Problem 280 - Calculus Volume 3

Question

For the following exercises, use a computer algebra system to approximate the area of the following surfaces using a parametric description of the surface. $$ \begin{aligned} &\begin{array}{llll} 280 . & {[\mathrm{T}]} & \text { Plane } & z=10-x-y & \text { above square } \end{array}\\\ &|x| \leq 2,|y| \leq 2 \end{aligned} $$

Step-by-Step Solution

Answer

The area of the surface is \(16\sqrt{3}\).

Step 1: Understand the Problem

We need to find the area of a surface described by the equation \(z = 10 - x - y\) over the domain of \(|x| \leq 2\) and \(|y| \leq 2\). This defines a plane above a square in the xy-plane.

Step 2: Parametrize the Surface

The given plane can be parametrized by the function \(r(x, y) = (x, y, 10 - x - y)\), where \(x\) and \(y\) vary within the given limits: \(-2 \leq x \leq 2\) and \(-2 \leq y \leq 2\).

Step 3: Compute Partial Derivatives

Calculate the partial derivatives of the parametrization: \(r_x = \left(1, 0, -1\right)\) and \(r_y = \left(0, 1, -1\right)\).

Step 4: Evaluate Cross Product

Find the cross product of \(r_x\) and \(r_y\): \(r_x \times r_y = (1, 1, 1)\).

Step 5: Compute Magnitude of Cross Product

Compute the magnitude of the cross product: \(\|r_x \times r_y\| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}\).

Step 6: Set Up the Area Integral

The area of the surface is given by the integral \(\int_{-2}^{2} \int_{-2}^{2} \sqrt{3} \, dy \, dx\).

Step 7: Evaluate the Integral

Calculate the integral: \(\int_{-2}^{2} \int_{-2}^{2} \sqrt{3} \, dy \, dx = \sqrt{3} \times 4 \times 4 = 16\sqrt{3}\).

Key Concepts

Surface Area Calculation

Calculating the surface area of a parametric surface can initially seem like a daunting task. However, it becomes manageable with a systematic approach. You start by understanding the surface description and identifying the region over which you calculate the area.

In our example, the surface is given by the equation \(z = 10 - x - y\) and lies over a square region defined by \(|x| \leq 2\) and \(|y| \leq 2\). To find the area, you first need the expression for the surface as a function of variables \(x\) and \(y\).

The area is then calculated using a double integral over this region. You need to determine the integration bounds for both variables, after which you'll set up the integral that will give the total surface area.

Parametrization

Parametrization is the process of defining a surface using parameters, typically denoted as \(u\) and \(v\). For the plane given by \(z = 10 - x - y\), a straightforward parametrization is \(r(x, y) = (x, y, 10 - x - y)\).

This method turns the equation of the surface into a vector function, making it simpler to apply calculus techniques.

• Identify the parameters: In this case, \(x\) and \(y\).
• Express the surface in terms of these parameters, forming a vector equation.
• Ensure the range of the parameters matches the defined limits, which in this exercise, is \(-2 \leq x \leq 2\) and \(-2 \leq y \leq 2\).

Efficient parametrization makes it possible to use vector calculus tools to explore surface geometry.

Cross Product

The cross product is integral to finding the surface area when using parametrization. It's a vector operation used to find a vector perpendicular to two given vectors.

Here, after parametrizing the surface as \(r(x, y)\), we compute the partial derivatives \(r_x\) and \(r_y\), resulting in vectors \((1, 0, -1)\) and \((0, 1, -1)\) respectively. The cross product \(r_x \times r_y\) yields \((1, 1, 1)\).

• Calculate partial derivatives of the parametrized function.
• Apply the cross product to these derivatives.
• This new vector is crucial as its magnitude is part of the surface area integral.

The cross product simplifies the process of acquiring a normal vector needed to find area elements.

Calculus

Calculus is a powerful tool for solving the problem of finding surface areas, particularly through integration. After parametrizing the surface and calculating the cross product, calculus allows us to set up and solve the integral to find the surface area.

In this particular exercise, the magnitude of the cross product vector, \(\sqrt{3}\), appears in the integral \(\int_{-2}^{2} \int_{-2}^{2} \sqrt{3} \, dy \, dx\).

• Integrate over the specified range, here \(-2\) to \(2\) for both variables.
• Double integrals are used because the surface is in a two-dimensional domain.

The result of this integration, a real number, gives the total surface area, illustrating the broad utility of calculus in mathematical modeling and problem-solving.

Computer Algebra System

A computer algebra system (CAS) is an invaluable resource for performing complicated symbolic operations like those involved in surface area calculations. These systems automate and simplify the process of integration, differentiation, and algebraic manipulation, and can be instrumental when solving complex calculus problems.

In exercises such as this, using a CAS helps to efficiently handle the computations, allowing for an approximation of the integral without manual, error-prone calculations.

• They provide precise numerical results for integrals.
• Facilitate visualization of parametric surfaces and results.

With a CAS, students and professionals can focus more on the understanding and interpretation of mathematical models, rather than the tedious number-crunching, which enhances learning and insight.