NAME____________________________________ SCHOOL__________________________________ TEST 4 Monday, February 4, 2008 Fax your answers to: Irvin Roy Hentzel at Fax 515-294-5454 Snail Mail your answers to: Irvin Roy Hentzel Department of Mathematics 432 Carver Hall Iowa State University Ames, Iowa 50011-2064 3 2 1. Find the volume when the portion of the curve y = x - x which lies below the x-axis, and to the right of the y-axis is rotated about the y axis. Answer: x=1 INT 2 Pi x (x^2-x^3) dx = 2 Pi INT x^3 - x^4 dx x=0 | | x=1 = 2 Pi | x^4/4 - x^5/5 | = 2 Pi (1/4-1/5) = 2 Pi/20 = Pi/10 | | x=0 f[x_] := x^3 - x^2; P1 = Plot[f[x],{x,-3,3}]; Integrate[ 2 Pi x f[x],{x,0,1}] 2. Find the volume when the portion of the curve 3 2 y = x - x which lies below the x-axis, is rotated about the x axis. Answer x=1 INT Pi (x^3-x^2)^2 dx = Pi INT x^6 - 2 x^5 + x^4 dx x=0 | | x=1 = Pi | x^7/7 -2 x^6/6 + x^5/5 | | | x=0 = Pi (1/7-2/6+1/5) = Pi/105 f[x_] := x^3 - x^2; P1 = Plot[f[x],{x,-3,3}]; Integrate[ Pi ( f[x])^2,{x,0,1}] 3. Find the x coordinate of the Centroid of the area under y = (x-1)(x-2) from x=1 to x = 2. x=2 x=2 My = INT x(x-1)(x-2) dx = INT x^3 - 3 x^2 + 2x dx x=1 x=1 | | x=2 = | x^4/4 - 3 x^3/3 + 2 x^2/2 | | | x=1 = (4 -8 +4 - (1/4 -1 + 1) = -1/4 x=2 x=2 A = INT (x-1)(x-2) dx = INT x^2 - 3 x + 2 dx x=1 x=1 | | x=2 = | x^3/3 -3x^2/2 + 2x | = (8/3 -6 + 4 -(1/3 -3/2 +2) | | x=1 = 2/3 -( 2-9+12)/6 = 2/3 - 5/6 = -1/6 _ x = My/A = (1/4)/(1/6) = 3/2 f[x_] := (x-1)(x-2); My = Integrate[ x f[x],{x,1,2}]; A = Integrate[ f[x],{x,1,2}]; Print[" ",My/A]; 3 2 4. Find the length of the curve r(t) = { t /3 , t /2 } from t = 0 to t = 2. Solution: t=2 INT Sqrt[ t^4 + t^2] dx t=0 t=2 INT t Sqrt[ t^2 + 1 ] dx t=0 | (t^2+1)^(3/2) | t=2 | ------------- | = 1/3 (5^(3/2) - 1) = 3.39345 <= Answer | 3/2 2 | t=0 ParametricPlot[ {x^3/3, x^2/2},{x,0,2}]; Integrate[ Sqrt[x^4 + x^2],{x,0,2}] 5. A spring requires 2 pounds to stretch it 5 inches. (a) How much will a force of 12 pounds stretch it? (b) How much work is required to stretch it 10 inches? Solution: k = 2/5; x 2/5 = 12; x = 30; (a) x = 30. x=10 | 2 | x=10 W = INT 2/5 x dx = 2/5 | x /2 | = 2/5 50 = 20 inch pounds x= 0 | | x=0 6. Find the Centroid of the following figure. |\ /| | \ / | | \ / | | \10 10/ | | \ / | | \ / | 12 | \ / | 12 | \ / | | \/ | | | | | | | | | |__________________| 12 Solution: Use the lower left hand corner as the origin. 144(6,6) - 48 ( 6,12-8/3) = {576, 416} (576, 416)/(144-48) = (576,416)/96 = (6,13/3) <== Answer: (0,12) (12,12) |\ /| | \ / | | \ / | | \10 10/ | | \ / | 8| * \ / * | 8 | \ / | | 6 \ / 6 | (0,4) |--------\/--------| (12,4) | (6,4) | 4| | | | 4 | | |__________________| 12 48{ 6,2 } + 24{2,4+8/3} + 24{10,4+ 8/3} = {576, 416} {576, 416}/(48+48) = (6,13/3) 48 {6,2}+24 ({0,12}+{6,4}+{0,4})/3+24({12,12}+{12,4}+{6,4})/3 = {576, 416} {576, 416}/(48+48) = (6,13/3) 7. Weights are located as shown. At what point measured from the left end of the beam will the system balance. 10 8 12 <-------->|<--------->|<------------>| ====================================== __|__ __|__ /|\ __|__ __|__ | | | | | | | | | | 18 | | 12 | | | 6 | | 2 | | | | | | | | | | |_____| |_____| | |_____| |_____| | <-------?----->| Solution. 18 0 + 12 10 + 6 18 + 2 30 288 ---------------------------- = ------ = 144/19 = 7.57895 <== Answer 18+12+6+2 38 18* 144/19 = 2592/19 12 {10-144/19} + 6{18-144/19} + 2{30-144/19} = 2592/19 8. A plane passes 5 inches from the center of a sphere of radius 6. What is the surface area of the slice. ... ... _________.______|_____.___________________ . | . . 5| . . | 6 . .---------+--------. . | . . | . . | . 2 2 . | . x + y = 36 ` ...|... ' Solution: y=6 INT 2 Pi Sqrt[36-y^2] Sqrt[ 1 + y^2/(36-y^2) ] dy y=5 y=6 2 Pi INT Sqrt[ 36-y^2 + y^2 ] dy y=5 | | y=6 12 Pi | y | = 12 Pi | | y=5 Integrate[ 2 Pi Sqrt[36-y^2] Sqrt[ 1 + y^2/(36-y^2)] ,{y,5,6}] 9. The parabola y = x^2 from x=0 to x=2 is rotated about the y axis to form a bowl. How much work is done pumping water from this bowl out over the top of the bowl. . . . . . . . . ....... y=4 W = INT 62.4 Pi x^2 dy (4-y) y=0 y=4 W = INT 62.4 Pi y dy (4-y) y=0 y=4 W = 62.4 Pi INT 4y - y^2 dy y=0 | | y=4 W = 62.4 Pi | 2 y^2 - y^3/3 | | | y=0 W = 62.4 Pi ( 32 - 64/3) = 62.4 Pi 32/3 = 665.6 Pi = 2091.04 Integrate[ 62.4 Pi y (4-y),{y,0,4}] 10. Find the area of the surface of revolution generated by revolving the curve y = Sqrt[x], 0 <= x <= 4 about the x-axis. Solution: y = Sqrt[x] dy/dx = 1/2 x^(-1/2) x=4 INT 2 Pi Sqrt[x] Sqrt[1+1/4x] dx x=0 x=4 2 Pi INT Sqrt[x+1/4] dx x=0 3/2 | ( x+1/4) | x=4 2 Pi | -------- | | 3/2 | x=0 4 Pi 3/2 3/2 ---- ( (17/4) - (1/4) ) 3 Pi 3/2 ----- ( 17 - 1 ) = 36.1769 <=== Answer 6 Integrate[ 2 Pi Sqrt[x] Sqrt[1+1/(4x)],{x,0,4}]