Find the volume of the solid of revolution formed by rotating about the x-axis the region bounded by the curves.I have read textbook but I still don't understand how to solve this problem.

Please help me

F(x)=(3x+2)^1/2, y = 0, x = 1, x = 5

Printable View

- Dec 1st 2008, 05:38 PMMatho[Calculus]volume of the solid of revolution
**Find the volume of the solid of revolution formed by rotating about the x-axis the region bounded by the curves.**I have read textbook but I still don't understand how to solve this problem.

Please help me

F(x)=(3x+2)^1/2, y = 0, x = 1, x = 5 - Dec 1st 2008, 08:29 PMnzmathman
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. - Dec 1st 2008, 08:50 PMMatho
- Dec 1st 2008, 09:05 PMnzmathman
Hi Matho,

I integrated (anti-differentiated) $\displaystyle 3x + 2$ to get $\displaystyle \frac{3}{2} x^2 + 2x$, as the formula requires you to do.