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 - x^{2}))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 - x^{2})dx

= 387.6[integral from -3 to 3] (sqrt(9 - x^{2}))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