Volume of solid/washer

**silencecloak** · February 2nd 2009, 03:13 PM

$y=2/x$
$y=0$
$x=1$
$x=3$
Rotate about $y = -1$

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

Inside radius = 1

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

Is this correct?

Thanks in advance

**Danny** · February 2nd 2009, 03:28 PM

Originally Posted by silencecloak

$y=2/x$
$y=0$
$x=1$
$x=3$
Rotate about $y = -1$

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

Inside radius = 1

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

Is this correct?

Thanks in advance

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.

**silencecloak** · February 2nd 2009, 03:32 PM

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

**Danny** · February 2nd 2009, 03:39 PM

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

$<br /> r_o = f(x) - L$ (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 $y= L$

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