For the intersections: cos x = 0 or sin x = 1 2

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1 Chapter 6 Set-up examples The purpose of this document is to demonstrate the work that will be required if you are asked to set-up integrals on an exam and/or quiz.. Areas () Set up, do not evaluate, any integrals needed to find the area of the region enclosed by the curves y = cos x, y = sin 2x, x π 2 For the intersections: cos x = sin 2x = 2 sin x cos x (double angle identity) cos x = or sin x = 2 x = π 2, π 6 (since x π/2) there are two regions, the first (A ) for x π/6: top curve: y = cos x bottom curve: y = sin 2x width of slice: area of slice: [cos x sin 2x] in the second region (A 2 ), for π/6 x π/2, they switch top curve: y = sin 2x bottom curve: y = cos x. width of slice: area of slice: [sin 2x cos x] A = A + A 2 = π/6 [cos x sin 2x] + π/2 π/6 [sin 2x cos x] c 28Arizona State University School of Mathematical & Statistical Sciences

2 (2) Set up, do not evaluate, an integral to find the area of the region enclosed by the curves y = e x, y = x 2, x =, x = 2 The graph shows that when x = the curve y = e x is at y = /e, which is positive, and the curve y = x 2 is at the x-axis. When x = the curve y = e x is at y = e and the curve y = x 2 is again at the x-axis. top curve: y = e x bottom curve: y = x 2 width of slice: area of slice: [e x (x 2 )] A = [e x (x 2 )] = [e x x 2 + ] c 28Arizona State University School of Mathematical & Statistical Sciences

3 (3) Set up, do not evaluate, a single integral to find the area of the region enclosed by the curves y = (x 2) 2, y = x. 3 For the intersections: (x 2) 2 = x x 2 4x + 4 = x x 2 5x + 4 = (x 4)(x ) = x =, 4 When x = the curve y = e x is at y = e and the curve y = x 2 is again at the x-axis. top curve: y = e x bottom curve: y = x 2 width of slice: area of slice: [x (x 2) 2 ] A = 4 [x (x 2) 2 ] = 4 [x (x 2 4x + 4)] = 4 [ x 2 + 5x 4)] c 28Arizona State University School of Mathematical & Statistical Sciences

4 (4) Set up, do not evaluate, a single integral to find the area of the region enclosed by the curves x = 2y 2, x = 4 + y 2. 4 (note: this graph would be provided) dy For the intersections: right curve: x = 4 + y 2 left curve: x = 2y 2 width of slice: dy area of slice: [4 + y 2 2y 2 ] dy 2y 2 = 4 + y 2 y 2 = 4 y = ±2 A = 2 2 [4 + y 2 2y 2 ] dy = 2 2 [4 y 2 ] dy c 28Arizona State University School of Mathematical & Statistical Sciences

5 2. Volumes of Revolution () Use the method of washers to set up, do not evaluate, an integral to find the volume of the solid obtained by rotating the region bounded by about the y-axis. y 2 = x, x = 2y 5 R r For the intersections: y 2 = 2y y 2 2y = y(y 2) = y =, 2 We want washers, so we need our slice to be perpendicular to the axis of rotation; dy thickness of the washer: dy big radius is determined by the x value for the line: 2y area of the big disc: π(2y) 2 volume of the big disc: π(2y) 2 dy small radius is determined by the x value for the parabola: y 2 area of the hole to be removed: π(y 2 ) 2 volume of the hole to be removed: π(y 2 ) 2 dy volume of the washer: π(2y) 2 dy π(y 2 ) 2 dy = π[(2y) 2 (y 2 ) 2 ] dy V = 2 π[(2y) 2 (y 2 ) 2 ] dy = π 2 [4y 2 y 4 ] dy c 28Arizona State University School of Mathematical & Statistical Sciences

6 (2) Use the method of washers to set up, do not evaluate, an integral to find the volume of the solid obtained by rotating the region bounded by about the line y = 4. x y =, y = x 2 4x (note: this graph would be provided) R r For the intersections: x 2 4x + 3 = x (since x y = = y = x ) x 2 5x + 4 = (x 4)(x ) = x =, 4 We want washers, so we need our slice to be perpendicular to the axis of rotation; thickness of the washer: big radius is determined by the line y = 4 and the x value for the parabola: 4 (x 2 4x + 3) = 4 x 2 + 4x 3 = x 2 + 4x + area of the big disc: π( x 2 + 4x + ) 2 volume of the big disc: π( x 2 + 4x + ) 2 small radius is determined by the line y = 4 and the x value for the line y = x : 4 (x ) = 5 x area of the hole to be removed: π(5 x) 2 volume of the hole to be removed: π(5 x) 2 volume of the washer: π( x 2 + 4x + ) 2 π(5 x) 2 V = π 4 [( x 2 + 4x + ) 2 (5 x) 2 ] = π 4 [( x 2 + 4x + ) 2 (5 x) 2 ] c 28Arizona State University School of Mathematical & Statistical Sciences

7 (3) Use the method shells to set up, do not evaluate, an integral to find the volume of the solid obtained by rotating the region bounded by about the x-axis. y = x 3, y = 8, x = 7 dy h r We want shells, hence we need our slice to be parallel to the axis of rotation; dy radius of the shell: y height of the shell: x for x = 3 y thickness of shell: dy volume of shell: 2πy( 3 y) V = 2π 8 y 4/3 dy c 28Arizona State University School of Mathematical & Statistical Sciences

8 (4) Use the method shells to set up, do not evaluate, an integral to find the volume of the solid obtained by rotating the region bounded by about the x-axis. y = 4x x 2, y = 3 8 (note: this graph would be provided) x r h For the intersections: 4x x 2 = 3 x 2 4x + 3 = (x 3)(x ) = x =, 3 We want shells, hence we need our slice to be parallel to the axis of rotation; radius of the shell: x height of the shell = top curve - bottom curve: 4x x 2 3 thickness of shell: volume of shell: 2π(x )(4x x 2 3) V = 2π 3 (x )(4x x 2 3) c 28Arizona State University School of Mathematical & Statistical Sciences

9 9 3. Arc Length As there is not a lot of work to be shown here, I will just do one example. You can let me know if you want another. () Find the exact length of the curve y = + 6x 3/2, x L = = = = 8 = 8 + ( ) 2 dy + (9x /2 ) 2 + 8x ] u 3/2 82 = 8 3/2 = u du u /2 du ( (82) 3/2 (u = + 8x, du = 8 ) ) 4. Work 4.. Springs. () A spring has a natural length of 2 cm. If a 25-N force is required to keep it stretched to a length of 3 cm, how much work is required to stretch it from 2 cm to 25 cm? We must first convert centimeters to meters: the natural length of the spring is.2 m, etc. When the spring is stretched to a length of.3 m, this is. m beyond the natural length, it takes 25=N of force, so 25 = f(.) =.k = k = 25. = 25 c 28Arizona State University School of Mathematical & Statistical Sciences

10 We want to stretch the spring from.2 m to.25 m, that is, from x = to x =.5, and the work done is W = =.5.5 f(x) 25x ].5 = 25 x2 2 = 25(.5) 2 (ok if do not simplify ) = 25(.25) J (2) Suppose that 2 J of work is needed to stretch a spring from its natural length of 3 cm to a length of 42 cm. How much work is needed to stretch the spring from 35 cm to 4 cm? We must change units from cm to m, so that the natural length of the spring is.3 m. We are given that the work done in stretching the spring from x = to x =.2 is 2, so 2 = W =.2 kx = k 2 (.2)2 and we can solve for k: k = 4 (.2). 2 Thus the work done in stretching the spring from x =.5 to x =. is W =..5 = 4 (.2) x 2. = 4 x 2 (.2) ]. x.5 = 2 (.2) 2 ( (.) 2 (.5) 2) J 4.2. Cables. () A heavy rope, 5 ft long, weighs.5 lb/ft and hangs over the edge of a building 2 ft high. How much work is done in pulling the rope to the top of the building? Note: the information that the building is 2 ft hight is not really relevant the only important thing is that the building is at least tall enough so that the rope hangs vertically. Introduce an x-axis pointing down, with x = at the top of the rope and x = 5 at the bottom of the rope. So, we have x 5. c 28Arizona State University School of Mathematical & Statistical Sciences

11 Consider a short segment of the rope x ft below the top: length: ft density:.5 lb/ft weight:.5 lb raised: x ft work on segment:.5x ft-lb Therefore the total work is W = 5 5.5x = x 2 = ( ) = 625 ft-lb 4.3. Pumping. () Consider a hemispherical tank of radius 3 m, with a pipe of length m extending vertically from the top of the tank, and with a spout at the top of this pipe. If the tank is full of water, find the work required to pump the water out of the spout. Acceleration due to gravity is 9.8 m/sec 2 ; the density of water is kg/m 3. ALWAYS draw your picture, put in axes and label them (if you do not do so, the problem cannot be graded. x h y Note that the tank is 3 m deep (the radius of the sphere). The equation of the circle that forms the sides of the tank (determines the relationship between x and y) is then (x 3) 2 +y 2 = 3 2 which can be solved for y to get y = 9 (x 3) 2 = 6x x 2. Thus a horizontal cross section of the tank is a circle of radius = 6x x 2 c 28Arizona State University School of Mathematical & Statistical Sciences

12 area of the cross section:π( 6x x 2 ) 2 = π(6x x 2 ) m 2 thickness of the cross section: m volume of a horizontal slice is: π(6x x 2 ) m 3 its mass is: π(6x x 2 ) kg the force on the slice is: 9.8()π(6x x 2 ) kg-m/sec 2 =N this slice is lifted: 4 x m so the work done for the slice is: 98π(4 x)(6x x 2 ) N-m=J Thus the total work is W = 3 = 98π 98π(4 x)(6x x 2 ) 3 ( 24x x 2 + x 3) [ = (98)π 2x 2 3 x3 + 4 x4 ( = (98)π ) 4 ( ) 53 = (98)π 4 = 245(53)π ] 3 2 Make sure you look at the lecture notes for more examples of pumping problems. the first person to report a typo will get $ for each typo c 28Arizona State University School of Mathematical & Statistical Sciences