# Thread: Cylindrical Shell Volume, why am I getting the wrong answer?

1. ## Cylindrical Shell Volume, why am I getting the wrong answer?

My work is attached and the correct final answer is 250pi/3.

Any help would be greatly appreciated!

2. Originally Posted by s3a
My work is attached and the correct final answer is 250pi/3.

Any help would be greatly appreciated!
I think you left out the radius of the shell. Just multiply each of your integrands by y.

3. Originally Posted by s3a
My work is attached and the correct final answer is 250pi/3.

Any help would be greatly appreciated!
Hi s3a,

I think you want to solve this using shells rather than washers.

You haven't correctly set up the surface areas of the cylinders.

For the cylinders whose heights go from one end of the curve to the other,
corresponding to y=0 to y=4,

$r=y,\ h=2x=\sqrt{y}$

Then the vor for that region is $2{\pi}\int_{0}^4y\sqrt{y}dy=2{\pi}\int_{0}^4y^{\fr ac{3}{2}}dy$

This works out to be about $25.6{\pi}$

For the upper region, where the cylinder height stretches between the line and curve

$r=y,\ h=x_{line}-x_{curve}$

x for the curve is negative here always.

$x_{curve}=-\frac{\sqrt{y}}{2}$

$x_{line}=\frac{6-y}{2}$

$h=\frac{6-y}{2}-\left(-\frac{\sqrt{y}}{2}\right)$

The vor of this region is $2{\pi}\int_{4}^9y\left(\frac{6-y}{2}+\frac{\sqrt{y}}{2}\right)dy={\pi}\int_{4}^9\ left(6y-y^2+y^{\frac{3}{2}}\right)dy$

Evaluating this yields about $57.73{\pi}$

The sum of the 2 volumes of revolution is $(57.73+25.6){\pi}=83.33{\pi}$