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Chapter 15: Problem 3

Sketch the described regions of integration. $$-2 \leq y \leq 2, \quad y^{2} \leq x \leq 4$$

Short Answer

Expert verified
The region is between the parabola \(x = y^2\) and the line \(x = 4\), bounded by \(-2 \leq y \leq 2\).

Step by step solution

01

Analyze the Inequality for y

The given inequality for y is \(-2 \leq y \leq 2\). This tells us the range of y values extends from -2 to 2 along the vertical axis. This implies horizontal strips from \(y = -2\) to \(y = 2\) on a graph.
02

Analyze the Inequality for x

The inequality for x is \(y^2 \leq x \leq 4\). The expression \(x = y^2\) represents a parabolic curve that opens to the right. The inequality \(x \leq 4\) places a vertical boundary at \(x = 4\). Thus, x ranges between the curve \(y^2\) and the line \(x = 4\).
03

Determine the Region of Integration

Combine the information from Steps 1 and 2. The region is bounded below by \(y = -2\) and above by \(y = 2\), from the curve \(x = y^2\) to the right at \(x = 4\). This creates a bounded area on the Cartesian plane between the parabola and vertical line.
04

Sketch the Region

On the graph, draw the parabolic curve \(x = y^2\) which opens to the right. Mark the boundary at \(x = 4\). Draw horizontal lines across the parabola and the vertical boundary starting from \(y = -2\) to \(y = 2\),which fills in the area of integration. The resulting sketch will show a vertically stretched, right-facing parabolic region bounded at \(x = 4\).

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Key Concepts

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

Cartesian plane
The Cartesian plane is a fundamental concept that we use to graphically represent mathematical equations and functions. It's a two-dimensional plane defined by two perpendicular axes: the horizontal axis called the x-axis and the vertical axis called the y-axis.
  • Origin: The point where these two axes intersect is the origin, marked as (0,0).
  • Quadrants: The plane is divided into four quadrants. The top right is the first quadrant, moving counter-clockwise ends in the fourth quadrant.
  • Locating Points: Every point on this plane can be defined by a pair of numbers (x, y), which describe its location relative to the origin.
On the Cartesian plane, we can graph various types of functions to visualize the regions they define. In the context of our problem, the Cartesian plane helps us understand how the inequalities for x and y create a region of integration. Here, the region exists between a parabola and a vertical line, all plotted on the plane.
parabola
A parabola is a symmetrical, curved shape that can open either upwards, downwards, or sideways depending on the equation defining it. In our exercise, we encounter a parabola of the form \(x = y^2\), which opens to the right. This implies that for each y value, the x values are non-negative.
  • Standard Form: The general form of a parabolic equation is \(y = ax^2 + bx + c\) (opens up or down) or \(x = ay^2 + by + c\) (opens left or right).
  • Vertex: The turning point of the parabola is known as its vertex. For \(x = y^2\), the vertex is at the origin (0, 0).
  • Axis of Symmetry: Parabolas are symmetric about their axis. In \(x = y^2\), the axis is the x-axis.
This parabola, \(x = y^2\), defines one boundary of our region of integration, directing us to slice the Cartesian plane from this curve to a vertical line.
inequalities
Inequalities are mathematical expressions that show the relationship of non-equalities between variables. They tell us which values satisfy an equation or create constraints. In the problem, we deal with two inequalities:
  • For y: \(-2 \leq y \leq 2\). This sets the vertical limits for the region, indicating that our region stretches from \(y = -2\) to \(y = 2\).
  • For x: \(y^2 \leq x \leq 4\). It confines x between the parabola \(x = y^2\) and the vertical line \(x = 4\).
Understanding inequalities is crucial to identifying regions on a graph, as they highlight where certain conditions are met. By representing these inequalities graphically on the Cartesian plane, we establish clear boundaries for our region of integration.
bounded area
The bounded area in a graph is essentially the space enclosed between curves and lines. In this exercise, the bounded area is described by the intersection of inequalities on the Cartesian plane.
  • Boundaries: This region is bounded vertically by \(y = -2\) and \(y = 2\), and horizontally between \(x = y^2\) and the line \(x = 4\).
  • Characteristics: It forms a closed shape, meaning its perimeter is completely defined by the boundaries mentioned above.
The bounded area is crucial for integration, as it simplifies determining where a function or equation operates. By clearly defining this region, one can easily sketch and understand the area involved, which in our case lies between a right-facing parabola and a vertical line.

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Most popular questions from this chapter

Sketch the region of integration and convert each polar integral or sum of integrals to a Cartesian integral or sum of integrals. Do not evaluate the integrals. $$\int_{0}^{\pi / 4} \int_{0}^{2 \sec \theta} r^{5} \sin ^{2} \theta d r d \theta$$

Sketch the region of integration and the solid whose volume is given by the double integral. $$\int_{0}^{3} \int_{0}^{2-2 x / 3}\left(1-\frac{1}{3} x-\frac{1}{2} y\right) d y d x$$

Sketch the region of integration and write an equivalent double integral with the order of integration reversed. $$\int_{0}^{\ln 2} \int_{e}^{2} d x d y$$

Sketch the region of integration, reverse the order of integration, and evaluate the integral. \(\iint_{R}\left(y-2 x^{2}\right) d A\) where \(R\) is the region bounded by the square \(|x|+|y|=1\)

Evaluate the integral $$\int_{0}^{\infty} \int_{0}^{\infty} \frac{1}{\left(1+x^{2}+y^{2}\right)^{2}} d x d y$$
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