Problem 46 - Calculus Volume 1

Question

For the following exercises, find the exact area of the region bounded by the given equations if possible. If you are unable to determine the intersection points analytically, use a calculator to approximate the intersection points with three decimal places and determine the approximate area of the region. $$ [T]x=4-y^{2} \text { and } x=1+3 y+y^{2} $$

Step-by-Step Solution

Answer

The exact area is about 10.321 square units.

Step 1: Equation Rewriting

First, rewrite the given equations to make them easier to compare. The first equation is given as \( x = 4 - y^2 \). The second equation is \( x = 1 + 3y + y^2 \). These equations both express \( x \) in terms of \( y \).

Step 2: Find Intersection Points

To find the intersection points of the curves, set the equations equal to each other: \( 4 - y^2 = 1 + 3y + y^2 \). Rearrange this equation to form the quadratic equation \( 2y^2 + 3y - 3 = 0 \).

Step 3: Solve the Quadratic Equation

Solve \( 2y^2 + 3y - 3 = 0 \) using the quadratic formula: \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2 \), \( b = 3 \), and \( c = -3 \). Compute the discriminant: \( b^2 - 4ac = 9 + 24 = 33 \). The roots are \( y = \frac{-3 \pm \sqrt{33}}{4} \). Compute these to find approximate points: \( y_1 = 0.686 \) and \( y_2 = -2.186 \).

Step 4: Set Up the Integral for Area

Integrate with respect to \( y \) from \( y_2 \) to \( y_1 \). The area between the curves is given by the formula \( \int_{y_2}^{y_1} [(4 - y^2) - (1 + 3y + y^2)] \ dy \). Simplify the integrand to \( 3 - 3y - 2y^2 \).

Step 5: Evaluate the Integral

Find the antiderivative of \( 3 - 3y - 2y^2 \), which is \( 3y - \frac{3}{2}y^2 - \frac{2}{3}y^3 \). Evaluate this antiderivative at the bounds \( y_1 = 0.686 \) and \( y_2 = -2.186 \): calculate \( 3(0.686) - \frac{3}{2}(0.686)^2 - \frac{2}{3}(0.686)^3 \) and \( 3(-2.186) - \frac{3}{2}(-2.186)^2 - \frac{2}{3}(-2.186)^3 \).

Step 6: Compute the Area

Compute the difference: subtract the lower bound evaluation from the upper bound evaluation to find the area. After calculation, the area is approximately \( 10.321 \).

Key Concepts

Intersection Points

Intersection points are where two curves meet on a graph. To find these points, we set the equations equal to each other. This gives us a single equation with one variable, in this case, \( y \).

• The first step involves writing each equation in a comparable form.
• In our problem, \( x = 4 - y^2 \) and \( x = 1 + 3y + y^2 \) are equalized.
• This creates the equation: \( 4 - y^2 = 1 + 3y + y^2 \).

The solution to this equation helps us locate where the curves intersect. These intersection points are crucial for determining the bounds for calculating the area between the curves.

Quadratic Equation

A quadratic equation is a polynomial equation of degree 2. It takes the standard form \( ax^2 + bx + c = 0 \).

• While finding intersection points, we rearrange the equation into this form.
• In our exercise: \( 2y^2 + 3y - 3 = 0 \) represents the quadratic equation.
• Each letter (\( a = 2, b = 3, c = -3 \)) represents a coefficient of the equation.

To solve it, you apply the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). This formula helps us calculate the values of \( y \) at which the intersection lies.

Integral Calculation

After finding the intersection points, the next step is to calculate the area. This involves setting up an integral.

• For our problem, we integrate \( 3 - 3y - 2y^2 \)
• The limits of integration are from \( y_2 = -2.186 \) to \( y_1 = 0.686 \).
• The integral describes the 'total' or 'accumulated' area between the curves over this interval.

The expression you integrate comes from subtracting one function from the other. This subtraction gives the height of the 'slice' of area at each point \( y \). Understanding how to set up the integral correctly is key to finding the exact area between curves.

Antiderivative Evaluation

The antiderivative, also known as the indefinite integral, is the reverse process of differentiation. To find the area, you evaluate the antiderivative at two bounds and subtract.

• For the function \( 3 - 3y - 2y^2 \), the antiderivative is \( 3y - \frac{3}{2}y^2 - \frac{2}{3}y^3 \).
• Evaluate this expression at \( y_1 = 0.686 \) and \( y_2 = -2.186 \).
• Subtract the value at the lower limit from the value at the upper limit to find the total area.

This calculated difference gives the exact or approximate area between the curves depending on the accuracy of the evaluation. Without this step, determining the precise area would not be possible.