Problem 35 - Calculus Early Transcendentals

Question

Evaluate the integral \(\iint_{\sigma} f(x, y, z) d S\) over the surface \(\sigma\) represented by the vector-valued function \(\mathbf{r}(u, v) .\) $$ \begin{array}{l}{f(x, y, z)=x y z ; \mathbf{r}(u, v)=u \cos v \mathbf{i}+u \sin v \mathbf{j}+3 u \mathbf{k}} \\ {(1 \leq u \leq 2,0 \leq v \leq \pi / 2)}\end{array} $$

Step-by-Step Solution

Answer

The integral evaluates to \(\frac{93 \sqrt{10}}{10}\).

Step 1: Parameterize the surface

The surface \(\sigma\) is given by the vector function \(\mathbf{r}(u, v) = u \cos v \mathbf{i} + u \sin v \mathbf{j} + 3u \mathbf{k}\), with \(1 \leq u \leq 2\) and \(0 \leq v \leq \frac{\pi}{2}\). This parameterizes a surface in 3D space where \(u\) controls the radial distance, and \(v\) controls the angular position.

Step 2: Compute partial derivatives

To find the tangent vectors, we need the partial derivatives \(\frac{\partial \mathbf{r}}{\partial u}\) and \(\frac{\partial \mathbf{r}}{\partial v}\).\[ \frac{\partial \mathbf{r}}{\partial u} = \cos v \mathbf{i} + \sin v \mathbf{j} + 3 \mathbf{k} \]\[ \frac{\partial \mathbf{r}}{\partial v} = -u \sin v \mathbf{i} + u \cos v \mathbf{j} \]

Step 3: Cross product to find normal vector

Calculate the cross product of the tangent vectors to find the normal vector \(\mathbf{N}\).\[ \mathbf{N} = \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \cos v & \sin v & 3 \ -u \sin v & u \cos v & 0 \end{vmatrix} \]This results in:\[ \mathbf{N} = -3u \cos v \mathbf{i} - 3u \sin v \mathbf{j} + u \mathbf{k} \]

Step 4: Magnitude of the normal vector

The differential area element \(dS\) is given by the magnitude of \(\mathbf{N}\):\[ dS = ||\mathbf{N}|| = \sqrt{(-3u \cos v)^2 + (-3u \sin v)^2 + (u)^2} = u\sqrt{10} \]

Step 5: Express f(x, y, z) using parameters

Express \(f(x, y, z) = x y z\) in terms of \(u\) and \(v\) using the parameterization:\[ x = u \cos v, \quad y = u \sin v, \quad z = 3u \]So,\[ f(x, y, z) = (u \cos v)(u \sin v)(3u) = 3u^3 \cos v \sin v \]

Step 6: Set up the double integral

We set up the double integral \(\iint_{\sigma} f(x, y, z) \, dS\) as:\[ \iint_{D} (3u^3 \cos v \sin v) \cdot (u\sqrt{10}) \, du \, dv \]which simplifies to:\[ 3 \sqrt{10} \iint_{D} u^4 \cos v \sin v \, du \, dv \] where \(D\) is the region given by \(1 \leq u \leq 2\) and \(0 \leq v \leq \frac{\pi}{2}\).

Step 7: Evaluate the integral

First integrate with respect to \(u\):\[ \int_{1}^{2} u^4 \, du = \left[ \frac{u^5}{5} \right]_{1}^{2} = \frac{32}{5} - \frac{1}{5} = \frac{31}{5} \]Then integrate with respect to \(v\):\[ \int_{0}^{\frac{\pi}{2}} \cos v \sin v \, dv = \frac{1}{2} \left[ \sin^2 v \right]_{0}^{\frac{\pi}{2}} = \frac{1}{2} \times 1 = \frac{1}{2} \]Combine results:\[ 3 \sqrt{10} \times \frac{31}{5} \times \frac{1}{2} = \frac{93 \sqrt{10}}{10} \]

Step 8: Final result

Thus, the integral \(\iint_{\sigma} f(x, y, z) \, dS\) evaluates to \(\frac{93 \sqrt{10}}{10}\).

Key Concepts

Cross Product

The cross product is a fundamental operation in vector calculus, particularly important for determining a vector normal to a surface. In this exercise, we start by taking the tangent vectors of the parameterization of the surface. Once we have these tangent vectors, we find their cross product. This helps us determine the normal vector to the surface, which is crucial when evaluating surface integrals.

The cross product involves the vectors \(\frac{\partial \mathbf{r}}{\partial u}\) and \(\frac{\partial \mathbf{r}}{\partial v}\). The cross product is defined as the determinant of a matrix formed by these tangent vectors, with the unit vectors \(\mathbf{i}, \mathbf{j}, \mathbf{k}\). The result \(\mathbf{N} = -3u \cos v \mathbf{i} - 3u \sin v \mathbf{j} + u \mathbf{k}\) shows how each component is influenced by both direction and magnitude.

• The cross product helps find a perpendicular (normal) unit vector to the surface.
• In surface integrals, this is pivotal as it defines the orientation of the surface.

Parameterization of Surfaces

Parameterization is the process of describing a surface using variables. It allows us to express the surface in terms of parameters \(u\) and \(v\). In this exercise, the surface is parameterized by the vector function \(\mathbf{r}(u, v) = u \cos v \mathbf{i} + u \sin v \mathbf{j} + 3u \mathbf{k}\).

Parameterization essentially maps a two-dimensional region in the \(uv\)-plane onto the surface in three-dimensional space. Here:

• \(u\) controls the radial distance from the origin, behaving like a radius in polar coordinates.
• \(v\) acts like an angle, controlling the direction in which this distance is measured.

By rearranging the original function \(f(x, y, z)\) in terms of \(u\) and \(v\), we transform the surface integral into a more manageable double integral over the region \(D\) in the parameter space.

Normal Vector

The normal vector is a key concept in evaluating surface integrals, representing a vector perpendicular to the surface at each point. To find this, we use the cross product of the tangent vectors derived from the partial derivatives \(\frac{\partial \mathbf{r}}{\partial u}\) and \(\frac{\partial \mathbf{r}}{\partial v}\) of the parameterized surface.

Here, the normal vector \(\mathbf{N} = -3u \cos v \mathbf{i} - 3u \sin v \mathbf{j} + u \mathbf{k}\) is crucial because:

• It determines the direction in which the surface is oriented.
• Its magnitude gives the differential area element \(dS = u\sqrt{10}\), which must be integrated over the surface.

Calculating the normal vector and its magnitude ensures that each differential area \(dS\) precisely covers the intended portion of the surface, factoring in both its orientation and inclination in space.

Double Integrals

Double integrals are used to compute the integral over a surface by evaluating iterated integrals over its parameterized region. In this exercise, after converting the function \(f(x, y, z) = x y z\) into terms of \(u\) and \(v\), we set up a double integral over the region \(D\):

\[3 \sqrt{10} \iint_{D} u^4 \cos v \sin v \, du \, dv\]

Here, the region \(D\), defined by \(1 \leq u \leq 2\) and \(0 \leq v \leq \frac{\pi}{2}\), means:

• First, we integrate with respect to \(u\), yielding the function of \(u\).
• Next, we integrate with respect to \(v\), incorporating the trigonometric functions \(\cos v\) and \(\sin v\).

The calculated double integral expresses the sum of tiny elements \(dS\) multiplied by the function \(f(u, v)\), efficiently summing over the entire parameterized surface. Such integrals generalize beyond simple areas, incorporating surface curvature.