/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 13 Let \(D\) be the region bounded ... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(D\) be the region bounded by the curves of equations \(y=x, y=-x\), and \(y=2-x^{2} .\) Explain why \(D\) is neither of Type Inor II.

Short Answer

Expert verified
Region \(D\) cannot be expressed as Type I or II due to disjoint intersections and boundaries.

Step by step solution

01

Understand the Types

Type I region refers to regions that can be expressed in the form \(a \leq x \leq b\), where \(\varphi_1(x) \leq y \leq \varphi_2(x)\) for two functions \(\varphi_1\) and \(\varphi_2\). Meanwhile, a Type II region refers to regions that can be expressed as \(c \leq y \leq d\), where \(\psi_1(y) \leq x \leq \psi_2(y)\) for functions \(\psi_1\) and \(\psi_2\). Let's determine if region \(D\) fits either description.
02

Analyze the Curves

The curves given are \(y = x\), \(y = -x\), and \(y = 2 - x^2\). These curves intersect to form bounds for the region \(D\). The parabola \(y = 2 - x^2\) is downward-opening and the lines \(y = x\) and \(y = -x\) are straight lines.
03

Check Intersections

To find the point of intersection between the parabola \(y = 2 - x^2\) and each line, set the equations equal. For \(y = x\), we solve \(x = 2 - x^2\). Solving \(x^2 + x - 2 = 0\) gives roots \(x = 1\) and \(x = -2\). For \(y = -x\), solve \(-x = 2 - x^2\), resulting in \(x^2 - x - 2 = 0\) with roots \(x = 2\) and \(x = -1\).
04

Identify Boundaries for Type I

The boundaries for \(x\) are from \(-2\) to \(2\). For each \(x\), the terminals \(y\) can drop from \(y = -x\) to \(y = x\), or \(y = 2 - x^2\) for overlapping segments. There's no single continuous function pair for all \(x\) from \(-2\) to \(2\) without contradiction, specifically around \(x = -1\) to \(x = 1\), failing the Type I structure.
05

Identify Boundaries for Type II

For \(y\) from \(-2\) to \(1\), evaluate different \(x\) bounds. When \(-2 \leq y \leq 0\), possible \(x\) values run from intersection of \(y = -x\) to \(y = x\). For \(0 \leq y \leq 1\), \(x\) runs from intersection with \(x=-1\) to \(x=1\). Above \(y=1\), \(x\) is bound by the parabolic arcs. The lack of unified span for all \(y\) means it fails Type II constraints.
06

Conclusion

Region \(D\) is neither Type I nor Type II because you can't find a continuous expression for \(y\) in terms of \(x\) satisfying Type I, and there is no continuous \(x\) in terms of \(y\) satisfying Type II across the whole region.

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Key Concepts

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

Type I Region
A Type I region in integral calculus is one that can be described with boundaries primarily defined along the x-axis. Specifically, it is expressed as
  • \(a \leq x \leq b\)
  • where the region between two curves or lines, \(\varphi_1(x) \leq y \leq \varphi_2(x)\), lies.
This implies that for every x-value between \(a\) and \(b\), the y-values are dependently bounded by the two curves or lines.
Such regions are simpler to integrate over as the functions specify clear upper and lower bounds for each x.
However, for the region \(D\) bounded by the curves \(y=x\), \(y=-x\), and \(y=2-x^2\), it cannot fit the Type I structure because there's no single pair of continuous functions \(\varphi_1\) and \(\varphi_2\) that spans from \(-2\) to \(2\) on the x-axis without logically conflicting at some x-values.
Specifically, between \(x=-1\) and \(x=1\), the vertical structure of the region changes, breaking the continual function pair requirement.
Type II Region
Type II regions focus on boundaries defined along the y-axis. They can be expressed as
  • \(c \leq y \leq d\)
  • where for every y in this range, the limits are from two functions \(\psi_1(y)\) to \(\psi_2(y)\) on the x-axis.
This scenario allows integration along y, where \(x\) is seen as a dependent variable.
However, in our bounded region \(D\), using stepwise segments like when \(-2 \leq y \leq 0\) or \(0 \leq y \leq 1\), shows varying x-bounds.
The limitation lies in a lack of a coherent description that is applicable across the entire range for \(y\).
Essentially, no single continuous description can satisfy the Type II region's structural requirement which implies constant x-bound limits for every y-span from \(-2\) to \(1\).
Curves Intersection
Understanding the intersection of curves is crucial as it determines the boundaries of a region.
The curves given in the exercise comprise the intersecting lines and parabola:
  • \(y = x\)
  • \(y = -x\)
  • \(y = 2 - x^2\)
To find where these curves meet, equate them two at a time.
For instance, solving \(x = 2 - x^2\) gives the intersection points \(x = 1\) and \(x = -2\).
Meanwhile, \(-x = 2 - x^2\) yields \(x = 2\) and \(x = -1\).
These intersection points effectively define edges of region \(D\) on the Cartesian plane.
In solving practical problems, it is this intersection logic that helps establish the framework over which regions can be calculated.
Bounded Region
A bounded region is an area confined by the intersection of curves or lines.
Region \(D\), in this instance, is framed by the equations: \(y=x\), \(y=-x\), and \(y=2-x^2\).
The enclosing of this area within these curves is what makes it *bounded*.
This region is neither Type I nor Type II, as checked, due to its complexity along both axes.
When analyzing such regions, one must consider whether it can simplify into a computable Type I or II. Otherwise, more complex multivariable calculus solutions might be necessary.
Integrating over bounded regions often involves scrutinizing these limits and structures to find feasible paths of calculation.

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