Problem 41 - College Algebra and Calculus: An Applied Approach

Question

Use a double integral to find the area of the region bounded by the graphs of the equations. $$ y=9-x^{2}, y=0 $$

Step-by-Step Solution

Answer

The area of the region bounded by the graphs of the given equations is 108 square units.

Step 1: Plotting the given functions

First, plot the given functions \(y=9-x^2\) and \(y=0\) on the same graph. Go along the x-axis from -3 to 3 and mark the points where these two curves intersect. They intersect at points (-3, 0) and (3, 0).

Step 2: Setting up the integral for the area

The next step is to set up the integral. The area A of the region between these two curves can be represented as a iterated double integral, where the outer integral goes from -3 to 3 (representing the x-bounds of our region) and the inner integral goes from 0 to \(9 - x^{2}\) (representing the y-bounds of our region, namely from the x-axis up to the curve of the parabola). With this, we have the double integral: ∫ from -3 to 3 [∫ from 0 to \(9 - x^{2}\) dy] dx.

Step 3: Calculate Integral

Now calculate the integral. The inner integral gives us the y-value between 0 and \(9 - x^2\), which is simplly \(9 - x^2\). So we have \(\int_{-3}\)^3(9 - x^2) dx. Now, we just solve this single integral, which can be separated into two integrals: 9∫ from -3 to 3 dx - ∫ from -3 to 3 \(x^2\) dx. Solving each of these separately, we end up with [9x] evaluated from -3 to 3 minus [1/3\(x^3\)] evaluated from -3 to 3. This gives us: (27 - (-27)) - [(27 - (-27))] which equals 108.

Key Concepts

Area Between Curves

When we are tasked with finding the area between two curves, we use the concept of integration. It begins with identifying the boundaries where the curves meet, which can easily be found by setting the equations equal to each other and solving for the points of intersection. In this exercise, the curves are given by the equations:

• The parabola: \( y = 9 - x^2 \)
• The x-axis, which is represented by \( y = 0 \)

These functions intersect at \( x = -3 \) and \( x = 3 \). The area between these curves is thus bound by these x-values and starts at the x-axis (\( y=0 \)).
Finding the area between the curves requires integrating the difference between the upper curve (parabola) and the lower curve (x-axis) over the range of the x-values determined by their points of intersection.

Definite Integral

A definite integral is pivotal for calculating the area under curves. In this exercise, it allows us to find the area beneath the curve of a parabola from \(x = -3\) to \(x = 3\) and only above the x-axis.
The area under a curve \(y = f(x)\) from \(x = a\) to \(x = b\) is given by the definite integral:

• \( \int_{a}^{b} f(x) \, dx \)

In the provided exercise, the definite integral \( \int_{-3}^{3} (9 - x^2) \, dx \) calculates the bounded area's measure.
The process involves integrating the expression \(9 - x^2\) to find the total area enclosed. The bounds here are the x-values from -3 to 3. After computing the definite integral, we arrive at a numerical value representing this area. The result in the example exercise was shown to be 108, illustrating the accumulated area measurements from left boundary to right.

Parabola

A parabola is a distinct, U-shaped curve found in many areas of mathematics and physics. It is represented by a quadratic function, such as \( y = 9 - x^2 \), used in our exercise. This curve opens downward due to the negative coefficient of \(x^2\).
Parabolas can be characterized by:

• Vertex: The highest or lowest point on the graph. Here, the vertex is at the point \((0, 9)\).
• Axis of Symmetry: A vertical line that divides the parabola into two mirror-image halves. For this equation, it is \(x = 0\).
• Direction: Determined by the sign of the \(x^2\) term; downward if negative, upward if positive.

Understanding the parabola's properties helps identify important factors such as the vertex and range, ensuring proper setup for integration when seeking areas bounded by it and another curve or line.