Rog_Sec_15_3

# Rog_Sec_15_3 - Math 20C Lecture Examples Section 15.3...

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Unformatted text preview: (8/28/10) Math 20C. Lecture Examples. Section 15.3. Triple integrals† 1 dx dy dz? What is the geometric signiﬁcance of Example 1 V 1 dx dy dz = [Volume of V ] Answer: V 3xy dx dy dz, where V is the solid between the xy-plane Evaluate Example 2 V and the hyperbolic paraboloid z = xy for 0 ≤ y ≤ x, 0 ≤ x ≤ 1. 3xyz dx dy dz = 1 6 Answer: V SOLUTION: The projection of V on the xy -plane is the triangle R in Figure A2. • The base of V is z = 0 and its top is z = xy . y 1 y=x R Figure A2 1 x x z =xy 3xy dx dy dz = V x=1 3xy dz dy dx R y =x z =0 z =xy 3xy dz dy dx = x=0 x=1 y =0 z =0 y =x = z =xy 3xyz x=0 x=1 dy dx z =0 y =0 y =x 3x2 y 2 dy dx = x=0 y =0 x=1 x2 y 3 = y =x dx y =0 x=0 x=1 x5 dx = x=0 16 6x = = † Lecture x=1 x=0 1 6 notes to accompany Section 15.3 of Calculus, Early Transcendentals by Rogawski 1 Math 20C. Lecture Examples. (8/28/10) Section 15.3, p. 2 The solid V in xyz-space with distances measured in meters is bounded by z = 0, z = y, y = x2 , and y = 1. Its density at (x, y, z) is ρ(x, y, z) = 8yz kilograms per cubic meter. What is its mass? Example 3 Answer: [Mass] = 16 kilograms 9 SOLUTION: [Mass] = ρ(x, y, z ) dx dy dz = V V 8yz dx dy dz • The projection of V on the xy -plane is the region R in Figure A3, which is bounded by the parabola y = x2 and the line y = 1 for −1 ≤ x ≤ 1. • The bottom of V is z = 0 (the xy -plane) and its top is the plane z = y . y y=1 R y = x2 −1 Figure A3 [Mass] = x 8yz dx dy dz V z =y = 8yz dx dy dz R z =0 y =1 x=1 z =y = 8yz dz dy dx z =−1 y =x2 x=1 y =1 z =0 4yz 2 = x=1 y =1 x=−1 dx dx z =0 y =x2 z =1 z =y y =x2 4y 3 dy dx = x=1 y4 = x+−1 y =1 y =x2 dx x=1 = x=−1 (1 − x8 ) dx x=8 =2 x=0 (1 − x8 ) dx = 2 x − 1 x9 9 = 16 9 x=1 x=0 kilograms 1 = 2(1 − 9 ) 1 x Section 15.3, p. 3 Example 4 Math 20C. Lecture Examples. (8/28/10) What is the average value of g(x, y, z) = x + y + z on the cube V: 0 ≤ x ≤ 2, 0 ≤ y ≤ 2, 0 ≤ z ≤ 2? Answer: [Average value] = 3 SOLUTION: 1 (x + y + z ) dx dy dz • [Volume of V ] V The projection of V is the square R in Figure A4. • The base of V is z = 0 and its top is z = 2. [Volume of V ] = 23 = 8 • [Average value ] = y 2 Figure A4 2 x x=2 [Average value ] = y =2 z =2 1 8 (x + y + z ) dz dy dx x=0 y =0 x=2 y =2 z =0 1 xz + yz + 2 z 2 = x=0 y =0 1 8 y =0 x=2 2xy + y 2 + 2y 1 8 1 8 (4x + 8) dx x=0 2x2 + 8x = 1 8 = 1 8 (8 + 16) =3 y =2 dx y =0 x=0 x=2 = dy dx z =0 (2x + 2y + 2) dy dx x=0 = z =2 y =2 x=2 = x x=2 x=0 Math 20C. Lecture Examples. (8/28/10) Section 15.3, p. 4 More practice Example 5 z dx dy dz with V bounded by the xy-plane, the cone Evaluate V z= x2 + y2 , and the vertical planes x = ±1, y = ±1. x2 +y 2 z dz dy dx = z dx dy dz = Answer: x=−1 V Example 6 √ z= y =1 x=1 y =−1 4 3 z =0 √ 15 yz dx dy dz if V is the solid bounded by What is the value of V z = 0, z = y, y = x2 , y = 1? x=1 y =1 x=−1 y =x2 z =y 15y 1/2 z 1/2 dz dy dx = 40 . 7 15 yz dx dy dz = √ Answer: V z =0 Interactive Examples Work the following Interactive Examples on Shenk’s web page, http//www.math.ucsd.edu/˜ashenk/:‡ Section 16.4: Examples 1–4 ‡ The chapter and section numbers on Shenk’s web site refer to his calculus manuscript and not to the chapters and sections of the textbook for the course. ...
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