Question

Set up the double integral that finds the surface area \(S\) of the given surface \(\mathcal{S},\) then use technology to approximate its value. \(\mathcal{S}\) is the plane \(z=5 x-y\) over the region enclosed by the parabola \(y=1-x^{2}\) and the \(x\) -axis.

Short answer

The surface area is approximately 6.928.

Step-by-step solution

Identify the region of integration

First, determine the region over which the surface is defined. The surface is bounded by the parabola \(y = 1 - x^2\) and the x-axis, specifically between the points where this parabola intersects the x-axis. Solving \(1 - x^2 = 0\) gives the points \(x = -1\) and \(x = 1\). Hence, the region of integration in the xy-plane is bounded by \(y = 0\), \(y = 1 - x^2\), \(x = -1\), and \(x = 1\).

Set up the double integral for surface area

The formula for the surface area \( S \) of a surface defined by \( z = f(x, y) \) is given by \[ S = \int \int_R \sqrt{1 + \left( \frac{\partial z}{\partial x} \right)^2 + \left( \frac{\partial z}{\partial y} \right)^2} \; dx \, dy \] where \(f(x, y) = 5x - y\). Calculate the partial derivatives: \(\frac{\partial z}{\partial x} = 5\) and \(\frac{\partial z}{\partial y} = -1\). Substitute into the formula to get \[ S = \int_{-1}^{1} \int_{0}^{1-x^2} \sqrt{1 + 5^2 + (-1)^2} \; dy \, dx \] Simplifying inside the square root gives: \(\sqrt{1 + 25 + 1} = \sqrt{27} = 3\sqrt{3}\).

Evaluate the double integral

Substitute the simplification into the integral: \[ S = \int_{-1}^{1} \int_{0}^{1-x^2} 3\sqrt{3} \; dy \, dx \] and evaluate the inner integral with respect to \( y \), results in \(3\sqrt{3}(y)\) from \(0\) to \(1-x^2\). Thus, \[ S = 3\sqrt{3} \int_{-1}^{1} (1-x^2) \, dx \].

Simplify and Solve the final integral

The integral \( \int_{-1}^{1} (1-x^2) \, dx \) is solved by splitting into two parts: \( \int_{-1}^{1} 1 \, dx - \int_{-1}^{1} x^2 \, dx \). The first part evaluates to \(2\) and the second part evaluates to \(\frac{2}{3}\). Thus, \( \int_{-1}^{1} (1-x^2) \, dx = 2 - \frac{2}{3} = \frac{4}{3} \).

Calculate the surface area

Finally, multiply this result by \(3\sqrt{3}\) to find the surface area: \[ S = 3\sqrt{3} \times \frac{4}{3} = 4\sqrt{3} \approx 6.928 \]

Key concepts

Surface Area

When we talk about surface area in the context of calculus, we're generally referring to the size of a specific surface in 3-dimensional space. Calculating the surface area using double integrals requires understanding the shape and boundaries of the surface. In simpler terms, imagine wrapping a surface with a thin layer of film, and the task is to find out how much film is needed to cover the surface completely.

In this exercise, the surface is defined by a plane equation, given as \( z = 5x - y \). The surface area of this plane over a specified region can be calculated by setting up a double integral. The double integral formula used in this case is:

• \[ S = \int \int_R \sqrt{1 + \left( \frac{\partial z}{\partial x} \right)^2 + \left( \frac{\partial z}{\partial y} \right)^2} \; dx \, dy \]

This integral accounts for the slopes of the surface in the \( x \) and \( y \) directions through the use of partial derivatives, which brings us to our next topic: partial derivatives.

Partial Derivatives

To fully understand surface area calculations, it's crucial to grasp the concept of partial derivatives. A partial derivative measures how a function changes as one of its variables changes, while the other variables are held constant. In the context of a surface described by \( z = f(x, y) \), the partial derivative with respect to \( x \) (\( \frac{\partial z}{\partial x} \)) tells us how the surface changes as we move along the \( x \)-axis, and similarly, \( \frac{\partial z}{\partial y} \) denotes changes along the \( y \)-axis.

For this specific problem, the partial derivatives are straightforward since the surface \( z = 5x - y \) is a plane:

• \( \frac{\partial z}{\partial x} = 5 \)
• \( \frac{\partial z}{\partial y} = -1 \)

These values are plugged into the surface area formula, \[ S = \int \int_R \sqrt{1 + (5)^2 + (-1)^2} \; dx \, dy \], simplifying the expression under the square root to \( \sqrt{27} = 3\sqrt{3} \). By understanding partial derivatives, we can handle the influence of each variable on the surface, a fundamental step in the integration process.

Region of Integration

The region of integration is a critical component in setting up any double integral. Essentially, it determines the boundaries over which the integration is performed. In this problem, the region is defined by a parabola \( y = 1 - x^2 \) and the \( x \)-axis.

To find the points of intersection between the parabola and the \( x \)-axis, we solve \( 1 - x^2 = 0 \), which gives us \( x = -1 \) and \( x = 1 \). This defines the limits for the \( x \)-integral. Meanwhile, for a fixed \( x \), the function \( y \) varies from \( y = 0 \) to \( y = 1 - x^2 \).

By defining these boundaries, we set up the integral over the correct area in the plane:

• \( \int_{-1}^{1} \int_{0}^{1-x^2} \; dy \, dx \)

Understanding the region of integration ensures that we are considering the entire area over which the surface itself lies. This is crucial to correctly calculating the surface area.

Plane Surface

A plane surface, in mathematical terms, is a flat, two-dimensional surface that extends infinitely in two directions. It can be defined by a linear equation in three variables, such as \( z = 5x - y \) in this problem. Unlike more complex surfaces, planes have the characteristic of uniformity and do not involve any curvature. This makes the mathematics simpler, as the slopes (partial derivatives) are constant.

In this exercise, our plane \( \mathcal{S} \) is intersected with the region defined by the parabola and \( x \)-axis. Thus, we only consider a section of this theoretically infinite plane. The task boils down to integrating over this restricted region to find the actual bounded area on the plane.

This piece of the plane forms our surface of study. By translating this problem into a double integral, students not only practice integration but also get a sense of how these abstract formulas apply to real-world surface areas.