Course Notes for Calculus IV

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

1. The rectangle can be parametrized by the variables \(x\) and \(y\) with \(z\) set to constant zero.

   \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.

   \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*}

2. The sphere can be parametrized by following Example 7.1.4. Here, the radius is one.

   \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.

   \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.

   \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*}

3. The cylinder can be parametrized following Example 7.1.5. The radius here is \(3\text{.}\) The range for \(z\) is \([0,2]\) as given in the question.

   \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.

   \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.

   \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*}

4. 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.

   \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.

   \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.

   \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.

   \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.

   \begin{equation*} \int_C F \cdot da = 27 \pi + 26 \pi + 0 = 53 \pi \end{equation*}

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

1. The rectangle can be parametrized by the variables \(x\) and \(y\) with \(z\) set to constant zero.

   \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.

   \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*}

2. The sphere can be parametrized by following Example 7.1.4. Here, the radius is one.

   \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.

   \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.

3. The cylinder can be parametrized following Example 7.1.5. The radius here is \(3\text{.}\) The range for \(z\) is \([0,2]\) as given in the question.

   \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.

   \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.

4. 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.

   \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.

   \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.

   Then I do the bottom disc. I reverse the variable order to get a downward pointing normal.

   \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.

   \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.

   \begin{equation*} \int_C F \cdot da = 0 + 0 + 0 = 0 \end{equation*}

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

1. The rectangle can be parametrized by the variables \(x\) and \(y\) with \(z\) set to constant zero.

   \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.

   \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*}

2. The sphere can be parametrized by following Example 7.1.4. Here, the radius is one.

   \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.

   \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.

   \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*}

3. The cylinder can be parametrized following Example 7.1.5. The radius here is \(3\text{.}\) The range for \(z\) is \([0,2]\) as given in the question.

   \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.

   \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.

4. 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.

   \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.

   \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.

   \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.

   \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.

   \begin{equation*} \int_C F \cdot da = 0 + 36\pi + 0 = 36\pi \end{equation*}