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6 Nov 2019
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Find area between the two curves y1 = (x - 2)3, y2 = x - 2, y1 = x3 - x, y2 = 3x y1 = x(x2 - 3x + 3), y2 = x2, y1 = x2 - 4x + 3, y2 = 3 + 4x - x2 y1 = cos x, y2 = 2 - cos x, 0 x 2pi, y1 = sin(x), y2 = cos(2x),-pi/pi x pi/6 Using appropriate method (disk, washer, shell) to find the volume of the solids generated by revolving the region bounded by the graph of equations about the given lines: (a) the x-axis, (b) the y-axis, c) the line y=8, d) the line x= 6 y = -x + 1, x = 0, x = 1; y1 = x2, y2 = x3 y = sin(x), y = 0, x = 0, x = pi, y1 = x2 - 4x + 3, y2 = 3 + 4x - x2 y1 = 1 + cos x, y2 = 3 - cos x, 0 x 2pi, y = root 25 - x2, x = 0, y = 0. y = 1/x, x = 0, x = 4, y = 0, y = root x + 2, x = 0, y = x. y = 1/2 x2 + 1, x = 0, x = 2, y = 0, (y - 2)2 = 4 - x, x = 0. Show transcribed image text
help with these problems
Find area between the two curves y1 = (x - 2)3, y2 = x - 2, y1 = x3 - x, y2 = 3x y1 = x(x2 - 3x + 3), y2 = x2, y1 = x2 - 4x + 3, y2 = 3 + 4x - x2 y1 = cos x, y2 = 2 - cos x, 0 x 2pi, y1 = sin(x), y2 = cos(2x),-pi/pi x pi/6 Using appropriate method (disk, washer, shell) to find the volume of the solids generated by revolving the region bounded by the graph of equations about the given lines: (a) the x-axis, (b) the y-axis, c) the line y=8, d) the line x= 6 y = -x + 1, x = 0, x = 1; y1 = x2, y2 = x3 y = sin(x), y = 0, x = 0, x = pi, y1 = x2 - 4x + 3, y2 = 3 + 4x - x2 y1 = 1 + cos x, y2 = 3 - cos x, 0 x 2pi, y = root 25 - x2, x = 0, y = 0. y = 1/x, x = 0, x = 4, y = 0, y = root x + 2, x = 0, y = x. y = 1/2 x2 + 1, x = 0, x = 2, y = 0, (y - 2)2 = 4 - x, x = 0.
Show transcribed image textSixta KovacekLv2
29 Jul 2019