Review-FinalSolutions

# Review-FinalSolutions - Math136 Solutions to Review for...

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Unformatted text preview: Math136 Solutions to Review for Final Exam Spring 2010 Please attempt each of the problems before looking at the solutions. Please let me know if you find any mistakes. 1. For each one of the regions R described below, write and evaluate an integral expressing: (a) the area or R . (b) the volume of the solid resulting when revolving R about the x- axis. (c) the volume of the solid resulting when revolving R about the y- axis. i. Enclosed by the curves y = ( x- 2) 2 and x + y = 4 . ii. cos x ≤ y ≤ sec 2 x, ≤ x ≤ π/ 4 iii. Below y = √ x , above the line y- x + 2 = 0 and the x- axis. (i) We first find where the two curves intersect. We set ( x- 2) 2 = 4- x , so x 2- 4 x + 4 = 4- x so x 2- 3 x = 0, x ( x- 3) = 0 and x = 0 , 3. We note that the curve y = 4- x is above the curve y = ( x- 2) 2 . Therefore, (a) A = ∫ 3 4- x- ( x- 2) 2 dx = ∫ 3 3 x- x 2 dx = 3 2 x 2- x 3 3 | 3 = 27 2- 9- 0 = 9 2 (b) Around the x-axis, if we use washers, we have V = π ∫ 3 (4- x ) 2- ( x- 2) 4 dx = π [- (4- x ) 3 3- ( x- 2) 5 5 ] 3 = π [- 1 3- 1 5 + 64 3- 32 5 ] = π [ 63 3- 33 5 ] = 72 π 5 If we instead use the cylindrical shells method, we have, V = 2 π [∫ 1 (2 + √ y- (2- √ y )) ydy + ∫ 4 1 (4- y- (2- √ y )) ydy ] = 2 π [∫ 1 2 y 3 / 2 dy + ∫ 4 1 2 y- y 2 + y 3 / 2 dy ] = 2 π [ 4 5 y 5 / 2 | 1 + ( y 2- y 3 3 + 2 5 y 5 / 2 ) 4 1 ] = 2 π [ 4 5 + 16- 64 3 + 64 5- 1 + 1 3- 2 5 ] = 2 π [ 15 + 66 5- 63 3 ] = 2 π [ 66 5- 6 ] = 2 π 36 5 = 72 π 5 The washer method is probably easier. (c) We now rotate around the y-axis. If we use cylindrical shell method, we have V = 2 π ∫ 3 x (4- x- ( x- 2) 2 ) dx = 2 π ∫ 3- x 3 + 3 x 2 dx = 2 π [- x 4 4 + x 3 ] 3 = 2 π [- 81 4 + 27 ] = 27 π 2 If we instead use the washers, we have V = π [∫ 1 (2 + √ y ) 2- (2- √ y ) 2 dy + ∫ 4 1 (4- y ) 2- (2- √ y ) 2 dy ] = π [∫ 1 8 y 1 / 2 dy + ∫ 4 1 12- 9 y + y 2 + 4 √ ydy ] = π [ 16 3 y 3 / 2 | 1 + ( 12 y- 9 2 y 2 + 1 3 y 3 + 8 3 y 3 / 2 ) 4 1 ] = π [ 16 3 + 48- 72 + 64 3 + 64 3- 12 + 9 2- 1 3- 8 3 ] = π [- 36 + 9 2 + 135 3 ] = 27 π 2 The shell method is probably the easier method. (ii) (a) A = ∫ π/ 4 sec 2 ( x )- cos xdx = [tan x- sin x ] π/ 4 = 1- 1 √ 2- 0 = 1- 1 √ 2 . (b) Using washers, we have V = π ∫ π/ 4 (sec 2 x ) 2- cos 2 ( x ) dx = π [ ∫ π/ 4 (sec 2 ( x ))(1 + tan 2 ( x )) dx- 1 2 ∫ π/ 4 cos(2 x ) + 1 dx ] Let u = tan x , so du = sec 2 xdx = π [ ∫ 1 (1 + u 2 ) du- 1 2 ∫ π/ 4 cos(2 x ) + 1 dx ] = π [ ( u + u 3 3 ) 1- 1 2 ( sin(2 x ) 2 + x ) π/ 4 ] = π [ 1 + 1 3- 1 2 ( 1 2 + π 4- )] = π [ 4 3- 1 4- π 8 ] = 13 12 π- 1 8 π 2 Using shells is not really feasible. (c) Using shells, we have V = 2 π ∫ π/ 4 x (sec 2 x- cos x ) dx We use integration by parts. Let u = x , dv = sec 2 x- cos x , so du = dx and v = tan x- sin x ....
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Review-FinalSolutions - Math136 Solutions to Review for...

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