Problem 38 - 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 \(\mathbf{r}(u, v) .\) $$ \begin{array}{l}{f(x, y, z)=e^{-z}} \\ {\mathbf{r}(u, v)=2 \sin u \cos v \mathbf{i}+2 \sin u \sin v \mathbf{j}+2 \cos u \mathbf{k}} \\ {(0 \leq u \leq \pi / 2,0 \leq v \leq 2 \pi)}\end{array} $$

Step-by-Step Solution

Answer

The integral evaluates to \(4\pi (1-e^{-2})\).

Step 1: Set Up the Surface Integral

The surface integral \( \iint_{\sigma} f(x, y, z) \, dS \) over the surface \( \sigma \) involves the function \( f(x, y, z) = e^{-z} \). We are given that the surface \( \sigma \) is parameterized by \( \mathbf{r}(u, v) = 2 \sin u \cos v \, \mathbf{i} + 2 \sin u \sin v \, \mathbf{j} + 2 \cos u \, \mathbf{k} \), with \( 0 \leq u \leq \frac{\pi}{2} \) and \( 0 \leq v \leq 2 \pi \). This represents the surface of a quarter of a sphere.

Step 2: Compute the Partial Derivatives

The partial derivatives of \( \mathbf{r}(u, v) \) with respect to \( u \) and \( v \) are:\[ \frac{\partial \mathbf{r}}{\partial u} = 2 \cos u \cos v \, \mathbf{i} + 2 \cos u \sin v \, \mathbf{j} - 2 \sin u \, \mathbf{k} \]\[ \frac{\partial \mathbf{r}}{\partial v} = -2 \sin u \sin v \, \mathbf{i} + 2 \sin u \cos v \, \mathbf{j} \]

Step 3: Find the Cross Product

Compute the cross product \( \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v} \):\[\frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2 \cos u \cos v & 2 \cos u \sin v & -2 \sin u \ -2 \sin u \sin v & 2 \sin u \cos v & 0 \end{vmatrix}\]Evaluating this determinant, we get:\[ 4 \sin^2 u \cos v \, \mathbf{i} + 4 \sin^2 u \sin v \, \mathbf{j} + 4 \sin u \cos u \, \mathbf{k} \]

Step 4: Calculate the Magnitude

The magnitude of \( \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v} \) is:\[\sqrt{(4 \sin^2 u \cos v)^2 + (4 \sin^2 u \sin v)^2 + (4 \sin u \cos u)^2}\]This simplifies to:\[4 |\sin u| \sqrt{\sin^2 u + \cos^2 u} = 4 |\sin u| \]

Step 5: Set Up the Integral Expression

The surface integral becomes:\[\iint_{\sigma} e^{-z} \, dS = \int_{0}^{\frac{\pi}{2}} \int_{0}^{2\pi} e^{-2 \cos u} \cdot 4 \sin u \, dv \, du \]where \( z = 2 \cos u \) from the parameterization of \( \sigma \).

Step 6: Evaluate the Inner Integral

Evaluate the integral with respect to \( v \):\[\int_{0}^{2\pi} e^{-2 \cos u} \cdot 4 \sin u \, dv = (2\pi) e^{-2 \cos u} \, 4 \sin u = 8\pi e^{-2 \cos u} \sin u\]

Step 7: Evaluate the Outer Integral

Finally, evaluate the integral with respect to \( u \):\[\int_{0}^{\frac{\pi}{2}} 8\pi e^{-2 \cos u} \sin u \, du\]Using the substitution \( t = \cos u \), \( dt = -\sin u \, du \), the limits change from \( u = 0 \) to \( u = \frac{\pi}{2} \) corresponding to \( t = 1 \) to \( t = 0 \). Thus, the integral becomes:\[\int_{1}^{0} 8\pi e^{-2t} \, (-dt) = 8\pi \int_{0}^{1} e^{-2t} \, dt = 8\pi \times \left[-\frac{1}{2} e^{-2t}\right]_{0}^{1}\]Evaluating this we get:\[8\pi \times \left( -\frac{1}{2}e^{-2} + \frac{1}{2} \right) = 4\pi(1-e^{-2})\]

Step 8: Conclusion

The value of the surface integral over the given surface \( \sigma \) is \( 4\pi (1-e^{-2}) \).

Key Concepts

Parameterization of Surfaces

Surface parameterization is a method used in multivariable calculus to describe surfaces using two parameters. This transforms a two-dimensional surface into a map from a region in the plane, making complex calculations more manageable. In our problem, the surface is represented by the function \( \mathbf{r}(u, v) = 2 \sin u \cos v \, \mathbf{i} + 2 \sin u \sin v \, \mathbf{j} + 2 \cos u \, \mathbf{k} \). Here, \( u \) and \( v \) are the parameters that vary over specified intervals, namely \( 0 \leq u \leq \frac{\pi}{2} \) and \( 0 \leq v \leq 2\pi \). This specific parameterization describes a quarter of a sphere, where the parameters navigate the spherical coordinates. Using parameterization, the two-dimensional integral over the surface can be approached with tools of single-variable calculus, after being transformed into a double integral over the \( u-v \) plane.

Vector Calculus

Vector calculus is a specialized field of calculus that handles vectors and vector fields, crucial for understanding dynamics and physical systems. It deals with operations that return vector quantities, such as differentiation and integration on vector fields. In this context, the vector-valued parameterized function \( \mathbf{r}(u, v) \) is explored using partial derivatives. These derivatives are understood and manipulated to facilitate more complex calculus, transforming the three-dimensional surface calculations into two-dimensional integrals over parameters. The interplay of vectors in this calculation underlies the geometrical interpretation of the resulting surface and its properties, like area and curvature.

Cross Product

The cross product is an operation on two vectors in three-dimensional space that produces another vector perpendicular to the plane of the inputs. This is vital in calculating surface normals, which are essential in surface integral problems. In our exercise, once the parameterization \( \mathbf{r}(u, v) \) is derived in terms of partial derivatives, the cross product \( \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v} \) is calculated. This resultant vector \(([4 \sin^2 u \cos v \, \mathbf{i} + 4 \sin^2 u \sin v \, \mathbf{j} + 4 \sin u \cos u \, \mathbf{k}])\) serves as a normal vector to the surface, which is crucial for evaluating the equation for surface area and in turn for integrating functions over the surface. The cross product's magnitude is used to transform the integrand of the surface integral from involving \(dS\) to involving the differential area on the \(uv\)-plane.

Multivariable Calculus

Multivariable calculus expands on single-variable calculus concepts, leveraging functions of several variables, which are common in real-world applications, like computing area, volume, and optimizations. It encompasses parameterization, integration, and differentiation in systems involving more than one variable. In our problem, this branch of mathematics helps solve the integral \( \iint_{\sigma} f(x, y, z) \, dS \) by working through transformations and substitutions to evaluate.

• Understanding the transformation from surface integrals on a parameterized surface to double integrals over a plane simplifies complex geometry to familiar integral calculus techniques.
• The method of substituting \( t = \cos u \) transforms the integration variable into a manageable form, making it easier to find the integral bounds and evaluate the integrals.

In the end, multivariable calculus allows handling the complex behavior of three-dimensional structures and their interactions with multivariable functions, allowing us to find the solution \( 4\pi (1-e^{-2}) \) as the result of our surface integral calculation.