Problem 13 - Calculus: Early Transcendentals

Question

Sketch the region enclosed by the given curves and find its area. \( y = 12 - x^2 \) , \( y = x^2 - 6 \)

Step-by-Step Solution

Answer

The area enclosed by the curves is 72 square units.

Step 1: Find Points of Intersection

To find the region enclosed by the curves, first determine where they intersect. Set the two equations equal to each other: \( 12 - x^2 = x^2 - 6 \). Simplify to find \( 12 + 6 = 2x^2 \), which gives \( 18 = 2x^2 \) or \( x^2 = 9 \). Solving this gives \( x = 3 \) and \( x = -3 \). So, the curves intersect at \((-3, 3^2 - 6 = 3\)) and \((3, 3^2 - 6 = 3\)).

Step 2: Set Up the Integral for Area

The area between the curves can be found by integrating the upper function minus the lower function from the left intersection point to the right intersection point. The upper curve is \( y = 12 - x^2 \) and the lower is \( y = x^2 - 6 \). Thus, set up the integral: \( \int_{-3}^{3} ((12 - x^2) - (x^2 - 6)) \, dx \).

Step 3: Simplify the Integral Expression

Simplify the function you're integrating: \( (12 - x^2) - (x^2 - 6) = 12 - x^2 - x^2 + 6 = 18 - 2x^2 \). The integral becomes \( \int_{-3}^{3} (18 - 2x^2) \, dx \).

Step 4: Compute the Integral

Evaluate the integral \( \int_{-3}^{3} (18 - 2x^2) \, dx \). First, find the antiderivative: \( 18x - \frac{2}{3}x^3 \). Then compute from \(-3\) to \(3\): \((18(3) - \frac{2}{3}(3)^3) - (18(-3) - \frac{2}{3}(-3)^3)\).

Step 5: Calculate and Verify the Area

Calculate the definite integral: \( 18(3) - \frac{2}{3}(27) = 54 - 18 = 36 \) and \( 18(-3) - \frac{2}{3}(-27) = -54 + 18 = -36 \). This results in \( 36 - (-36) = 72 \). The area between the curves is \( 72 \) square units.

Key Concepts

Points of Intersection

To determine where two curves intersect, we equate their expressions and solve for the values of the variable. In this exercise, the curves given are described by equations \( y = 12 - x^2 \) and \( y = x^2 - 6 \). By setting these two equations equal to each other, \( 12 - x^2 = x^2 - 6 \), and simplifying, we find:\

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• The equation reduces to \( 12 + 6 = 2x^2 \), leading to \( 18 = 2x^2 \).
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• This simplifies further to \( x^2 = 9 \), giving solutions \( x = 3 \) and \( x = -3 \).
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\Thus, the curves intersect at the points \((-3, 3)\) and \((3, 3)\), which are crucial in defining the interval for integrating to find the area between these curves.

Definite Integral

The definite integral is a powerful tool for computing the area between two curves. To find the area enclosed by the curves \( y = 12 - x^2 \) and \( y = x^2 - 6 \), you must compute the integral over the interval defined by the points of intersection found previously, \([-3, 3]\). Define the upper and lower curves over this interval. In the provided problem:\

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• The upper curve is \( y = 12 - x^2 \).
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• The lower curve is \( y = x^2 - 6 \).
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• The integral concerning area is set as \( \int_{-3}^{3} ((12 - x^2) - (x^2 - 6)) \, dx \).
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\This integral will compute the accumulated area between these two curves over the specified interval, taking into account their relative positions.

Antiderivative

Finding the antiderivative is a crucial step in the integration process, especially when calculating definite integrals to find area. The antiderivative gives us a function that, when derived, yields the integrand, i.e., the expression within the integral. In this example, we simplify the integrand \((12 - x^2) - (x^2 - 6) = 18 - 2x^2\).\

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• The antiderivative of each term then becomes \( 18x - \frac{2}{3}x^3 \).
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• This expression, evaluated at the endpoints of the interval, will give the desired net area.
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\Understanding the concept of finding an antiderivative is key to computing integrals and ultimately solving these types of problems.

Integral Evaluation

The final step in finding the area between two curves using definite integrals involves evaluating the integral at the points of intersection. For the integral \( \int_{-3}^{3} (18 - 2x^2) \, dx \), the antiderivative \( 18x - \frac{2}{3}x^3 \) is used.\

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• First, substitute \( x = 3 \): \( 18(3) - \frac{2}{3}(3)^3 = 54 - 18 = 36 \).
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• Then, substitute \( x = -3 \): \( 18(-3) - \frac{2}{3}(-3)^3 = -54 + 18 = -36 \).
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• The difference between these results gives the total area: \( 36 - (-36) = 72 \).
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\By following these steps of evaluating the integral, students can accurately find the area between the specified curves. This foundational process is essential in calculus and helps expand understanding of integrals and their applications.