Find the volume of the solid generated by revolving the region bounded by the gi

Question

Find the volume of the solid generated by revolving the region bounded by the given lines and curves about the x-axis; y=sqrt(2x+3), y=0, x=0, x=1

Explanation / Answer

First we find x-coordinates of points of intersection of y = 2 + x - x² y = 2 - x 2 - x = 2 + x - x² x² - 2x = 0 x (x - 2) = 0 x = 0, x = 2 So we integrate from x = 0 to x = 2 On this interval 2 + x - x² > 2 - x Axis of rotation = y-axis (line x = 0) Surface area of cylindrical shell = 2p r h where r = distance from axis of rotation = x and h = height of cylinder = (2 + x - x²) - (2 - x) = 2x - x² V = 2p ?0² r h dx V = 2p ?0² x (2x - x²) dx V = 2p ?0² (2x² - x³) dx Solving, we get V = 8p/3 y = 8vx intersects x-axis (line y=0) at point (0, 0) We integrate from x = 0 (intersection) to x = 1 (given) Axis of rotation: line = -4 Surface area of cylindrical shell = 2p r h where r = distance from axis of rotation = x - (-4) = x + 4 and h = height of cylinder = y = 8vx V = 2p ?0¹ r h dx V = 2p ?0¹ (x + 4) 8vx dx V = 2p ?0¹ (8x^(3/2) + 32x^(1/2)) dx Solving, we get V = 736p/15 Points of intersection: (0,0) and (4,4) y² = 4x ----> y = 2vx Cylinder has radius = x, height = 2vx - x V = 2p ?04 x (2vx - x) dx V = 2p ?04 (2x^(3/2) - x²) dx Solving, we get: V = 128p/15 Points of intersection (0,0) and (1,5) Since we are rotating about a horizontal axis, we will have to integrate with respect to y We integrate from y = 0 to y = 5 y = 5x² ------> x = v(y/5) y = 5x -------> x = y/5 Cylinder has radius = y, height = v(y/5) - y/5 V = 2p ?05 y (v(y/5) - y/5) dy V = 2p ?05 (1/v5 y^(3/2) - 1/5 y²) dy V = 10p/3

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