Originally Posted by
nzmathman We are given $\displaystyle f(x) = {(3x+2)}^ \frac{1}{2} $
Formula for the volume when rotated about the x-axis is: $\displaystyle V = \pi \int \limits_1^5 y^2 dx$
$\displaystyle f(x) = y$, so $\displaystyle y^2 = (3x+2)$
$\displaystyle V = \pi \int \limits_1^5 (3x+2) dx$
$\displaystyle V = \pi [ \frac{3}{2} x^2 +2x]_1^5$
$\displaystyle V = \pi [( \frac{3}{2} 5^2 +2*5) - ( \frac{3}{2} 1^2 +2*1)]$
$\displaystyle V = \pi ( \frac{75}{2} +10 + \frac{3}{2} - 1)$
And you can solve from there onwards.