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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 xyplane 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 xyzspace 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 xyplane, 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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This note was uploaded on 08/31/2011 for the course MATH 20C taught by Professor Helton during the Spring '08 term at UCSD.
 Spring '08
 Helton
 Calculus, Integrals

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