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

option 1 is correct.

step-by-step explanation:

y varies inversely with x and constant of variation is 4

the equation will be:

y = k/x

where k =4 so,

y = 4/x

if we put values of x of option 1 in the above equation and find values of y

y = 4/-2 => y =-2

y = 4/-1 => y = -4

y = 4/1 => y = 4

y = 4/2 => y =2

so, all the values of x and y are correct.

if we put values of x given in other options and find value of y we cannot find whole table satisfying the equation.

hence option 1 is correct.

false

step-by-step explanation: