Problem 59 - Calculus: Early Transcendental Functions

Question

Find the surface area extending from the given curve in the \(x y\) -plane to the given surface. Above the unit square \(x \in[0,1], y \in[0,1]\) up to the plane \(z=4-x-y.\)

Step-by-Step Solution

Answer

The surface area extending from the given curve in the \(xy\)-plane up to the plane \(z = 4 - x - y\) is \(\sqrt{3}\).

Step 1: Compute the Partial Derivatives

The first step is to compute the partial derivatives of \(z = 4 - x - y\) with respect to x and y. These are \(\dfrac{\partial{z}}{\partial{x}} = -1\) and \(\dfrac{\partial{z}}{\partial{y}} = -1\).

Step 2: Substitute into the Surface Area Formula

Substitute these into the formula for surface area to get \(SA = \int\int_S \sqrt{1 + (-1)^2 + (-1)^2} \,dS = \int\int_S \sqrt{3} \,dS\), where S is the unit square in the \(xy\)-plane.

Step 3: Solve the Double Integral

The next step is to solve the double integral. The limits of integration for both x and y are from 0 to 1 as defined by the unit square. The double integral therefore simplifies to \(SA = \int_0^1 \int_0^1 \sqrt{3} \,dx\,dy = \sqrt{3}\).

Step 4: Compute the Surface Area

The surface area is then the result of this integral multiplied by the measure of the unit square, which is 1 since it's a unit square. Therefore, \(SA = \sqrt{3} \times 1 = \sqrt{3}\).

Key Concepts

Partial Derivatives

Understanding partial derivatives is fundamental in surface area calculation for surfaces above certain regions. When you have a function of multiple variables, partial derivatives help you explore how the function changes as you tweak one variable while keeping the others fixed. In the given problem, we need to compute partial derivatives for the function defining the surface:

• For the function \(z = 4 - x - y\), the partial derivative with respect to \(x\), denoted \(\frac{\partial{z}}{\partial{x}}\), is \(-1\).
• The partial derivative with respect to \(y\), \(\frac{\partial{z}}{\partial{y}}\), is also \(-1\).

These derivatives tell us how the surface slopes or tilts when moving parallel to each axis. Calculating these correctly is pivotal before integrating to find the surface area.

Double Integral

A double integral is a way to extend single variable integration to functions of two variables, making it perfect for computing areas or volumes. It's crucial here because it helps us calculate the surface area over a region in the plane, specifically over the unit square bounded by \(x \in[0,1]\) and \(y \in[0,1]\).
The surface area formula we derived uses the double integral:

• It is given by \(\int_0^1 \int_0^1 \sqrt{3} \,dx\,dy\).

This expression simply means you evaluate the integral over the domain of \(x\) and \(y\), taking one variable at a time, starting with \(x\) from 0 to 1, and similarly for \(y\). The result is the total accumulated "weight" of \(\sqrt{3}\) across the region, representing the surface area.

Surface Integral

While the exercise at hand primarily deals with a double integral, the concept of a surface integral expands on this by integrating over a surface in 3D space. Surface integrals are a powerful tool in vector calculus for evaluating different physical properties such as fluid flow or electric fields over surfaces. Although in our exercise the integration simplifies due to the uniform layout of the plane \(z = 4 - x - y\):

• The integrand is constant (\( \sqrt{3} \)) due to our earlier calculation of the expression under the square root.

Understanding surface integrals in broader contexts involves integrating vector fields, where the surface is parametric or more complex. Yet, the same principles of setup, computation, and comprehension apply.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, provides the tools to represent and analyze geometric figures using coordinates and equations. For solving surface area problems, it allows us to translate geometric scenarios into algebraic expressions. Here, the surface of interest is a plane represented by the equation \(z = 4 - x - y\), intersecting with the unit square in the xy-plane.
Using coordinate geometry:

• You visualize each point within the unit square \([0, 1] \times [0, 1]\), which forms the base of our surface.
• You can "look up" to find the point on the surface directly above, thus constructing a section of the surface between the curve and the defined plane.

The coordination between geometry and algebraic expressions bridges the comprehension gap when transitioning from physical to mathematical problem-solving.