Sketch The Region Enclosed By The Given Curves. Decide Whether To Integrate With Respect To X Or Y. Draw
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Introduction
Understanding how to sketch the region enclosed by curves and determine the best way to perform integration is a fundamental skill in calculus. This process not only helps in visualizing the problem but also simplifies the computation of areas, volumes, and other related quantities. When dealing with multiple curves, selecting the appropriate variable of integration—either x or y—is crucial for setting up the integral efficiently. This article provides a comprehensive guide on how to sketch the enclosed region, decide whether to integrate with respect to x or y, and illustrate the process with practical examples.
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Why Sketching the Enclosed Region Matters
Sketching the enclosed region formed by given curves is essential because:
- It provides a clear visual understanding of the limits of integration.
- It helps identify the shape and boundaries of the region.
- It reveals intersections and overlaps of curves, which are crucial in setting up integrals.
- It aids in choosing the most straightforward variable of integration.
Without a proper sketch, setting up integrals can become confusing, especially for complex regions.
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Understanding the Problem: Given Curves
In many calculus problems, you are given two or more curves, and the goal is to find the area enclosed between them. Typical scenarios include:
- Finding the area between two functions, such as \( y = f(x) \) and \( y = g(x) \).
- Enclosed regions bounded by curves in different quadrants.
- Regions enclosed by curves that intersect at multiple points.
The key steps involve:
- Identifying the curves involved.
- Finding the intersection points.
- Analyzing the shape of the region.
- Deciding the best variable of integration.
- Sketching the region accordingly.
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Step-by-Step Guide to Sketching and Deciding the Variable of Integration
- Identify the Curves and Their Equations
Begin by carefully noting the equations of the curves involved. For example:
- Lines: \( y = mx + c \)
- Parabolas: \( y = ax^2 + bx + c \)
- Circles: \( (x - h)^2 + (y - k)^2 = r^2 \)
- Other functions: \( y = \sqrt{x} \), \( y = \sin x \), etc.
Understanding the nature of these curves helps predict the shape of the enclosed region.
- Find the Intersection Points
Calculate the points where the curves intersect by solving their equations simultaneously. These points determine the limits of integration. For example:
- Set \( y = f(x) \) and \( y = g(x) \) equal to find the intersection \( x \)-values.
- For curves in terms of \( y \), set their \( y \)-equations equal to find corresponding \( y \)-values.
Having the intersection points helps in sketching the region accurately.
- Analyze the Region's Shape and Boundaries
Based on the intersection points and the equations:
- Sketch the curves on a coordinate plane.
- Shade the region enclosed by the curves.
- Observe whether the region spans more in the x-direction or y-direction.
This analysis will inform whether the integral should be with respect to \( x \) or \( y \).
- Decide on the Variable of Integration
Choosing whether to integrate with respect to \( x \) or \( y \) depends on:
- The orientation of the region.
- The ease of expressing the bounds.
- The nature of the curves.
When to integrate with respect to \( x \):
- The region is bounded between curves that can be expressed as \( y = f(x) \) and \( y = g(x) \).
- The limits of \( x \) are clear from the intersection points.
- The region's shape is more straightforward horizontally.
When to integrate with respect to \( y \):
- The region is bounded between curves that can be expressed as \( x = f(y) \) and \( x = g(y) \).
- The limits of \( y \) are clear from the intersection points.
- The region's shape is more straightforward vertically.
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Practical Examples
Example 1: Enclosed Region Between Two Functions
Given Curves:
- \( y = x^2 \)
- \( y = 4x \)
Step 1: Find intersection points.
Set \( x^2 = 4x \):
\[
x^2 - 4x = 0 \Rightarrow x(x - 4) = 0
\]
Solutions:
\[
x = 0 \quad \text{or} \quad x = 4
\]
Corresponding \( y \)-values:
- At \( x=0 \), \( y=0 \).
- At \( x=4 \), \( y=16 \).
Step 2: Sketch the curves.
- \( y = x^2 \) is a parabola opening upwards.
- \( y=4x \) is a straight line passing through the origin with slope 4.
Step 3: Determine the region.
Between \( x=0 \) and \( x=4 \):
- \( y=4x \) is above \( y=x^2 \).
Step 4: Decide variable of integration.
Since the region is bounded between \( x=0 \) and \( x=4 \), and the curves are expressed as functions of \( x \):
- It’s easier to integrate with respect to \( x \).
Step 5: Set up the integral.
Area \( A \):
\[
A = \int{0}^{4} [\, \text{top curve} - \text{bottom curve} \,] \, dx = \int{0}^{4} (4x - x^2) \, dx
\]
Step 6: Sketch the region.
- Draw the parabola \( y=x^2 \).
- Draw the line \( y=4x \).
- Shade the area between \( x=0 \) and \( x=4 \).
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Example 2: Enclosed Region Between Two Curves in Terms of \( y \)
Given Curves:
- \( y = \sqrt{x} \)
- \( y = x/4 \)
Step 1: Find intersection points.
Express both \( y \)-curves:
- \( y = \sqrt{x} \Rightarrow x = y^2 \)
- \( y = x/4 \Rightarrow x=4y \)
Set equal:
\[
\sqrt{x} = \frac{x}{4}
\]
Express in terms of \( x \):
\[
\sqrt{x} = \frac{x}{4}
\]
Multiply both sides by 4:
\[
4 \sqrt{x} = x
\]
Let \( t = \sqrt{x} \Rightarrow x = t^2 \):
\[
4t = t^2
\]
Rearranged:
\[
t^2 - 4t = 0 \Rightarrow t(t - 4) = 0
\]
Solutions:
\[
t=0 \Rightarrow x=0
\]
\[
t=4 \Rightarrow x=16
\]
Corresponding \( y \):
- For \( x=0 \), \( y=0 \).
- For \( x=16 \), \( y=\sqrt{16}=4 \).
Step 2: Sketch the curves.
- \( y= \sqrt{x} \) is a rightward-opening curve.
- \( y= x/4 \) is a straight line.
Step 3: Find the bounds in \( y \).
Express \( x \) in terms of \( y \):
- \( x= y^2 \)
- \( x=4 y \)
In the \( y \)-direction:
- \( y \) varies from 0 to 4.
Step 4: Determine which curve is to the right or left.
At a particular \( y \), the \( x \)-values are:
- \( x= y^2 \) (left boundary)
- \( x= 4 y \) (right boundary)
Between \( y=0 \) and \( y=4 \), the region is bounded by these two curves.
Step 5: Choose to integrate with respect to \( y \).
- Since \( x \) is expressed as functions of \( y \), and the bounds are in \( y \), it's more efficient to integrate with respect to \( y \).
Step 6: Set up the integral.
Area \( A \):
\[
A = \int{0}^{4} [\, \text{right curve} - \text{left curve} \,] \, dy = \int{0}^{4} (4 y - y^2) \, dy
\]
Step 7: Sketch the region.
- Draw the parabola \( y= \sqrt{x} \).
- Draw the line \( y= x/4 \).
- Shade the region between \( y=0 \) and \( y=4 \), bounded by the curves.
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Tips for Deciding the Best Variable of Integration
- Plot the curves first: Visualizing the region helps in understanding its shape.
- Express the boundaries explicitly: Know whether the curves are functions of \( x \) or \( y \).
- Check the intersection points: Limits of integration are based on these points.
- Compare complexity: Choose the
