Problem 47 - Calculus: Early Transcendentals

Question

Sketch the region enclosed by the given curves and calculate its area. \( y = 4 - x^2 \), \( y = 0 \)

Step-by-Step Solution

Answer

The area enclosed by the curves is \( \frac{32}{3} \) square units.

Step 1: Identify the Curves

We have two curves to consider: the parabola given by the equation \( y = 4 - x^2 \) and the line given by \( y = 0 \). The parabola opens downwards with its vertex at (0, 4), and the line \( y = 0 \) is the x-axis.

Step 2: Determine the Intersection Points

To find where the curves intersect, set the equations equal: \( 4 - x^2 = 0 \). This simplifies to \( x^2 = 4 \). Taking the square root of both sides gives \( x = \pm 2 \). Thus, the curves intersect at the points (2, 0) and (-2, 0).

Step 3: Sketch the Region

The region enclosed by the curves lies between the parabola and the x-axis (\( y = 0 \)) from \( x = -2 \) to \( x = 2 \). Sketch the parabola opening downwards with its vertex at (0, 4) and mark the intersection points at x = -2 and x = 2. Shade the region between the parabola and the x-axis.

Step 4: Set Up the Integral for Area

The area between the curves from \( x = -2 \) to \( x = 2 \) can be found by integrating the difference between the function \( y = 4 - x^2 \) and \( y = 0 \). The integral is: \[ \int_{-2}^{2} (4 - x^2) \, dx \]

Step 5: Calculate the Integral

Calculate the antiderivative of the function \( 4 - x^2 \). The antiderivative is \( 4x - \frac{x^3}{3} \). Evaluate this antiderivative from \( x = -2 \) to \( x = 2 \):\[ \left[ 4x - \frac{x^3}{3} \right]_{-2}^{2} = \left( 4(2) - \frac{2^3}{3} \right) - \left( 4(-2) - \frac{(-2)^3}{3} \right) \]This simplifies to:\[ (8 - \frac{8}{3}) - (-8 + \frac{8}{3}) \]Calculate each part to find the total area.

Step 6: Simplify Calculations

Simplify the expression by calculating each term:\[ 8 - \frac{8}{3} = \frac{24}{3} - \frac{8}{3} = \frac{16}{3} \] \[ -8 + \frac{8}{3} = \frac{-24}{3} + \frac{8}{3} = \frac{-16}{3} \]The total area is:\[ \frac{16}{3} - (\frac{-16}{3}) = \frac{16}{3} + \frac{16}{3} = \frac{32}{3} \]

Step 7: Conclusion on Area

The area of the region enclosed by the curves \( y = 4 - x^2 \) and \( y = 0 \) is \( \frac{32}{3} \) square units.

Key Concepts

Definite Integral

A definite integral is a fundamental concept in calculus that allows us to calculate the area under a curve over a specified interval. When we calculate an integral, we essentially sum up infinitesimally small rectangles that fit under the curve, from a starting point to an ending point on the x-axis.

For this exercise, the area between the parabola and the x-axis is found using the definite integral of the function that represents the parabola, specifically:

• The function: \(4 - x^2\)
• The limits: from \(x = -2\) to \(x = 2\)

The integral setup looks like this: \[\int_{-2}^{2} (4 - x^2) \, dx\]We find the antiderivative, or the original function from which the derivative was taken, evaluate it at these limits, and subtract the results to get the total enclosed area.

This technique allows us to precisely calculate areas even if the shapes formed by the curves are irregular, making definite integrals incredibly useful in a wide range of applications.

Intersection Points

Finding intersection points between curves is critical in determining the bounds of the enclosed area that we need to calculate. The process involves setting the equations of the curves equal to each other and solving for the x-values where they intersect. These x-values are where the curves meet.

In our example, the intersection is found by solving:\[4 - x^2 = 0\]This simplifies to \(x^2 = 4\), giving us the intersection points \(x = \pm 2\).

These points, \((2, 0)\) and \((-2, 0)\), define the interval for our definite integral. Understanding where curves intersect helps us correctly set up the limits for integrations and accurately determine the areas we are interested in.

If curves intersect at more than two points, the process can be extended by breaking down the integration into multiple intervals, ensuring the correct regions are considered in the area calculation.

Parabola

A parabola is a U-shaped curve described by a quadratic equation of the form \(y = ax^2 + bx + c\). The curve either opens upwards or downwards depending on the sign of the leading coefficient \(a\).

In this exercise, the parabola is given by the equation \(y = 4 - x^2\). Here, \(-x^2\) indicates a parabola that opens downwards. The vertex, which is the highest point since it opens downwards, is located at \((0, 4)\). This means for every unit step left or right from the vertex, the y-value decreases by \(-x^2\).

Parabolas are symmetric with respect to their vertical axis, which is called the line of symmetry. For \(y = 4 - x^2\), the symmetry is around the y-axis, making the parabola an ideal function for integration when calculating areas. Understanding these characteristics helps in sketching and visualizing the parabola.

Sketching Curves

Sketching curves is a crucial skill for visualizing the region whose area we need to calculate. It involves drawing the graphs of the equations involved to identify regions of intersection and enclosed areas.

To sketch the parabola \(y = 4 - x^2\), note the vertex at \((0, 4)\) and that the parabola opens downwards due to the \(-x^2\) term. Plot the vertex and mark the intersection points at \(x = -2\) and \(x = 2\), where the parabola crosses the x-axis. Draw a smooth curve connecting these points, forming a U-shape.

Also, sketch the x-axis, which is given by \(y = 0\), a horizontal line across the graph. The region we are interested in is the space between the downward-opening parabola and the x-axis between the points \(x = -2\) and \(x = 2\). Shading this region helps visually confirm what the integral calculation represents.

By sketching, we ensure clarity in setting up integrals and the ability to identify any additional features of the curves that might affect our calculations.