Answer :
Answer:
[tex]V = 153.45[/tex]
Explanation:
We are required to find the volume of the solid generated by revolving the shaded region about x-axis on the interval [a, b] = [2, 12] for the following function
[tex]y = \frac{15}{\sqrt{15x -x^2}}[/tex]
To find the volume we integrate its area
[tex]V = \int\limits^b_a A \, dx[/tex]
where A = πr² and
[tex]r^2 = y^2= (\frac{15}{\sqrt{15x -x^2}})^2 = \frac{225}{15x -x^2}}[/tex]
So the volume becomes
[tex]V = 225\pi\int\limits^b_a \frac{1}{15x -x^2}}\, dx[/tex]
Integrating yields
[tex]V = 225\pi[\frac{1}{15}(lnx - ln(15-x))][/tex]
Evaluating the limits a = 2 and b = 12
[tex][\frac{1}{15}(ln(12) - ln(15-12))] - [\frac{1}{15}(ln(2) - ln(15-2))] = 0.0924 - (-0.1247)[/tex]
[tex]V = 225\pi(0.0924+0.1247) = 225\pi((0.2171)[/tex]
[tex]V = 153.45[/tex]