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# Homework5Solns - SHORT CALCULUS Math 16C Sec 1 Fall 2008...

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Unformatted text preview: SHORT CALCULUS Math 16C Sec 1 Fall 2008 Homework #5 Solutions Peter Malkin Section 7.8 Question 2 x2 x 1 y2 y dy = · x 2x x2 = x 1 2 (x2 )2 (x)2 − x x = 13 x −x . 2 Question 6 √ x2 x √ x x +y 2 2 1 dy = y x + y 3 3 x2 √ 1 1√ 3 = ( x)x2 + ( x) − (x2 )x2 + (x2 )3 3 3 3 17 16 = x 2 + x 2 − x4 − x . 3 3 2 Question 12 2 0 0 2 2 0 2 6 − x2 dydx = = 6y − x2 y 2 0 dx 0 (12 − 2x2 )dx 2 2 = 12x − x3 3 0 16 56 23 = = 12(2) − (2) = 24 − 3 3 3 Question 16 2 0 2y −y 2 3y 2 −6y 2 3ydxdy = 0 2 [3xy ]2y−y6y dy 3y 2 − 3(2y − y 2 )y − 3(3y 2 − 6y )y dy = 2 0 4 2 0 2 = 0 (24y 2 − 12y 3 )dy = 8y 3 − 3y 4 = 8(2)3 − 3(2) = 16 1 Question 20 4 0 0 x 2 dydx = (x + 1)(y + 1) = 4 0 4 0 4 2 ln |y + 1| x+1 x 0 2 2 ln |x + 1| − ln |1| dx x+1 x+1 2 ln |x + 1|dx x+1 = 0 Now, (integration by substitution) let u = ln |x + 1|; then, du = and u = ln(5) when x = 5. So, 4 0 1 x+1 dx, and u = 0 when x = 0 2 ln |x + 1|dx = x+1 ln(5) 2udu 0 ln(5) 0 2 = u2 = (ln(5)) The above integral can also be solved by the integration by parts technique. Question 28 The region for which we are computing the area is √ R = {(x, y ) : 0 ≤ y ≤ x, 0 ≤ x ≤ 4} = {(x, y ) : y 2 ≤ x ≤ 4, 0 ≤ y ≤ 2}. So, the are of R is 4 0 0 √ x 4 dydx = 0 [y ]0 dx = √ x 4 0 √ xdx = 23 x2 3 4 = 0 3 2 16 (4) 2 = , 3 3 and also, the area of R is 2 0 4 2 dxdy = y2 0 [x]y2 dy = 4 2 0 1 4 − y 2 dy = 4y − y 3 3 2 0 8 16 1 . = 4(2) − (2)3 = 8 − = 3 3 3 2 Question 32 The region for which we are computing the area is √ √ R = {(x, y ) : −2 ≤ y ≤ 2, 0 ≤ x ≤ 4 − y 2 } = {(x, y ) : 0 ≤ x ≤ 4, − 4 − x ≤ y ≤ + 4 − x}. So, the area of R is 2 −2 0 4−y 2 2 dxdy = −2 [x]0 4−y 2 dy = 1 (4−y 2)dy = 4y − y 3 3 −2 2 2 = −2 8− 8 8 − −8 + 3 3 = 32 , 3 and also, the area of R is 4 0 √ 4−x 4 − 4−x √ dydx = 0 √ √ ( 4 − x) − (− 4 − x) dx = 4 0 √ 3 4 2 4 − x dx = − (4 − x) 2 3 4 = 0 42 32 43 = . 3 3 Question 38 The region is R = {(x, y ) : −2 ≤ x ≤ 1, x + 2 ≤ y ≤ 4 − x2 }. The area of this region is 1 −2 4−x2 x+2 1 dydx = −2 1 [y ]x+2 dx (4 − x2 ) − (x + 2) dx (2 − x − x2 )dx 1 −2 4−x2 = −2 1 = −2 1 1 = 2x − x2 − x3 2 3 = = 9 2 11 2− − 23 − −4 − 2 + 8 3 3 Section 7.9 Question 4 The region of integration is R = {(x, y ) : 0 ≤ y ≤ 6, y ≤ x ≤ 3}. 2 6 0 3 6 (x + y )dxdy = y 2 0 6 12 x + xy 2 3 6 = y 2 0 12 1 3 + 3y − 2 2 y 2 2 − 6 0 y · y dy 2 = 0 9 5 3 5 9 + 3y − y 2 dy = y + y2 − y3 2 8 2 2 24 = (27 + 54 − 45) − (0) = 36 Question 18 The region of integration is R = {(x, y ) : 0 ≤ x ≤ 2, 0 ≤ y ≤ x} = {(x, y ) : 0 ≤ y ≤ 2, y ≤ x ≤ 2}. The volume of the speciﬁed solid is 2 x 2 V= 0 0 6dydx = 0 [6y ]x = 0 2 0 6xdx = 3x2 2 0 = 12 or 2 2 2 V= 0 y 6dxdy = 0 [6x]2 dy = y 2 0 (12 − 6y )dy = 12y − 3y 2 2 0 = (24 − 12) − (0) = 12. 4 Question 24 The region of integration is R = {(x, y ) : 0 ≤ x, 0 ≤ y }. The diagram in the book is a little misleading since the x and y variables are not both bounded above by 2. Note that we need to use improper integrals to solve this problem. V= 0 ∞ 0 ∞ 0 ∞ 1 dxdy (x + 1)2 (y + 1)2 b = = 0 b→∞ ∞ lim 1 (x + 1)2 (y + 1)2 b dx dy dy 0 b→∞ ∞ lim −1 (x + 1)(y + 1)2 0 = 0 b→∞ ∞ lim 1 −1 + (b + 1)(y + 1)2 (y + 1)2 dy = 0 1 dy (y + 1)2 b 0 = lim = lim b→∞ 1 dy (y + 1)2 b −1 b→∞ (y + 1) 0 −1 = lim +1 b→∞ (b + 1) =1 If we reverse the order of integration, then the working out is the same after swapping x and y . Question 26 ∞ 0 0 ∞ 0 ∞ 0 b b→∞ b→∞ ∞ V= = = = 0 e−(x+y)/2 dydx e−(x+y)/2 dydx b 0 lim 0 lim −2e−(x+y)/2 dx ∞ 2e−x/2 dx b = lim b→∞ b→∞ 2e−x/2 dx 0 b 0 = lim −4e−x/2 =4 If we reverse the order of integration, then the working out is the same after swapping x and y . 5 Question 34 We must ﬁnd the average value of the function f (x, y ) = xy over the region R = {(x, y ) : 0 ≤ x ≤ 4, 0 ≤ y ≤ 2}. The area of R is 8. So, the average value of f over R is as follows: 1 8 2 0 0 4 xydxdy = = 1 8 2 0 12 xy 2 4 dy 0 12 8ydy 80 1 2 4y 2 0 = 8 1 = 16 = 2. 8 So, the average value of f over R is 2. Question 38 We want to ﬁnd the average value of the function P (x1 , x2 ) = 192x1 + 576x2 − x2 − 5x2 − 2x1 x2 − 5000 1 2 over the region R = {(x1 , x2 ) : 40 ≤ x1 ≤ 50, 45 ≤ x2 ≤ 50}. Firstly, since R is a rectangular region, its area is 10 × 5 = 50. Then, the average value of P over R is 1 50 = = 1 50 1 50 50 45 50 45 50 45 50 40 (192x1 + 576x2 − x2 − 5x2 − 2x1 x2 − 5000)dx1 dx2 1 2 50 1 96x2 + 576x1 x2 − x3 − 5x1 x2 − x2 x2 − 5000x1 1 2 1 31 48200 + 4860x2 − 50x2 dx2 2 3 50 45 dx2 40 1 48200 50 = x2 + 2430x2 − x3 2 50 3 32 =13400. 6 ...
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## This note was uploaded on 01/25/2011 for the course MAT 16C taught by Professor Kouba during the Spring '08 term at UC Davis.

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