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

pretty sure 750 each month!

step-by-step explanation:

the answer is:

the answer is:

to solve the problem, we need to remember the quotient of power property, it's defined by the following relation:

if we have a quotienf of powers that have the same base, we need to keep the same base and subtract the exponent of the denominator power to the exponent of the numerator power.

so, we are given the expression:

then, calculating we have:

hence, the answer is:

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