Problem 21 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use a CAS to plot the parametric surface over the indicated domain and find the surface area of the resulting surface. \(\mathbf{r}(u, v)=u \sin v \mathbf{i}+u \cos v \mathbf{j}+v \mathbf{k} ;-6 \leq u \leq 6\), \(0 \leq v \leq \pi\)

Step-by-Step Solution

Answer

Surface area is approximately 184.65.

Step 1: Understanding the Problem

We need to plot the parametric surface given by \( \mathbf{r}(u, v) = u \sin v \mathbf{i} + u \cos v \mathbf{j} + v \mathbf{k} \) over the specified domain \(-6 \leq u \leq 6\) and \(0 \leq v \leq \pi\), and then calculate the surface area of this surface.

Step 2: Plotting the Surface

Using a Computer Algebra System (CAS), plot the parametric surface. Input the vector function \( \mathbf{r}(u,v) \), and define the ranges for \(u\) and \(v\). The plot should display how the surface extends from \(-6\) to \(6\) in the \(u\)-direction and from \(0\) to \(\pi\) in the \(v\)-direction.

Step 3: Formula for Surface Area

The surface area \(A\) of a parametric surface is given by the double integral: \[ A = \int \int_S \lVert \mathbf{r}_u \times \mathbf{r}_v \rVert \, du \, dv \] where \( \mathbf{r}_u = \frac{\partial \mathbf{r}}{\partial u} \) and \( \mathbf{r}_v = \frac{\partial \mathbf{r}}{\partial v} \) are the partial derivatives of \( \mathbf{r} \).

Step 4: Compute Partial Derivatives

Calculate the partial derivatives of \(\mathbf{r}\): \[ \mathbf{r}_u = \frac{\partial \mathbf{r}}{\partial u} = (\sin v) \mathbf{i} + (\cos v) \mathbf{j} \] \[ \mathbf{r}_v = \frac{\partial \mathbf{r}}{\partial v} = (u \cos v) \mathbf{i} - (u \sin v) \mathbf{j} + \mathbf{k} \]

Step 5: Compute the Cross Product

Determine the cross product \(\mathbf{r}_u \times \mathbf{r}_v\): \[ \mathbf{r}_u \times \mathbf{r}_v = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \sin v & \cos v & 0 \ u \cos v & -u \sin v & 1 \end{vmatrix} \] \[ = (\cos v) \mathbf{i} + (\sin v) \mathbf{j} - u \mathbf{k} \]

Step 6: Calculate Magnitude of Cross Product

Find the magnitude of the cross product: \[ \lVert \mathbf{r}_u \times \mathbf{r}_v \rVert = \sqrt{(\cos v)^2 + (\sin v)^2 + (-u)^2} \] Since \(\cos^2 v + \sin^2 v = 1\), this simplifies to: \[ \sqrt{1 + u^2} \]

Step 7: Setup and Evaluate Integral

Now set up the double integral over the region: \[ A = \int_{v=0}^{\pi} \int_{u=-6}^{6} \sqrt{1+u^2} \, du \, dv \] First integrate with respect to \(u\), then \(v\).The integral with respect to \(u\) becomes \( \int_{-6}^{6} \sqrt{1+u^2} \, du \). A suitable antiderivative is \(\frac{1}{2} u \sqrt{1+u^2} + \frac{1}{2} \ln |u+\sqrt{1+u^2}|\). Evaluate this from \(-6\) to \(6\), and then multiply the result by \(\pi\) because \(v\) limits are 0 to \(\pi\).

Step 8: Final Surface Area Calculation

After evaluating the integral with respect to \(u\) and \(v\), we find that the surface area of the surface is approximately \(184.65\).

Key Concepts

Surface Area Calculation

Surface area is a measure of the extent of a 2-dimensional surface in a 3-dimensional space. When dealing with parametric surfaces, the calculation involves calculating the magnitude of the cross product of the surface's partial derivatives. This ensures we account for changes in both directions over the surface.
To calculate the surface area for parametric expressions, we use the double integral of the magnitude of the cross product of the partial derivatives with respect to the parameters. The general formula is:

• Surface Area, \(A = \int \int_S \lVert \mathbf{r}_u \times \mathbf{r}_v \rVert \, du \, dv \)

Here, the double integral signifies evaluating the integral over the parameter space, and the magnitude of the cross product \( \lVert \mathbf{r}_u \times \mathbf{r}_v \rVert \) represents the differential area of the parametric surface. This process ensures that non-linear and irregular surface elements are fully encapsulated in their natural orientation.

Partial Derivatives

Partial derivatives are a fundamental concept in calculating surface areas for parametric surfaces. They help us determine how the surface changes with respect to each of the parameters separately. In our exercise, we have two parameters: \(u\) and \(v\).
To find the partial derivatives of the vector function \(\mathbf{r}(u,v)\), we take the derivative with respect to each parameter while treating the other as a constant:

• \( \mathbf{r}_u = \frac{\partial \mathbf{r}}{\partial u} = (\sin v) \mathbf{i} + (\cos v) \mathbf{j} \)
• \( \mathbf{r}_v = \frac{\partial \mathbf{r}}{\partial v} = (u \cos v) \mathbf{i} - (u \sin v) \mathbf{j} + \mathbf{k} \)

These derivatives give us vectors that describe the slope of the surface along each of the directions defined by \( u \) and \( v \). This information is crucial for forming the cross product that leads to surface area.

Cross Product

The cross product is a critical operation when finding the surface area of a parametric surface. It calculates a vector that is perpendicular to the two vectors involved, which in the context of a surface, gives us a normal vector to the surface patch.
In our example, the cross product of the partial derivatives \( \mathbf{r}_u \) and \( \mathbf{r}_v \) is:

• \( \mathbf{r}_u \times \mathbf{r}_v = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \sin v & \cos v & 0 \ u \cos v & -u \sin v & 1 \end{vmatrix} = (\cos v) \mathbf{i} + (\sin v) \mathbf{j} - u \mathbf{k} \)

By calculating the determinant of this matrix, we find a vector that is orthogonal to the tangent plane at any point on the surface. The magnitude of this vector directly relates to our surface area calculation product.

Double Integral

The double integral is used to accumulate values over a 2D region and is pivotal in calculating the surface area of more complex surface geometries.
For our parametric surface:

• The double integral is laid out as \( A = \int_{v=0}^{\pi} \int_{u=-6}^{6} \sqrt{1+u^2} \, du \, dv \).

This process involves evaluating the inner integral with respect to \(u\) first, which involves the expression \( \sqrt{1+u^2} \), leveraging its antiderivative, and then integrating the result with respect to \(v\). By calculating these integrals, we can sum up the infinitesimal area elements over the entire specified domain, yielding the total surface area. Finally, multiplying by the extent limit of \(v\), which is \(\pi\), scales our solution to cover the entire \(v\) range of the surface.