Respuesta :
Answer:
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
[tex]A = 2\pi\int\limits^b_a f(x){\sqrt{1+f'(x)} } \, dx[/tex]
where f(x) = y = √(2x -x²)
Integrating the f(x) yields
f'(x) = (1 - x)/√(2x -x²)
so the above equation becomes
[tex]A = 2\pi\int\limits^b_a \sqrt{2x - x^{2} } {\sqrt{1+(\frac{1-x}{\sqrt{2x-x^2} } } })^2 \, dx[/tex]
The second term can be simplified to
[tex]{\sqrt{1+(\frac{1-x}{\sqrt{2x-x^2} } } })^2 ={\sqrt{\frac{1}{{2x-x^2} } } }={\frac{1}{{\sqrt{2x-x^2}} }[/tex]
Now equation reduces to
[tex]A = 2\pi\int\limits^b_a \sqrt{2x - x^{2} } \frac{1}{\sqrt{2x-x^2} } \, dx[/tex]
The term √(2x -x²) cancels out
[tex]A = 2\pi\int\limits^b_a \, dx = [x][/tex]
Evaluating the limits
[tex]A = 2\pi(1.25 - 0.25)[/tex]
[tex]A = 2\pi(1) = 2\pi[/tex]
