Problem 14 - Thomas Calculus

Question

The integrals and sums of integrals in Exercises 9–14 give the areas of regions in the xy-plane. Sketch each region, label each bounding curve with its equation, and give the coordinates of the points where the curves intersect. Then find the area of the region. $$ \int_{0}^{2} \int_{x^{2}-4}^{0} d y d x+\int_{0}^{4} \int_{0}^{\sqrt{x}} d y d x $$

Step-by-Step Solution

Answer

The total area is \( \frac{32}{3} \) square units.

Step 1: Understand the Double Integral Setup

The given expression consists of two double integrals:1. \( \int_{0}^{2} \int_{x^2-4}^{0} dy \, dx \) Here, the inner integral \( \int_{x^2-4}^{0} dy \) implies that for any fixed \( x \) within [0, 2], \( y \) ranges from \( x^2 - 4 \) to 0. The outer integral \( \int_{0}^{2} \, dx \) then sweeps through x-values from 0 to 2. 2. \( \int_{0}^{4} \int_{0}^{\sqrt{x}} dy \, dx \) For this integral, \( y \) ranges from 0 to \( \sqrt{x} \) for \( x \) in the interval [0, 4]. This defines two regions in the xy-plane, which we need to visualize and calculate their areas.

Step 2: Sketch the Regions

1. For the first integral, the curve \( y = x^2 - 4 \) is a parabola opening upwards, shifted downwards by 4 units from \( y = x^2 \). It intersects the x-axis at: - Set \( y = 0 \): \( 0 = x^2 - 4 \) gives \( x = \pm 2 \). In the range \( x \) from 0 to 2, this defines a region below the x-axis and above the line from \( y = x^2 - 4 \) to \( y = 0 \).2. For the second integral, the curve \( y = \sqrt{x} \) is the top boundary, and \( y = 0 \) (x-axis) is the bottom boundary. Thus, this region is above the x-axis from \( x = 0 \) to \( x = 4 \).

Step 3: Identify Intersection Points

For potential intersection points among the curves:1. The parabola from the first integral does not intersect with the vertical boundaries of the second integral.2. The intersection of \( y = \sqrt{x} \) (curve from the second integral) with the x-axis occurs at \( x = 0 \) and \( x = 4 \).There are no intersections involving the region bound by \( y = x^2 - 4 \) and \( y = \sqrt{x} \) in the prescribed limits. Each region is distinctly non-overlapping within the described bounds.

Step 4: Calculate First Area

Evaluate the first integral: \[ \int_{0}^{2} \int_{x^2-4}^{0} dy \, dx = \int_{0}^{2} [0 - (x^2 - 4)] \, dx = \int_{0}^{2} (4 - x^2) \, dx \]Integrate with respect to \( x \):\[ = \left[ 4x - \frac{x^3}{3} \right]_{0}^{2} = \left( 8 - \frac{8}{3} \right) = \frac{24}{3} - \frac{8}{3} = \frac{16}{3} \]

Step 5: Calculate Second Area

Evaluate the second integral:\[ \int_{0}^{4} \int_{0}^{\sqrt{x}} dy \, dx = \int_{0}^{4} [\sqrt{x} - 0] \, dx = \int_{0}^{4} \sqrt{x} \, dx \]Change the integrand into a power function:\[ = \int_{0}^{4} x^{1/2} \, dx \]Integrate with respect to \( x \):\[ = \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{4} = \frac{2}{3}[x^{3/2}]_{0}^{4} = \frac{2}{3}[8] = \frac{16}{3} \]

Step 6: Sum the Areas

The total area is the sum of the areas obtained from both integrals:\[ \text{Total Area} = \frac{16}{3} + \frac{16}{3} = \frac{32}{3} \]

Step 7: Conclusion: Provide Total Area

The area of the combined regions described by the integrals is \( \frac{32}{3} \) square units.

Key Concepts

Area Calculation

To understand the area calculation using double integrals, imagine peeling layers from an onion. Each "layer" corresponds to a tiny slice of area in the region of interest.
Double integrals allow us to calculate the total area seamlessly by summing these slices over the entire region.
The given expression consists of two separate double integrals. Each double integral calculates the area of a specific region in the xy-plane. The final total area is simply the sum of these two areas.
By evaluating each integral, we slice through the region step-by-step:

• First, we integrate in the y-direction, adding vertical slices of the region.
• Then, we integrate the accumulated y data in the x-direction, sweeping across the x-range.

This sequential approach provides the precise sum of all tiny areas within the specified bounds, resulting in the total area of the region.

xy-plane Region

The xy-plane is where all the action happens for double integrals and our area calculations. It acts as a 2D canvas where we map and analyze different regions defined by mathematical equations. In the given problem, two distinct regions are carved out on this plane:

• The region for the first integral is defined by the area between the parabola curve "below the x-axis." Specifically from the curve equation, its lower boundary is given by \( y = x^2 - 4 \), while the x-axis \( y = 0 \) constitutes the upper boundary.
• The next region for the second integral lies "above the x-axis," framed by the square root function \( y = \sqrt{x} \). This indicates it covers an area extending all the way from \( x = 0 \) to \( x = 4 \).

In essence, these regions are areas on the plot where function limits create a space through their mathematical relationships. This visual representation helps us understand where the calculated areas are found in the plane.

Curve Intersection Points

Intersection points between curves are like meeting spots on our xy-plane canvas. They provide crucial information about where regions start or end.In this exercise, identifying the intersection points allows us to determine the exact bounds for our integration: - For the first region, the parabola \( y = x^2 - 4 \) intersects the x-axis at \( x = \pm 2 \). In our range from \( x = 0 \) to \( x = 2 \), the parabola sits below the x-axis.- The second curve \( y = \sqrt{x} \) intersects the x-axis at \( x = 0 \) and \( x = 4 \), which are boundaries for the second integration region.These intersections don't overlap with each other or encompass multiple areas, which simplifies our task. Each bound sets clear limits for evaluating the double integrals for distinct and separate regions.

Parabolic Boundary

A parabola can be thought of as a bell-shaped curve, commonly appearing in mathematics and real-world phenomena. For this exercise, the parabolic boundary of interest comes from the equation \( y = x^2 - 4 \).The feature of this equation is:

• It exhibits a U-shape curve.
• Begins to plot its curve just below the x-axis between \( x = -2 \) and \( x = 2 \), dictating one region's limits.

Understanding the parabola's effects is crucial because it informs us how the region beneath the curve is formed. The portion between the x-axis and the parabola counts toward the calculated area within the confined segment. Adjusting the equation of the parabola shifts its position, thereby directly influencing the parabolic region and the resulting area in calculations.