Multivariate Calculus - Bruce E. Shapiro
Multivariate Calculus - Bruce E. Shapiro Multivariate Calculus - Bruce E. Shapiro
140 LECTURE 17. DOUBLE INTEGRALS OVER GENERAL REGIONS Example 17.5 Find the iterated integral ∫ 12 ∫ x+3 −2 −x 2 (x + 3y)dxdy Solution. Separating the integral into two parts and factoring what we can gives ∫ 12 ∫ x+3 −2 (x + 3y)dydx = = = = = ∫ 12 ∫ x+3 −2 ∫ 12 −2 ∫ 12 −2 ∫ 12 −2 ∫ 12 −2 x −x 2 xdydx + ∫ x+3 −x 2 dydx + 3 x(y| x+3 −x 2 )dx + 3 ∫ 12 ∫ x+3 −2 −x ∫ 2 12 ∫ x+3 −2 ∫ 12 −2 x(x + 3 + x 2 )dx + 3 2 (x 2 + 3x + x 3 )dx + 3 2 −x 2 3ydydx ydydx ( 1 2 y2 | x+3 )dx −x 2 ∫ 12 −2 ∫ 12 −2 [(x + 3) 2 − (−x 2 ) 2 ]dx (x + 3) 2 dx − 3 2 ∫ 12 −2 x 4 dx Integrating and plugging in numbers, we find that ∫ 12 ∫ x+3 −2 −x 2 (x + 3y)dydx = ( 1 3 x3 + 3 2 x2 + 1 4 x4 )| 12 −2 + 3 1 2 3 (x + 3)3 | 12 −2 − 3 2 1 5 x5 | 12 = 1 3 (12)3 + 3 2 (12)2 + 1 4 (12)4 − 1 3 (−8) − 3 2 (4) − 1 2 (16) + 1 2 (153 − (−8)) − 3 10 (125 − (−32)) = 576 + 216 + 5184 + 8 − 6 − 8 + 1691.5 − 124432 3 ≈ −116775.83 −2 Figure 17.7: The plane z = 6 − 2x − 3y and its cross-section on the xy plane. Example 17.6 Find the volume of the tetrahedron bounded by the coordinate planes and the plane z = 6 − 2x − 3y. Revised December 6, 2006. Math 250, Fall 2006
LECTURE 17. DOUBLE INTEGRALS OVER GENERAL REGIONS 141 Solution Dividing the equation of the plane by 6 and rearranging gives x 3 + y 2 + z 6 = 1 This tells us that the plane intersects the coordinate axes at x= 3, y=2, and z=6 as illustrated in figure 17.7. We can find the intersection of the given plane with the xy-plane by setting z = 0. The cross-section is illustrated on the right-had side of figure 17.7, and the intersection is the line 2x + 3y = 6 ⇒ x 3 + y 2 = 1 Solving for y, y = − 2 3 x + 2 If we integrate first over y (in the inner integral), then going from the bottom to the top of the triangle formed by the x and y axes and the line y = (2/3)x + 2, our limits are y = 0 (on the bottom) to y = −(2/3)x + 2 (on the top).. Then we integrate over x (in the outer integral) going from left to right, with limits x = 0 to x = 3, V = = = ∫ 3 ∫ −(2/3)x+2 0 ∫ 3 0 ∫ 3 0 0 (6 − 2x − 3y)dx (6y − 2xy − 3 )∣ ∣∣∣ −(2/3)x+2 2 y2 dx 0 [ 6 (− 2 ) 3 x + 2 − 2x (− 2 3 x + 2 ) − 3 2 (− 2 3 x + 2 ) 2 ] dx Expanding the factors in the integrand gives V = = = = ∫ 3 0 ∫ 3 0 ∫ 3 0 [ −4x + 12 + 4 3 x2 − 4x − 3 ( 4 2 9 x2 − 8 )] 3 x + 4 dx [ −4x + 12 + 4 3 x2 − 4x − 2 ] 3 x2 + 4x − 6 dx [−4x + 6 + 2 3 x2 ] dx (−2x 2 + 6x + 2 9 x3 )∣ ∣∣∣ 3 = −2(9) + 6(3) + 2 (27) = −18 + 18 + 6 = 6 9 0 Example 17.7 Find the volume of the tetrahedron bounded by the coordinate planes and the plane 3x + 4y + z − 12 = 0. Math 250, Fall 2006 Revised December 6, 2006.
- Page 101 and 102: LECTURE 12. THE CHAIN RULE 89 Solut
- Page 103 and 104: LECTURE 12. THE CHAIN RULE 91 Examp
- Page 105 and 106: Lecture 13 Tangent Planes Since the
- Page 107 and 108: LECTURE 13. TANGENT PLANES 95 Solut
- Page 109 and 110: LECTURE 13. TANGENT PLANES 97 Multi
- Page 111 and 112: LECTURE 13. TANGENT PLANES 99 Accor
- Page 113 and 114: Lecture 14 Unconstrained Optimizati
- Page 115 and 116: LECTURE 14. UNCONSTRAINED OPTIMIZAT
- Page 117 and 118: LECTURE 14. UNCONSTRAINED OPTIMIZAT
- Page 119 and 120: LECTURE 14. UNCONSTRAINED OPTIMIZAT
- Page 121 and 122: LECTURE 14. UNCONSTRAINED OPTIMIZAT
- Page 123 and 124: LECTURE 14. UNCONSTRAINED OPTIMIZAT
- Page 125 and 126: LECTURE 14. UNCONSTRAINED OPTIMIZAT
- Page 127 and 128: LECTURE 14. UNCONSTRAINED OPTIMIZAT
- Page 129 and 130: Lecture 15 Constrained Optimization
- Page 131 and 132: LECTURE 15. CONSTRAINED OPTIMIZATIO
- Page 133 and 134: LECTURE 15. CONSTRAINED OPTIMIZATIO
- Page 135 and 136: LECTURE 15. CONSTRAINED OPTIMIZATIO
- Page 137 and 138: LECTURE 15. CONSTRAINED OPTIMIZATIO
- Page 139 and 140: Lecture 16 Double Integrals over Re
- Page 141 and 142: LECTURE 16. DOUBLE INTEGRALS OVER R
- Page 143 and 144: LECTURE 16. DOUBLE INTEGRALS OVER R
- Page 145 and 146: LECTURE 16. DOUBLE INTEGRALS OVER R
- Page 147 and 148: Lecture 17 Double Integrals over Ge
- Page 149 and 150: LECTURE 17. DOUBLE INTEGRALS OVER G
- Page 151: LECTURE 17. DOUBLE INTEGRALS OVER G
- Page 155 and 156: LECTURE 17. DOUBLE INTEGRALS OVER G
- Page 157 and 158: Lecture 18 Double Integrals in Pola
- Page 159 and 160: LECTURE 18. DOUBLE INTEGRALS IN POL
- Page 161 and 162: LECTURE 18. DOUBLE INTEGRALS IN POL
- Page 163 and 164: LECTURE 18. DOUBLE INTEGRALS IN POL
- Page 165 and 166: LECTURE 18. DOUBLE INTEGRALS IN POL
- Page 167 and 168: Lecture 19 Surface Area with Double
- Page 169 and 170: LECTURE 19. SURFACE AREA WITH DOUBL
- Page 171 and 172: LECTURE 19. SURFACE AREA WITH DOUBL
- Page 173 and 174: Lecture 20 Triple Integrals Triple
- Page 175 and 176: LECTURE 20. TRIPLE INTEGRALS 163 Fi
- Page 177 and 178: LECTURE 20. TRIPLE INTEGRALS 165 so
- Page 179 and 180: LECTURE 20. TRIPLE INTEGRALS 167 Tr
- Page 181 and 182: Lecture 21 Vector Fields Definition
- Page 183 and 184: LECTURE 21. VECTOR FIELDS 171 Defin
- Page 185 and 186: LECTURE 21. VECTOR FIELDS 173 Examp
- Page 187 and 188: Lecture 22 Line Integrals Suppose t
- Page 189 and 190: LECTURE 22. LINE INTEGRALS 177 Exam
- Page 191 and 192: LECTURE 22. LINE INTEGRALS 179 ener
- Page 193 and 194: LECTURE 22. LINE INTEGRALS 181 The
- Page 195 and 196: LECTURE 22. LINE INTEGRALS 183 so t
- Page 197 and 198: LECTURE 22. LINE INTEGRALS 185 wher
- Page 199 and 200: LECTURE 22. LINE INTEGRALS 187 The
- Page 201 and 202: LECTURE 22. LINE INTEGRALS 189 The
LECTURE 17. DOUBLE INTEGRALS OVER GENERAL REGIONS 141<br />
Solution Dividing the equation of the plane by 6 and rearranging gives<br />
x<br />
3 + y 2 + z 6 = 1<br />
This tells us that the plane intersects the coordinate axes at x= 3, y=2, and z=6 as<br />
illustrated in figure 17.7. We can find the intersection of the given plane with the<br />
xy-plane by setting z = 0. The cross-section is illustrated on the right-had side of<br />
figure 17.7, and the intersection is the line<br />
2x + 3y = 6 ⇒ x 3 + y 2 = 1<br />
Solving for y,<br />
y = − 2 3 x + 2<br />
If we integrate first over y (in the inner integral), then going from the bottom to<br />
the top of the triangle formed by the x and y axes and the line y = (2/3)x + 2,<br />
our limits are y = 0 (on the bottom) to y = −(2/3)x + 2 (on the top).. Then we<br />
integrate over x (in the outer integral) going from left to right, with limits x = 0 to<br />
x = 3,<br />
V =<br />
=<br />
=<br />
∫ 3 ∫ −(2/3)x+2<br />
0<br />
∫ 3<br />
0<br />
∫ 3<br />
0<br />
0<br />
(6 − 2x − 3y)dx<br />
(6y − 2xy − 3 )∣ ∣∣∣<br />
−(2/3)x+2<br />
2 y2 dx<br />
0<br />
[<br />
6<br />
(− 2 )<br />
3 x + 2 − 2x<br />
(− 2 3 x + 2 )<br />
− 3 2<br />
(− 2 3 x + 2 ) 2<br />
]<br />
dx<br />
Expanding the factors in the integrand gives<br />
V =<br />
=<br />
=<br />
=<br />
∫ 3<br />
0<br />
∫ 3<br />
0<br />
∫ 3<br />
0<br />
[<br />
−4x + 12 + 4 3 x2 − 4x − 3 ( 4<br />
2 9 x2 − 8 )]<br />
3 x + 4 dx<br />
[<br />
−4x + 12 + 4 3 x2 − 4x − 2 ]<br />
3 x2 + 4x − 6 dx<br />
[−4x + 6 + 2 3 x2 ]<br />
dx<br />
(−2x 2 + 6x + 2 9 x3 )∣ ∣∣∣<br />
3<br />
= −2(9) + 6(3) + 2 (27) = −18 + 18 + 6 = 6 <br />
9<br />
0<br />
Example 17.7 Find the volume of the tetrahedron bounded by the coordinate planes<br />
and the plane 3x + 4y + z − 12 = 0.<br />
Math 250, Fall 2006 Revised December 6, 2006.
Change language