Problem 269 - Calculus Volume 3

Question

For the following exercises, determine whether the statements are true or false. $$ \begin{aligned} &\begin{array}{llllll} 269 . & \text { If } & \text { surface } & S & \text { is } & \text { given } & \text { by } \end{array}\\\ &\mid(x, y, z): 0 \leq x \leq 1,0 \leq y \leq 1, z=10\\}, \quad \text { then }\\\ &\iint_{S} f(x, y, z) d S=\int_{0}^{1} \int_{0}^{1} f(x, y, 10) d x d y \end{aligned} $$

Step-by-Step Solution

Answer

True, the integrals are equivalent on the surface \( z=10 \).

Step 1: Understand the Surface

The surface \( S \) is defined by the set of points \( (x, y, z) \) where \( 0 \leq x \leq 1 \), \( 0 \leq y \leq 1 \), and \( z = 10 \). This describes a plane in 3D space, parallel to the \( xy \)-plane at \( z = 10 \), confined within the square defined by \( 0 \leq x, y \leq 1 \).

Step 2: Surface Integral Interpretation

The surface integral \( \iint_{S} f(x, y, z) \, dS \) over this plane \( S \) is the same as the double integral \( \int_{0}^{1} \int_{0}^{1} f(x, y, 10) \ \, dx \, dy \). Here, the function \( f \) is evaluated at the fixed \( z = 10 \) over the unit square in the \( xy \)-plane.

Step 3: Two-dimensional Integral Setup

Since \( z = 10 \) is constant for the surface \( S \), substituting \( z \) in \( f(x, y, z) \) with 10 simplifies the scenario to a two-dimensional problem over the region \( 0 \leq x \leq 1 \), \( 0 \leq y \leq 1 \). The surface integral simplifies to a standard double integral.

Step 4: Evaluation of the Statement

The setup of the surface \( S \) and the integral as described ensures that both the given surface integral and the double integral are equivalent. Therefore, the statement \( \iint_{S} f(x, y, z) \, dS = \int_{0}^{1} \int_{0}^{1} f(x, y, 10) \, dx \, dy \) is true.

Key Concepts

Double Integrals

In the realm of multivariable calculus, double integrals play a crucial role when working with functions of two variables. They provide a way to calculate volumes under surfaces across a given region in the plane. Double integrals extend the concept of a single-variable integral, but instead of summing over intervals on a line, they sum over a two-dimensional area. For instance, consider a surface above a region like a square or rectangle. The double integral helps find the total mass, if the function represents density, or any other quantity dependent on area.
Double integrals are usually denoted as \( \int \int_R f(x, y) \, dx \, dy \), meaning we're integrating the function \( f(x, y) \) over region \( R \). In our example, the region is defined by \( 0 \leq x \leq 1 \) and \( 0 \leq y \leq 1 \). This essentially creates a unit square in the coordinate plane where each tiny square contributes to the total integral.

• A key aspect is substituting limits of integration appropriately to fit the region’s boundaries.
• Also, the function \( f(x, y) \) is evaluated over every point within that region.

By transforming a surface integral into a double integral, as shown in the solution, we simplify calculations, maintaining focus on just two variables, thanks to the constancy of \( z \) with respect to the surface \( S \).

Multivariable Calculus

Multivariable calculus extends traditional calculus concepts into higher dimensions. Instead of dealing with functions concerning just one variable, it explores functions that depend on two or more variables, like \( f(x, y) \) or \( f(x, y, z) \). This extension is particularly useful in physics, engineering, and any field dealing with space and motion in three dimensions.
Some of the key topics in multivariable calculus include partial derivatives, gradient vectors, and integrals covering areas and volumes in multi-dimensional space. Partial derivatives, for example, allow us to understand how a function changes with respect to one variable while keeping others constant.

• Double integrals: As discussed, they are pertinent to finding areas and volumes.
• Partial derivatives: These help analyze the rate of change in different spatial directions.
• Surface integrals: These are a generalization of double integrals for curved surfaces in 3D space.

Within multivariable calculus, one learns to navigate these complex interactions, using tools that simplify dealing with multi-dimensional functions. The transition from using surface integrals to double integrals, as illustrated in the solution provided, exemplifies how multivariable calculus empowers us to solve complex spatial problems by converting them into more intuitive, two-dimensional analyses.

Coordinate Planes

In mathematics, particularly in calculus and geometry, understanding coordinate planes is foundational. The coordinate plane allows us to define spaces in which every point is identified by coordinates. In three-dimensional spaces, these coordinates are \( (x, y, z) \). Each of these represents a specific point in 3D space.
The concept of coordinate planes becomes particularly important in multivariable calculus, where we deal with functions of two or more variables. The Cartesian coordinate system lays out three planes:

• The \( xy \)-plane: where \( z = 0 \)
• The \( xz \)-plane: where \( y = 0 \)
• The \( yz \)-plane: where \( x = 0 \)

These planes help in visualizing how functions behave across different dimensions. For instance, in the original exercise, the surface \( S \) was parallel to the \( xy \)-plane at \( z = 10 \). This means that every vertical line intersecting this plane stays at \( z = 10 \).
Understanding these planes isn't just about locating points. It's about recognizing relationships and constraints that help in defining integrals and interpreting results, just like transforming a surface integral into a double integral by visualizing the intersection within these planes.