Find the volume of the solid generated by revolving the regi

Solution

1. 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 ========================= 2. 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 ========================= 3. 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 ========================= 4. 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 -------------------- Of course, this would have been easier to solve using washer method Integrate from x = 0 to x = 1 Outer radius: R = 5x Inner radius: r = 5x² V = p ?0¹ (R² - r²) dx V = p ?0¹ (25x² - 25x4) dx V = 10p/3