# Volume of solid/washer

• Feb 2nd 2009, 03:13 PM
silencecloak
Volume of solid/washer
$\displaystyle y=2/x$
$\displaystyle y=0$
$\displaystyle x=1$
$\displaystyle x=3$
Rotate about $\displaystyle y = -1$

Outside radius = $\displaystyle (-1-2/x)^2$

$\displaystyle \int_1^3 ((-1-2/x)^2-1)dx = 22.1831495912$

Is this correct?

• Feb 2nd 2009, 03:28 PM
Jester
Quote:

Originally Posted by silencecloak
$\displaystyle y=2/x$
$\displaystyle y=0$
$\displaystyle x=1$
$\displaystyle x=3$
Rotate about $\displaystyle y = -1$

Outside radius = $\displaystyle (-1-2/x)^2$

$\displaystyle \int_1^3 ((-1-2/x)^2-1)dx = 22.1831495912$

Is this correct?

Outside radius = $\displaystyle (2/x + 1)$ although the square will fix that. Just need a pie

$\displaystyle \pi \int_1^3 ((2/x+1)^2-1)dx = 22.1831495912$

although judging from the answer you put it in.
• Feb 2nd 2009, 03:32 PM
silencecloak
So I did this problem correct? Or that number is just the correct number for the integral i set up?
• Feb 2nd 2009, 03:39 PM
Jester
Quote:

Originally Posted by silencecloak
So I did this problem correct? Or that number is just the correct number for the integral i set up?

Yes, just make sure you have the outer radius

$\displaystyle r_o = f(x) - L$ (for $\displaystyle f(x)>L$) where $\displaystyle y = f(x)$ the curve you're rotating and $\displaystyle y = L$ the line you're rotating about. Switch f and L if f is under $\displaystyle y= L$