigma_\theta \times \sigma_r (\theta,r) \amp = (0,0,-r) \end{align*}

Then I calculate the flux integral.
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\begin{align*} \int_\sigma F \cdot dA \amp = \int_D F(\sigma(r,\theta)) \cdot (\sigma_r \times \sigma_\theta) dr d\theta\\ \amp = \int_0^3 \int_0^{2\pi} (r^2 \cos^2 \theta, r^2 \sin^2 \theta, 0) \cdot (0,0,-r) d\theta dr \\ \amp = \int_0^3 \int_0^{2\pi} 0 d\theta dr = 0 \end{align*}

Flux over the entire closed cylinder is the sum of the flux other the three pieces.
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\begin{equation*} \int_C F \cdot da = 0 + 36\pi + 0 = 36\pi \end{equation*}

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Subsection 11.3.3 Conceptual Review Questions

What is a parametric surface?
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How does the normal to a parametric surface determine its properties?
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What is a flux integral? What does it measure?

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Why does a flux integral need a parametrized surface instead of just a locus?

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