Question

Find the volume of the solid generated by revolving the shaded region about the​ x-axis. The curve is y equals StartFraction 15 Over StartRoot 15 x minus x squared EndRoot EndFraction ​; x 1 equals 2​, x 2 equals 12.

Answer 1

Answer:

$V = 153.45$

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

$y = \frac{15}{\sqrt{15x -x^2}}$

To find the volume we integrate its area

$V = \int\limits^b_a A \, dx$

where A = πr² and

$r^2 = y^2= (\frac{15}{\sqrt{15x -x^2}})^2 = \frac{225}{15x -x^2}}$

So the volume becomes

$V = 225\pi\int\limits^b_a \frac{1}{15x -x^2}}\, dx$

Integrating yields

$V = 225\pi[\frac{1}{15}(lnx - ln(15-x))]$

Evaluating the limits a = 2 and b = 12

$[\frac{1}{15}(ln(12) - ln(15-12))] - [\frac{1}{15}(ln(2) - ln(15-2))] = 0.0924 - (-0.1247)$

$V = 225\pi(0.0924+0.1247) = 225\pi((0.2171)$

$V = 153.45$