Question

Sketch the region bounded by the graphs of the given equations, show a typical slice, approximate its area, set up an integral, and calculate the area of the region. Make an estimate of the area to confirm your answer. \(y=3-\frac{1}{3} x^{2}, y=0\), between \(x=0\) and \(x=3\)

Short answer

The area of the region is 6 square units.

Step-by-step solution

Sketch the Graphs

To begin, plot the equations on a graph. The first equation is a downward-opening parabola: \( y = 3 - \frac{1}{3}x^2 \). The second equation is a horizontal line: \( y = 0 \), which is the x-axis. Sketch these graphs between \( x = 0 \) and \( x = 3 \). The region to be considered is bounded above by the parabola and below by the x-axis.

Identify a Typical Slice

Choose a vertical slice through this region, located at a position \( x \) between \( 0 \) and \( 3 \). The height of this slice is given by the difference in the functions of the curves, which is \( y = 3 - \frac{1}{3}x^2 - 0 = 3 - \frac{1}{3}x^2 \).

Approximate the Slice Area

The area of a thin vertical slice of width \( \Delta x \) can be approximated by the formula for the area of a rectangle: \( \text{Height} \times \text{Width} = (3 - \frac{1}{3}x^2) \Delta x \).

Set Up the Integral for Total Area

To find the total area under the curve from \( x = 0 \) to \( x = 3 \), integrate the height function of the typical slice with respect to \( x \). The integral is: \[\int_{0}^{3} \left(3 - \frac{1}{3}x^2\right) \, dx.\]

Calculate the Area

Calculate the definite integral:1. Integrate term by term: \[\int 3 \, dx = 3x,\quad \int \frac{1}{3}x^2 \, dx = \frac{1}{9}x^3.\]2. Evaluate from \( 0 \) to \( 3 \): \[\left[3x - \frac{1}{9}x^3\right]_{0}^{3} = \left(3 \times 3 - \frac{1}{9}(3)^3\right) - \left(3 \times 0 - \frac{1}{9}(0)^3\right) = 9 - 3 = 6.\]

Confirm the Area Estimation

To confirm, return to the graph. The region is a typical segment under a parabola, and the estimated area of 6 aligns with the geometric appearance. Given the symmetry, the area computation looks reasonable.

Key concepts

Definite Integrals

Definite integrals are used to find the total accumulation of quantities, such as areas, over an interval. It involves integrating a function within specified limits, which are the endpoints of the interval. Definite integrals are important tools in calculus for solving real-world problems. In the given exercise, the purpose of the definite integral is to calculate the area between the curve of the function and the x-axis. This interval is defined from where the parabola starts at point \(x = 0\) and ends at \(x = 3\).

When setting up a definite integral, you follow these steps:

• Identify the function to integrate. Here, it is \(3 - \frac{1}{3}x^2\).
• Determine the limits of integration, which in this case are from 0 to 3.
• Integrate the function. This involves finding the antiderivative.
• Evaluate the antiderivative at the upper and lower limits and subtract the lower limit result from the upper limit result.

Evaluating this definite integral gives us the total area under the parabola between \(x = 0\) and \(x = 3\). This whole process allows us to calculate the size of the space bounded by functions, which is very valuable in various physical and geometric problems.

Area Under a Curve

The area under a curve can be found using integration, which sums an infinite number of infinitesimally small areas. This concept is crucial because it not only helps in finding physical areas but also in determining quantities like distance, volume, and even probabilities. In the context of our exercise, the area under the curve represents the space bounded by the function \(y = 3 - \frac{1}{3}x^2\) and the x-axis.

To compute this area:

• Think of each tiny section as a thin rectangle with a very small width denoted as \(\Delta x\).
• The height of each rectangle corresponds to the value of the function at that point, \(3 - \frac{1}{3}x^2\).
• We sum the areas of these tiny rectangles over the interval from 0 to 3.
• In calculus, this sum is represented as an integral \(\int_{0}^{3} (3 - \frac{1}{3}x^2) \, dx\), which determines the total area.

This geometric and analytical way of determining the area is foundational in calculus, serving numerous practical purposes beyond mere geometry.

Parabola

A parabola is a U-shaped curve and a common graph of a quadratic equation. The standard form of a parabola is \(y = ax^2 + bx + c\), and it exhibits reflective symmetry about a vertical line called the axis of symmetry. In this exercise, the function \(y = 3 - \frac{1}{3}x^2\) describes a downward facing parabola with its peak, or vertex, at the point where \(x = 0\), \(y = 3\).

Understanding parabolas involves:

• Identifying the direction it opens (upwards or downwards). A negative coefficient in front of \(x^2\) like \(-\frac{1}{3}\) indicates it opens downward.
• Finding the vertex which is the highest or lowest point on the graph, depending on its direction.
• Analyzing its standard intersection points, such as the x-intercepts. For the parabola \(y = 3 - \frac{1}{3}x^2\), it intersects the x-axis at the points solved by setting \(3 - \frac{1}{3}x^2 = 0\).

The properties of the parabola in the exercise help us bound and define the region whose area under the curve is calculated using definite integrals. It's these inherent properties of parabolas that make them central objects of study in algebra and calculus.