Howard L. answered 04/29/19
Calculus Specialist with 10+ Years of Tutoring Experience
(a) h(x) = 3 - x; a(x) =L(x)dx = (2)(sqrt(9 - x2))dx
F(x) = (w)[integral from -3 to 3] (h(x)a(x))dx
= (64.6)[integral from -3 to 3] (3-x)(2)(sqrt(9 - x2)dx
= 387.6[integral from -3 to 3] (sqrt(9 - x2))dx -
129.2[integral from -3 to 3] (x)sqrt(9 - x^2)dx
= 387.6(9pi/2) = 5480 lb
(b) h(x) = x; a(x) = (2)(sqrt(9 - x^2)dx
F(x) = (64.6)[integral from 0 to 3] (x)(2)(sqrt(9 - x^2)dx
= (129.2)(9) = 1162.8 lb