Yefim S. answered 03/20/23
Math Tutor with Experience
a. v = π∫-12[(x + 3)2 - (x2 + 1)2]dx = π∫-12((6x - x2 - x4 + 8)dx = π(3x2 - x3/3 - x5/5 + 8x)-12 = π[(12 - 8/3 - 32/5 +
16) - (3 + 1/3 + 1/5 - 8)] = 117π/5; x2 + 1 = x + 3; x = -1 or x = 2∫
b. v = π∫01(22 - 4x)dx = π(4x - 2x2)01 = 2π; 2√0 = 0; 2√x = 2; x = 1