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Step 5 When r = 0, we have u = 6, and when r = V5, we have u = Thus, V5 re2 + 6 dr 1 ´eu du = 2 ---

Step 5 When r = 0, we have u = 6, and when r = V5, we have u = Thus, V5 re2 + 6 dr 1 ´eu du = 2 ---

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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15.7.513

Evaluate the integral, where E is enclosed by the paraboloid z = 6 + x² + y2, the cylinder x2 + y2 = 5, and the xy-plane. Use cylindrical coordinates. ez dV Step 1 In cylindrical coordinates, the paraboloid z = 6 + x2 + y2 has the equation z = 6+r2 and the cylinder x2 + y2 = 5 has the equation 6+72 V5 r = V5 Step 2 Therefore, the region E enclosed by the paraboloid, the cylinder, and the xy-plane is described by E = {(r, 0, z)| 0 szs 6 + r2 0<rs V5 V5 0 < 0 6+72 < 2n Step 3 -2x V5 (6 + 2 To evaluate ez dV = e?r dz dr de, we first calculate the innermost integral. Jo *6 + r2 16 + r2 re² H²+6 = 1 1 - 4ی(- e?r dz = Step 4 Next, we have V5 + 6 - r) dr = re? + 6 dr - r dr The first integral requires the substitution u = + 6 and 1 du = 2r dr, which means that r dr = 2 du. 2r Step 5 When r = 0, we have u = 6, and when r = V5, we have u = . Thus, V5 rer + 6 dr = eu du = - IN
Transcribed Image Text:Evaluate the integral, where E is enclosed by the paraboloid z = 6 + x² + y2, the cylinder x2 + y2 = 5, and the xy-plane. Use cylindrical coordinates. ez dV Step 1 In cylindrical coordinates, the paraboloid z = 6 + x2 + y2 has the equation z = 6+r2 and the cylinder x2 + y2 = 5 has the equation 6+72 V5 r = V5 Step 2 Therefore, the region E enclosed by the paraboloid, the cylinder, and the xy-plane is described by E = {(r, 0, z)| 0 szs 6 + r2 0<rs V5 V5 0 < 0 6+72 < 2n Step 3 -2x V5 (6 + 2 To evaluate ez dV = e?r dz dr de, we first calculate the innermost integral. Jo *6 + r2 16 + r2 re² H²+6 = 1 1 - 4ی(- e?r dz = Step 4 Next, we have V5 + 6 - r) dr = re? + 6 dr - r dr The first integral requires the substitution u = + 6 and 1 du = 2r dr, which means that r dr = 2 du. 2r Step 5 When r = 0, we have u = 6, and when r = V5, we have u = . Thus, V5 rer + 6 dr = eu du = - IN
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