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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

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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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).
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