Problem 23 - Calculus Early Transcendentals

Question

Find the flux of \(\mathbf{F}\) across the surface \(\sigma\) by expressing \(\sigma\) parametrically. \(\mathbf{F}(x, y, z)=\mathbf{i}+\mathbf{j}+\mathbf{k} ;\) the surface \(\sigma\) is the portion of the cone \(z=\sqrt{x^{2}+y^{2}}\) between the planes \(z=1\) and \(z=2\) oriented by downward unit normals.

Step-by-Step Solution

Answer

The flux of \(\mathbf{F}\) across \(\sigma\) is \(-3\pi\).

Step 1: Parameterize the Surface

The given surface is part of the cone \(z = \sqrt{x^2 + y^2}\) bounded by \(z = 1\) and \(z = 2\). We'll parameterize the cone using cylindrical coordinates: let \(x = r \cos \theta\), \(y = r \sin \theta\), and \(z = r\), where \(1 \leq r \leq 2\) and \(0 \leq \theta < 2\pi\).

Step 2: Calculate the Normal Vector

To find the flux, we need the downward unit normal. The surface parameterization is \(\mathbf{r}(r, \theta) = (r \cos \theta, r \sin \theta, r)\). Compute the partial derivatives \(\mathbf{r}_r = (\cos \theta, \sin \theta, 1)\) and \(\mathbf{r}_\theta = (-r \sin \theta, r \cos \theta, 0)\). The cross product \(\mathbf{r}_r \times \mathbf{r}_\theta\) gives the normal vector: \((-r \cos \theta, -r \sin \theta, r)\). The unit normal is \(-\frac{1}{r}(\cos \theta, \sin \theta, -1)\).

Step 3: Evaluate the Flux Integral

The flux through the surface is given by the integral \(\iint_\sigma \mathbf{F} \cdot d\mathbf{S}\). Substitute \(\mathbf{F}(x, y, z) = (1, 1, 1)\), resulting in \((1, 1, 1) \cdot \left(\frac{-1}{r} (r \cos \theta, r \sin \theta, -r)\right) = \frac{-1}{r}(r \cos \theta + r \sin \theta - r)\). Simplifying gives \(- (\cos \theta + \sin \theta - 1)\).\ The flux integral becomes \( \int_0^{2\pi} \int_1^2 -(\cos \theta + \sin \theta - 1) r\, dr \, d\theta \).

Step 4: Calculate the Integral

Break down the integral: \( \int_1^2 r\, dr = \left[ \frac{r^2}{2} \right]_1^2 = \frac{3}{2}\). For the angular part, use properties of sine and cosine about their \(\theta\) integration over \([0, 2\pi]\), where their individual integrals over a full period equal zero, leaving: \(-\int_0^{2\pi} \int_1^2 (r)\, dr \, d\theta = - \int_0^{2\pi} \frac{3}{2} \, d\theta = -\frac{3}{2} \times 2\pi = -3\pi\).

Step 5: Conclusion of the Flux Calculation

Hence, the flux of \(\mathbf{F}\) through the surface \(\sigma\), downward, is \(-3\pi\).

Key Concepts

Parameterization

Parameterization is a crucial step in solving problems involving surfaces, especially when calculating flux integrals. It involves expressing a surface in terms of two parameters, typically denoted as \(u\) and \(v\). For the given exercise, the surface is part of a cone defined by \(z = \sqrt{x^2 + y^2}\).
To parameterize this cone, we use cylindrical coordinates, which naturally suit problems with circular symmetry. We choose \(r\) (radius) and \(\theta\) (angle) as parameters:

• \(x = r \cos \theta\)
• \(y = r \sin \theta\)
• \(z = r\)

This parameterization nests by letting \(1 \leq r \leq 2\) and \(0 \leq \theta < 2\pi\). Such an expression simplifies the subsequent computation of the flux over the conical surface.

Cylindrical Coordinates

Cylindrical coordinates are a three-dimensional coordinate system extending polar coordinates by adding a vertical \(z\)-axis.
This system is ideal for surfaces or volumes that possess rotational symmetry around one axis, like a cylinder or cone.
By using cylindrical coordinates, calculations become more straightforward; they align with the natural geometry of the object. The coordinates \((r, \theta, z)\) relate to Cartesian coordinates \((x, y, z)\) as follows:

• \(x = r \cos \theta\)
• \(y = r \sin \theta\)
• \(z = z\)

In the problem at hand, \(z\) is described as \(z = r\), showing the surface is part of a cone. This simplicity aids in breaking down the problem and focusing only on needed integrations.

Unit Normal Vector

A unit normal vector is essential in calculating flux as it indicates the direction perpendicular to the surface at any given point,
ensuring proper computation of the integral.Given the parameterized surface \(\mathbf{r}(r, \theta) = (r \cos \theta, r \sin \theta, r)\), we find the unit normal vector by first computing two derivatives:

• \(\mathbf{r}_r = (\cos \theta, \sin \theta, 1)\)
• \(\mathbf{r}_\theta = (-r \sin \theta, r \cos \theta, 0)\)

The cross product \(\mathbf{r}_r \times \mathbf{r}_\theta\) yields the normal vector:
\((-r \cos \theta, -r \sin \theta, r)\).
To convert this into a unit normal vector, we divide by its magnitude:
\(-\frac{1}{r}\)(\(\cos \theta, \sin \theta, -1\)).
This correctly aligns with the cone's downward orientation as required.

Cross Product

The cross product is a vector operation occurring between two vectors, producing a third vector perpendicular to both. This operation is fundamental when you need to find normal vectors to surfaces, such as in flux calculations.
When you have two tangent vectors on the surface, their cross product results in a vector normal to the surface. This is vital for applying the concept of the flux integral.
In the exercise, computing the cross product of the partial derivatives of the surface parameterization \(\mathbf{r}(r, \theta)\) gave us:

• \(\mathbf{r}_r = (\cos \theta, \sin \theta, 1)\)
• \(\mathbf{r}_\theta = (-r \sin \theta, r \cos \theta, 0)\)

Their cross product:
\((-r \cos \theta, -r \sin \theta, r)\) is utilized to find a suitable normal vector before scaling into a unit normal vector.