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Chapter 14: Problem 18

The integrals and sums of integrals in Exercises give the areas of regions in the \(x y\) -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. $$\int_{0}^{2} \int_{x^{2}-4}^{0} d y d x+\int_{0}^{4} \int_{0}^{\sqrt{x}} d y d x$$

Short Answer

Expert verified
The total area of the regions is \(\frac{32}{3}\).

Step by step solution

01

Understand the Problem

We need to find the areas of two different regions in the XY-plane and sum them. The first integral covers the area under the curve \(y = x^2 - 4\) and above \(y = 0\) from \(x = 0\) to \(x = 2\), while the second integral covers the area under \(y = \sqrt{x}\) and above \(y = 0\) from \(x = 0\) to \(x = 4\).
02

Sketch the Regions

Plot the graphs of the equations. The parabola \(y = x^2 - 4\) opens upwards, crossing the y-axis at \(-4\). The square root function \(y = \sqrt{x}\) intersects the x-axis at \(x = 0\). Identify the relevant parts between the given limits of integration.
03

Find Intersection Points

To find the area, it's important to identify where these two curves intersect. Set \(x^2 - 4 = 0\) and \(y = \sqrt{x}\) equal to find intersections: \(x^2 = 4\) gives \(x = 2\). The parabola intersects the y-axis at \((0, -4)\) while \(y = \sqrt{x}\) starts at \((0, 0)\).
04

Calculate the First Integral

Evaluate the integral \(\int_{0}^{2} \int_{x^{2}-4}^{0} dy \, dx\). Integrate with respect to \(y\) first, which yields \([- (x^2 - 4)] = 4 - x^2\) from the outer limits from 0 to 2:\[\int_{0}^{2} (4 - x^2) \, dx\]Computing this gives:\[\left[4x - \frac{x^3}{3}\right]_{0}^{2} = \left(8 - \frac{8}{3}\right) = \frac{16}{3}.\]
05

Calculate the Second Integral

Evaluate \(\int_{0}^{4} \int_{0}^{\sqrt{x}} dy \, dx\). First, integrate with respect to \(y\), yielding \(\sqrt{x}\), then integrate with respect to \(x\):\[\int_{0}^{4} \sqrt{x} \, dx = \left[\frac{2}{3}x^{3/2}\right]_{0}^{4} = \frac{2}{3} \times 8 = \frac{16}{3}.\]
06

Add the Areas

Sum the results of the two integrals to find the total area of the regions. The first area is \( \frac{16}{3}\) and the second is also \( \frac{16}{3}\). Thus, the total area is:\[\frac{16}{3} + \frac{16}{3} = \frac{32}{3}.\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Regions in XY-plane
When working with double integrals, we often calculate areas of regions in the XY-plane. The exercise involves two different regions. Each region is bounded by specific curves: one integral describes the area under a parabola, and the other under a square root curve. Understanding these bounding curves is crucial. They define where the integration occurs in the XY-plane. In this problem:
  • The first region is bounded by the parabola curve \( y = x^2 - 4 \), spanning horizontally from \( x=0 \) to \( x=2 \).
  • In the second region, the boundary is formed by the curve \( y = \sqrt{x} \), extending from \( x=0 \) to \( x=4 \).
Visualizing these helps in setting up the limits for integration. Accurately sketching these regions with the curves labeled assists in better understanding how the region looks geometrically.
Intersection Points
Intersection points are where the curves meet in the XY-plane. These intersections can change how or where we set limits in our integrals. They are crucial for determining the integration bounds:
  • The curve \( y = x^2 - 4 \) intersects the y-axis at \( (0, -4) \).
  • Both curves need to be considered on the interval \( x = 0 \) to \( x = 2 \), as calculated by setting \( x^2 - 4 = 0 \), resulting in \( x = 2 \).
Identifying these points is key in splitting complex integration tasks into manageable sections, ensuring that integrations cover the correct portions of the XY-plane.
Area Calculation
Calculating the area using double integrals involves integrating over a defined region in the XY-plane. Each integral in this problem delineates an area:
  • For the first integral, the calculation is done under the curve \( y = x^2 - 4 \), spanning vertically from the curve to \( y = 0 \) over \( x = 0 \) to \( x = 2 \).
  • The second integral calculates the area under \( y = \sqrt{x} \) starting from \( y = 0 \) to the curve, extending from \( x = 0 \) to \( x = 4 \).
These integrals describe the size and shape of regions precisely. Integration within specific bounds minimizes errors and provides an exact measurement of the area.
Integral Evaluation
Evaluating an integral involves computing the actual numbers representing the area determined by our setup. Let's examine how these were computed:
  • The first integral \( \int_{0}^{2} \int_{x^2 - 4}^{0} dy \, dx \) yields the area \( \frac{16}{3} \). By integrating with respect to \( y \) and then \( x \), we evaluate it accurately.
  • The second integral \( \int_{0}^{4} \int_{0}^{\sqrt{x}} dy \, dx \) also results in an area of \( \frac{16}{3} \), first integrating with \( y \), followed by \( x \).
After evaluating, these areas are added up for a total \( \frac{32}{3} \). Mastery of integral evaluation is crucial, enabling you to solve area problems efficiently using mathematical integration techniques.

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Most popular questions from this chapter

Find the mass of the solid region bounded by the parabolic surfaces \(z=16-2 x^{2}-2 y^{2}\) and \(z=2 x^{2}+2 y^{2}\) if the density of the solid is \(\delta(x, y, z)=\sqrt{x^{2}+y^{2}}\)

Evaluate the spherical coordinate integrals. $$\int_{0}^{2 \pi} \int_{0}^{\pi / 4} \int_{0}^{2}(\rho \cos \phi) \rho^{2} \sin \phi d \rho d \phi d \theta$$

Find the average value of the function \(f(r, \theta, z)=r\) over the region bounded by the cylinder \(r=1\) between the planes \(z=-1\) and \(z=1\)

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The integrals we have seen so far suggest that there are preferred orders of integration for cylindrical coordinates, but other orders usually work well and are occasionally easier to evaluate. Evaluate the integrals. $$\int_{0}^{2 \pi} \int_{0}^{3} \int_{0}^{z / 3} r^{3} d r\quad d z\quad d \theta$$
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