Homework Help is Here – Start Your Trial Now!
Math
Calculus
806 Chapter 14 Multiple Integrals It does not apply to regions with complicated holes through them, although sometimes such regions can be subdivided into simpler regions for which the procedure does apply. We illustrate this method of finding the limits of integration in our first example. Let S be the sphere of radius 5 centered at the origin, and let D be the region under the sphere that lies above the plane z = 3. Set up the limits of integration for EXAMPLE 1 evaluating the triple integral of a function F(x, y, z) over the region D. Solution The region under the sphere that lies above the plane z = 3 is enclosed by the surfaces x² + y² + z² = 25 and z = 3. To find the limits of integration, we first sketch the region, as shown in Figure 14.31. The "shadow region" R in the xy-plane is a circle of some radius centered at the origin. By considering a side view of the region D, we can determine that the radius of this circle is 4; see Figure 14.32a. If we fix a point (x, y) in R and draw a vertical line M above (x, y), then we see that this line enters the region D at the height z = 3 and leaves the region at the height z = √25 - x² - y2; see Figure 14.31. This gives us the z-limits of integration. To find the y-limits of integration, we consider a line L that lies in the region R, passes through the point (x, y), and is parallel to the y-axis. For clarity we have separately pictured the region R and the line L in Figure 14.32b. The line L enters R when y = -√16-2 and exits when y = V16 - x². This gives us the y-limits of integration. Finally, as L sweeps across R from left to right, the value of x varies from x = -4 to x = 4. This gives us the x-limits of integration. Therefore, the triple integral of F over the region D is given by [ F(x, y F(x, y, z) dz dy dx = y dx = [ Surface x² + y2 +2²= 25 2= 3 > Re •V16-12 -V16-gữ, M √25-²-² F(x, y, z) dz dy dx. Leaves at 2 = √25-1²-1² Enters at

806 Chapter 14 Multiple Integrals It does not apply to regions with complicated holes through them, although sometimes such regions can be subdivided into simpler regions for which the procedure does apply. We illustrate this method of finding the limits of integration in our first example. Let S be the sphere of radius 5 centered at the origin, and let D be the region under the sphere that lies above the plane z = 3. Set up the limits of integration for EXAMPLE 1 evaluating the triple integral of a function F(x, y, z) over the region D. Solution The region under the sphere that lies above the plane z = 3 is enclosed by the surfaces x² + y² + z² = 25 and z = 3. To find the limits of integration, we first sketch the region, as shown in Figure 14.31. The "shadow region" R in the xy-plane is a circle of some radius centered at the origin. By considering a side view of the region D, we can determine that the radius of this circle is 4; see Figure 14.32a. If we fix a point (x, y) in R and draw a vertical line M above (x, y), then we see that this line enters the region D at the height z = 3 and leaves the region at the height z = √25 - x² - y2; see Figure 14.31. This gives us the z-limits of integration. To find the y-limits of integration, we consider a line L that lies in the region R, passes through the point (x, y), and is parallel to the y-axis. For clarity we have separately pictured the region R and the line L in Figure 14.32b. The line L enters R when y = -√16-2 and exits when y = V16 - x². This gives us the y-limits of integration. Finally, as L sweeps across R from left to right, the value of x varies from x = -4 to x = 4. This gives us the x-limits of integration. Therefore, the triple integral of F over the region D is given by [ F(x, y F(x, y, z) dz dy dx = y dx = [ Surface x² + y2 +2²= 25 2= 3 > Re •V16-12 -V16-gữ, M √25-²-² F(x, y, z) dz dy dx. Leaves at 2 = √25-1²-1² Enters at

Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)
Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)
6th Edition
ISBN:9781337111348
Author:Bruce Crauder, Benny Evans, Alan Noell
Publisher:Bruce Crauder, Benny Evans, Alan Noell
ChapterA: Appendix
SectionA.2: Geometric Constructions
Problem 9P: A soda can is made from 40 square inches of aluminum. Let x denote the radius of the top of the can,...
See similar textbooks
icon
Related questions
Question

Can you please explain why the radius decreases to 4 for the x and y limits

806 Chapter 14 Multiple Integrals regions can be subdivided into simpler regions for which the procedure does apply. It does not apply to regions with complicated holes through them, although sometimes such We illustrate this method of finding the limits of integration in our first example. EXAMPLE 1 Let S be the sphere of radius 5 centered at the origin, and let D be the region under the sphere that lies above the plane z = 3. Set up the limits of integration for evaluating the triple integral of a function F(x, y, z) over the region D. Solution The region under the sphere that lies above the plane z = 3 is enclosed by the surfaces x² + + z² = 25 and z = 3. To find the limits of integration, we first sketch the region, as shown in Figure 14.31. The "shadow region" R in the xy-plane is a circle of some radius centered at the origin. By considering a side view of the region D, we can determine that the radius of this circle is 4; see Figure 14.32a. If we fix a point (x, y) in R and draw a vertical line M above (x, y), then we see that this line enters the region D at the height z = 3 and leaves the region at the height z = √25 - x² - y²; see Figure 14.31. This gives us the z-limits of integration. To find the y-limits of integration, we consider a line L that lies in the region R, passes through the point (x, y), and is parallel to the y-axis. For clarity we have separately pictured the region R and the line L in Figure 14.32b. The line L enters R when y = -√16-² and exits when y V16x². This gives us the y-limits of integration. Finally, as L sweeps across R from left to right, the value of x varies from x = -4 to x = 4. This gives us the x-limits of integration. Therefore, the triple integral of F over the region D is given by J F(x, y, z) dz dy dx Surface x² + y² + z² = 25 z = 3 V16−x2 dz dy dx = [² + √ -4 J -V16-r2 J3 D R 5 4 Z M *(x, y) √25-x²-y² F(x, y, z) dz dy dx. Leaves at z = √25-x² - y² 4 Enters at 2=3 5 FIGURE 14.31 Finding the limits of integration for evaluating the triple integral of a function defined over the portion of the sphere of radius 5 that lies above the plane z = 3 (Example 1).
Transcribed Image Text:806 Chapter 14 Multiple Integrals regions can be subdivided into simpler regions for which the procedure does apply. It does not apply to regions with complicated holes through them, although sometimes such We illustrate this method of finding the limits of integration in our first example. EXAMPLE 1 Let S be the sphere of radius 5 centered at the origin, and let D be the region under the sphere that lies above the plane z = 3. Set up the limits of integration for evaluating the triple integral of a function F(x, y, z) over the region D. Solution The region under the sphere that lies above the plane z = 3 is enclosed by the surfaces x² + + z² = 25 and z = 3. To find the limits of integration, we first sketch the region, as shown in Figure 14.31. The "shadow region" R in the xy-plane is a circle of some radius centered at the origin. By considering a side view of the region D, we can determine that the radius of this circle is 4; see Figure 14.32a. If we fix a point (x, y) in R and draw a vertical line M above (x, y), then we see that this line enters the region D at the height z = 3 and leaves the region at the height z = √25 - x² - y²; see Figure 14.31. This gives us the z-limits of integration. To find the y-limits of integration, we consider a line L that lies in the region R, passes through the point (x, y), and is parallel to the y-axis. For clarity we have separately pictured the region R and the line L in Figure 14.32b. The line L enters R when y = -√16-² and exits when y V16x². This gives us the y-limits of integration. Finally, as L sweeps across R from left to right, the value of x varies from x = -4 to x = 4. This gives us the x-limits of integration. Therefore, the triple integral of F over the region D is given by J F(x, y, z) dz dy dx Surface x² + y² + z² = 25 z = 3 V16−x2 dz dy dx = [² + √ -4 J -V16-r2 J3 D R 5 4 Z M *(x, y) √25-x²-y² F(x, y, z) dz dy dx. Leaves at z = √25-x² - y² 4 Enters at 2=3 5 FIGURE 14.31 Finding the limits of integration for evaluating the triple integral of a function defined over the portion of the sphere of radius 5 that lies above the plane z = 3 (Example 1).
Expert Solution
steps

Step by step

Solved in 2 steps with 1 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Recommended textbooks for you
Functions and Change: A Modeling Approach to Coll…
Functions and Change: A Modeling Approach to Coll…
Algebra
ISBN:
9781337111348
Author:
Bruce Crauder, Benny Evans, Alan Noell
Publisher:
Cengage Learning
Calculus For The Life Sciences
Calculus For The Life Sciences
Calculus
ISBN:
9780321964038
Author:
GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:
Pearson Addison Wesley,
Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
Algebra
ISBN:
9781133382119
Author:
Swokowski
Publisher:
Cengage
College Algebra
College Algebra
Algebra
ISBN:
9781938168383
Author:
Jay Abramson
Publisher:
OpenStax
SEE MORE TEXTBOOKS