Volume around x axis

**silencecloak** · February 8th 2009, 11:59 AM

$x+y=1$
$x=5-(y-2)^2$

Rotate around the x-axis, using the cylindrical shell method

This is what ive came up with but its wrong

$2\pi\int_0^5 ((5-(y-2)^2+(1-y))(1-y) = \frac{145\pi}{6}$

Any help appreciated, thanks

**Soroban** · February 8th 2009, 12:34 PM

Hello, silencecloak!

$x+y\:=\:1$
$x\:=\:5-(y-2)^2$

Rotate around the x-axis, using the cylindrical shell method.

Your limits are correct . . .

The line has intercepts (1,0) and (0,1).
The parabola opens to the left with vertex at (5,2).
They intersect at (5,0) and (0,5).

The formula is: . $V \;=\;2\pi\int^b_a \text{(radius})\text{(height)}\,dy$

The radius is $y.$
The height is: . $\bigg[5-(y-2)^2\bigg] - \bigg[1 - y\bigg] \:=\:5y - y^2$

Then: . $V \;=\;2\pi\int^5_0y(5y-y^2)\,dy$

Got it?

**silencecloak** · February 8th 2009, 12:44 PM

Got it.

Thanks

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