Volume of solid/washer

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

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

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

Is this correct?

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

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

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

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

Is this correct?

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

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

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

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