Problem 34 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

AREA BETWEEN CURVES In Exercises 31 through 38 , sketch the indicated region \(R\) and find its area by integration. \(R\) is the region under the curve \(y=\sqrt{9-5 x^{2}}\) over the interval \(0 \leq x \leq 1\).

Step-by-Step Solution

Answer

The area is found by integrating \(y=\sqrt{9-5x^{2}}\) from 0 to 1.

Step 1: Understand the region to be integrated

The region R is bounded by the curve defined by the function \(y = \sqrt{9 - 5x^{2}}\) from \(x = 0\) to \(x = 1\).

Step 2: Sketch the curve and region

To sketch the curve \(y = \sqrt{9 - 5x^{2}}\), recognize that it is a part of an ellipse. Plot points for various values of \(x\) from 0 to 1 and draw the curve. The region R lies under this curve between these x-values.

Step 3: Set up the integral

The area under the curve can be found by integrating the function \(y = \sqrt{9 - 5x^{2}}\) with respect to \(x\) over the interval \([0, 1]\). The integral is \[ \text{Area} = \int_{0}^{1} \sqrt{9 - 5x^{2}} \, dx \].

Step 4: Solve the integral

To solve the integral \(\int_{0}^{1} \sqrt{9 - 5x^{2}} \, dx\), perform a trigonometric substitution or recognize it as a standard form requiring definite integration. For this integral, use the substitution \(x = \sqrt{9/5} \sin(\theta)\), which transforms the integral. Then, integrate and evaluate the bounds after the substitution.

Step 5: Evaluate the integral

After substitution and simplification, the indefinite integral will be evaluated and followed by applying the limits from 0 to 1. The solution involves calculating trigonometric values and constants to get the exact area.

Key Concepts

Integration

Integration is a fundamental concept in calculus. It represents the accumulation of quantities, which can be visualized as the area under a curve. In this exercise, integration helps us find the area of region R, which is bounded by the ellipse portion described by the function:
\(y=\sqrt{9-5x^{2}}\). The integral we need to solve for this is: \[ \text{Area} = \int_{0}^{1} \sqrt{9 - 5x^{2}} \,dx \].
By integrating, we effectively sum up the infinitesimally small areas under the curve across the interval [0, 1].

• Definite integrals help calculate exact areas.
• Setting up the correct integral based on given limits is crucial.
• Solving the integral often involves manipulation and substitution.

Trigonometric Substitution

Sometimes, standard algebraic methods are insufficient for solving integrals. That’s where trigonometric substitution comes in handy. It is a technique to simplify integrals involving \(\sqrt{a^2 - x^2}\) by transforming them into trigonometric forms. In our case, to solve \[\int_{0}^{1} \sqrt{9 - 5x^{2}} \,dx \], we use the substitution:
\( x = \sqrt{9/5} \sin(\theta) \).
Doing this changes the variable from \(x\) to \(\theta\), allowing us to leverage trigonometric identities, making the integral easier to solve.
Key steps in trigonometric substitution:

• Identify the suitable substitution based on the integral form.
• Change the variable and adjust the limits of integration accordingly.
• Simplify the transformed integral using trigonometric identities.
• Integrate and finally revert back to the original variable.

This method is powerful for transforming complex integrals into simpler forms that are more manageable.

Definite Integral

A definite integral calculates the exact area under a curve between specified limits. For the mentioned exercise, we are interested in the area from \(x = 0\) to \(x = 1\), which means integrating y = \sqrt{9 - 5x^{2}}\ over this interval.
The process involves:

• Setting up the integral with appropriate limits: \[ \int_{0}^{1} \sqrt{9 - 5x^{2}} \,dx. \].
• Choosing an appropriate method, such as trigonometric substitution, to solve the integral.
• Evaluating the integral in its transformed form.
• Applying the limits of the variable after transformation.
• Calculating the value to obtain the exact area.

Definite integrals give us a concrete numerical value representing the region’s area, distinguishing them from indefinite integrals, which result in functions plus a constant.

Ellipse

Understanding the function \( y = \sqrt{9 - 5x^{2}}\) requires recognizing its geometric form. This curve represents a portion of an ellipse, which is defined by the equation \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), where \ a, b\ are the semi-major and semi-minor axes. For our function:

• \(\frac{x^2}{\frac{9}{5}} + \frac{y^2}{9} = 1 \) connects the given function to its elliptical form.
• Knowing it’s an ellipse helps visualize and sketch it accurately.
• The region under the curve from \(x = 0\) to \(x = 1\) is a segment of this ellipse.

Ellipses are important in many areas of mathematics and physics, making understanding their properties fundamental.