Multivariate Calculus - Bruce E. Shapiro
Multivariate Calculus - Bruce E. Shapiro Multivariate Calculus - Bruce E. Shapiro
164 LECTURE 20. TRIPLE INTEGRALS Since we are interested in the first octant, we chose the positive solution. z = ± 1 8√ 4 − y 2 The ellipse where the cylinder crosses the yz-plane intersects the y-axis at y = 2 and the z-axis at z = 1/4. The base of the object in the xy-plane is a triangle formed by the lines y = x, y = 2, and the y-axis.Therefore the domain of integration is the set { √ } 4 − y 2 V = (x, y, z) : 0 ≤ z ≤ , x ≤ y ≤ 2, 0 ≤ x ≤ 2 8 The volume of this set is ∭ V (1) × dV , namely V = = 1 8 ∫ 2 ∫ 2 ∫ (4−y 2 ) 1/2 /8 0 x 0 ∫ 2 ∫ 2 0 x √ 4 − y 2 dydx dzdydx From integral formula (54) in the inside back cover ∫ √a 2 − y 2 dy = y 2√ a 2 − y 2 + a2 2 sin−1 y a Setting a = 2, ∫ √4 − y 2 dy = y 2 √ 4 − y 2 + 2 sin −1 y 2 Therefore ∫ 2 √ 4 − y 2 dy = [ y √ 4 − y 2 2 + 2 sin −1 y 2 x = ( 2 √ 4 − 2 2 2 + 2 sin −1 2 2 ] 2 x ) ( x √ − 4 − x 2 2 + 2 sin −1 x ) 2 = π − x 2 √ 4 − x 2 − 2 sin −1 x 2 and hence the volume is V = 1 ∫ 2 ∫ 2 √ 4 − y 8 2 dydx = 1 ∫ 2 ( π − x √ 4 − x 0 x 8 0 2 2 − 2 sin −1 x ) dx 2 = 1 [ ∫ 2 π dx − 1 ∫ 2 x √ ∫ 2 4 − x 8 2 2 dx − 2 sin −1 x ] 2 dx = π 4 − 1 16 0 ∫ 2 0 0 x √ 4 − x 2 dx − 1 4 ∫ 2 0 0 sin −1 x 2 dx The first integral we can solve with the substitution u = 4 − x 2 , which implies that du = −2dx. Furthermore, when x = 0, u = 4, and when x = 2, u = 0. Thus ∫ 2 x √ 4 − x 2 dx = − 1 ∫ 0 u 1/2 du = 1 ∫ 4 4 u 1/2 du = u3/2 = (4)3/2 = 8 2 2 3 ∣ 3 3 0 4 Revised December 6, 2006. Math 250, Fall 2006 0 0
LECTURE 20. TRIPLE INTEGRALS 165 so that V = π 4 − 1 8 16 3 − 1 4 ∫ 2 0 sin −1 x 2 dx = π 4 − 1 6 − 1 4 Using formula (69) in the textbook’s integral table, ∫ sin −1 udu = u sin −1 u + √ 1 − u 2 ∫ 2 0 sin −1 x 2 dx so that ∫ 2 0 sin −1 (x/2)dx = [ 2 (x/2) sin −1 (x/2) + √ ] 2 1 − x 2 /4 0 = 2 ( sin −1 1 + √ 1 − 1 − 0 sin −1 0 − √ 1 − 0 ) = π − 2 so that the desired volume is V = π 4 − 1 6 − 1 4 (π − 2) = π 4 − 1 6 − π 4 + 1 2 = 1 3 Math 250, Fall 2006 Revised December 6, 2006.
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LECTURE 20. TRIPLE INTEGRALS 165<br />
so that<br />
V = π 4 − 1 8<br />
16 3 − 1 4<br />
∫ 2<br />
0<br />
sin −1 x 2 dx = π 4 − 1 6 − 1 4<br />
Using formula (69) in the textbook’s integral table,<br />
∫<br />
sin −1 udu = u sin −1 u + √ 1 − u 2<br />
∫ 2<br />
0<br />
sin −1 x 2 dx<br />
so that<br />
∫ 2<br />
0<br />
sin −1 (x/2)dx =<br />
[<br />
2 (x/2) sin −1 (x/2) + √ ] 2<br />
1 − x 2 /4<br />
0<br />
= 2 ( sin −1 1 + √ 1 − 1 − 0 sin −1 0 − √ 1 − 0 )<br />
= π − 2<br />
so that the desired volume is<br />
V = π 4 − 1 6 − 1 4 (π − 2) = π 4 − 1 6 − π 4 + 1 2 = 1 3 <br />
Math 250, Fall 2006 Revised December 6, 2006.
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