[Calculus]volume of the solid of revolution

**Matho** · December 1st 2008, 05:38 PM

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

**nzmathman** · December 1st 2008, 08:29 PM

We are given $f(x) = {(3x+2)}^ \frac{1}{2}$

Formula for the volume when rotated about the x-axis is: $V = \pi \int \limits_1^5 y^2 dx$

$f(x) = y$ , so $y^2 = (3x+2)$

$V = \pi \int \limits_1^5 (3x+2) dx$

$V = \pi [ \frac{3}{2} x^2 +2x]_1^5$

$V = \pi [( \frac{3}{2} 5^2 +2*5) - ( \frac{3}{2} 1^2 +2*1)]$

$V = \pi ( \frac{75}{2} +10 + \frac{3}{2} - 1)$

And you can solve from there onwards.

**Matho** · December 1st 2008, 08:50 PM

Originally Posted by nzmathman

We are given $f(x) = {(3x+2)}^ \frac{1}{2}$

Formula for the volume when rotated about the x-axis is: $V = \pi \int \limits_1^5 y^2 dx$

$f(x) = y$ , so $y^2 = (3x+2)$

$V = \pi \int \limits_1^5 (3x+2) dx$

$V = \pi [ \frac{3}{2} x^2 +2x]_1^5$

$V = \pi [( \frac{3}{2} 5^2 +2*5) - ( \frac{3}{2} 1^2 +2*1)]$

$V = \pi ( \frac{75}{2} +10 + \frac{3}{2} - 1)$

And you can solve from there onwards.

Thank you so much this really helps.
But how did you get this from 3x + 2?
$V = \pi [ \frac{3}{2} x^2 +2x]_1^5$

I still have one more question. If 3x is x^2, what's gonna be?

**nzmathman** · December 1st 2008, 09:05 PM

Hi Matho,

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

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