Find the area of the surface generated by revolving the curve y= (2x-x^2)^1/2, 0.25<=x<=1.25, about the X-Axis

\(y=\sqrt{2x-x^{2}}\) 0.25 <= x <= 1.25

A = 2π

Step-by-step explanation:

To calculate the area of the surface generated by revolving the curve

y = √(2x -x²) about the x-axis for the interval 0.25 ≤ x ≤ 1.25 can be found by

where f(x) = y = √(2x -x²)

Integrating the f(x) yields

f'(x) = (1 - x)/√(2x -x²)

so the above equation becomes

The second term can be simplified to

Now equation reduces to

The term √(2x -x²) cancels out

Evaluating the limits

3 is greater then both of those numbers

step-by-step explanation: