Problem 36 - 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 \(\mathbf{r}(u, v) .\) $$ \begin{array}{l}{f(x, y, z)=\frac{x^{2}+z^{2}}{y} ; \mathbf{r}(u, v)=2 \cos v \mathbf{i}+u \mathbf{j}+2 \sin v \mathbf{k}} \\ {(1 \leq u \leq 3,0 \leq v \leq 2 \pi)}\end{array} $$

Step-by-Step Solution

Answer

The value of the integral is \( 16\pi \ln 3 \).

Step 1: Parametrize the Surface

The surface \( \sigma \) is parametrized by the vector function \( \mathbf{r}(u, v) = 2 \cos v \mathbf{i} + u \mathbf{j} + 2 \sin v \mathbf{k} \). The parameters vary according to the bounds \(1 \leq u \leq 3 \) and \(0 \leq v \leq 2 \pi \). This represents a surface where \(x=2\cos v\), \(y=u\), and \(z=2\sin v\).

Step 2: Compute Partial Derivatives

Calculate the partial derivatives of \( \mathbf{r}(u, v) \) with respect to \( u \) and \( v \).\( \mathbf{r}_u = \frac{\partial \mathbf{r}}{\partial u} = 0 \mathbf{i} + 1 \mathbf{j} + 0 \mathbf{k} = \mathbf{j} \)\( \mathbf{r}_v = \frac{\partial \mathbf{r}}{\partial v} = -2 \sin v \mathbf{i} + 0 \mathbf{j} + 2 \cos v \mathbf{k} \).

Step 3: Compute the Cross Product

The cross product \( \mathbf{r}_u \times \mathbf{r}_v \) gives the normal vector to the surface.\[ \mathbf{r}_u \times \mathbf{r}_v = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \0 & 1 & 0 \-2 \sin v & 0 & 2 \cos v \\end{vmatrix} = 2 \cos v \mathbf{i} + 0 \mathbf{j} + 2 \sin v \mathbf{k} \].

Step 4: Compute the Magnitude of the Cross Product

Find the magnitude of the vector \( \mathbf{r}_u \times \mathbf{r}_v \).\[ \|\mathbf{r}_u \times \mathbf{r}_v\| = \sqrt{(2 \cos v)^2 + 0^2 + (2 \sin v)^2} = \sqrt{4\cos^2 v + 4\sin^2 v} = 2 \].

Step 5: Plug into the Integral

Substitute the parameterized function and the magnitude of the cross product into the surface integral.\[ \iint_{\sigma} f(x, y, z) \, dS = \int_{0}^{2 \pi} \int_{1}^{3} \frac{(2\cos v)^2 + (2\sin v)^2}{u} \times 2 \, du \, dv \]. This simplifies to:\[ \int_{0}^{2 \pi} \int_{1}^{3} \frac{4}{u} \times 2 \, du \, dv = 8 \int_{0}^{2 \pi} \int_{1}^{3} \frac{1}{u} \, du \, dv \].

Step 6: Evaluate the Integral

Evaluate the integral:\[ 8 \int_{0}^{2 \pi} \left [ \ln |u| \right ]_{1}^{3} \, dv = 8 \int_{0}^{2 \pi} (\ln 3 - \ln 1) \, dv = 8 \int_{0}^{2 \pi} \ln 3 \, dv \].Since \( \ln 1 = 0 \), it simplifies to:\[ 8 \ln 3 \int_{0}^{2 \pi} 1 \, dv = 8 \ln 3 \times 2\pi = 16\pi \ln 3 \].

Key Concepts

Multiple Integrals

In mathematics, multiple integrals extend the concept of integration from functions of a single variable to functions of multiple variables. These are crucial when dealing with physical applications, where quantities depend on more than one factor. In the given problem, a double integral is used to evaluate a continuous function over a surface. Here, the function integrand is \(f(x, y, z) = \frac{x^2 + z^2}{y}\). This represents how we can integrate a function across a two-dimensional surface in three-dimensional space.

Double integrals can be visualized as summing an infinite number of infinitesimally small products throughout a region of interest. In the exercise, the region involves variables \(u\) and \(v\), which parametrize the surface. This results in integrating over a domain defined by \(1 \leq u \leq 3\) and \(0 \leq v \leq 2\pi\).

• Calculate partial derivatives and surface area elements when performing surface integration.
• Ensure the function is integrated correctly by following the limits precisely.
• Remember that the result has a specific physical or geometrical meaning.

Parametric Surfaces

Parametric surfaces are representations of surfaces in space using parameters. In the exercise, the given surface \(\sigma\) is described using a vector-valued function \(\mathbf{r}(u, v) = 2 \cos v \, \mathbf{i} + u \, \mathbf{j} + 2 \sin v \, \mathbf{k}\)\. This involves two parameters, \(u\) and \(v\), defining every point on the surface through a vector function.

These surfaces are advantageous because they allow flexible modeling of complex shapes. The parameters have specific ranges, ensuring they cover the desired section of space. In our problem:

• extit{x}-component: \(2\cos v\)
• extit{y}-component: \(u\)
• extit{z}-component: \(2\sin v\)

By understanding these components, we can visualize how the parameter values change across the surface. This creates a cohesive approach to studying complex geometries in calculus.

Remember, using parametric equations simplifies the computation of integrals and creates a bridge toward efficient and precise geometric analysis.

Vector Calculus

Vector calculus extends the techniques of calculus to vector fields and differentiable manifolds. It embraces \(n\)-dimensional spaces, making it a critical tool in fields like physics and engineering. Our problem involves vector functions and the calculation of the cross-product, a fundamental operation in vector calculus.

In vector calculus, working with vector-valued functions involves:

• Calculating derivatives and integrals of vectors.
• Understanding vector fields representing quantities such as force or fluid flow.
• Applying operations such as divergence, curl, and gradient to vector functions.

The calculus of vector-valued functions helps solve problems involving work, energy, and other phenomena across curves and surfaces. Here, we compute the cross product of two partial derivatives to find the normal vector crucial for surface integration.

The practice of vector calculus is thus crucial in analyzing behavior across multi-dimensional environments in practical and theoretical contexts.

Cross Product in Calculus

The cross product in calculus provides a means to find a vector perpendicular to two given vectors. This operation is vital when working with surfaces in three-dimensional space. Let's consider the cross product computed in the exercise: \[\mathbf{r}_u \times \mathbf{r}_v = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 0 & 1 & 0 \ -2 \sin v & 0 & 2 \cos v \end{vmatrix} = 2 \cos v \, \mathbf{i} + 0 \, \mathbf{j} + 2 \sin v \, \mathbf{k} \, .\]\This is the resulting normal vector used in calculating the magnitude relevant for the surface integration.

The cross product, denoted as \(\times\), is anti-commutative and follows the right-hand rule. It's essential for:

• Determining torque in physics applications.
• Finding orthogonal vectors to surfaces in geometry and calculus.
• Calculating vector areas of parallelograms defined by vectors.

Once you find the normal vector using the cross product, you can proceed to determine the integration element for the surface. This is crucial to solving integral equations involving surfaces.