Find the area of the surface generated by revolving x4sqrt1y

Solution

The area is found be integrating the above equation with limits y=0 to =15/16 integral 4 sqrt(1-y) dy Factoring out constants: = 4 integral sqrt(1-y) dy For the integrand sqrt(1-y), substitute u = 1-y and du = - dy: = -4 integral sqrt(u) du The integral of sqrt(u) is (2 u^(3/2))/3: = -(8 u^(3/2))/3+constant Substitute back for u = 1-y: = -8/3 (1-y)^(3/2)+constant after putting the limits of y we have, area = 21/8