the region bounded by the graphs of y=2sqrt(x), y=0, x=3, and x=8 is revolved about the x-axis. find the surface area of the solid generated.
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each cross section perpendicular to the axis of rotation will be a circle with a radius of the function defining its outmost limit, y=2x^(1/2). thus, each cross section, dV, is equal to π(2x^(1/2))^2dx, or 4πxdx. Now you need to integrate between the limits, 3 and 8.
∫ from 3 to 8 of 4πxdx
2πx^2 from 3 to 8
2π(8)^2-2π(3)^2
2(64)π-2(9)π
128π-18π
110π
the answer is 110π cubic units
∫ from 3 to 8 of 4πxdx
2πx^2 from 3 to 8
2π(8)^2-2π(3)^2
2(64)π-2(9)π
128π-18π
110π
the answer is 110π cubic units