My work is attached and the correct final answer is 250pi/3.
Any help would be greatly appreciated!
Thanks in advance!
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,
$\displaystyle r=y,\ h=2x=\sqrt{y}$
Then the vor for that region is $\displaystyle 2{\pi}\int_{0}^4y\sqrt{y}dy=2{\pi}\int_{0}^4y^{\fr ac{3}{2}}dy$
This works out to be about $\displaystyle 25.6{\pi}$
For the upper region, where the cylinder height stretches between the line and curve
$\displaystyle r=y,\ h=x_{line}-x_{curve}$
x for the curve is negative here always.
$\displaystyle x_{curve}=-\frac{\sqrt{y}}{2}$
$\displaystyle x_{line}=\frac{6-y}{2}$
$\displaystyle h=\frac{6-y}{2}-\left(-\frac{\sqrt{y}}{2}\right)$
The vor of this region is $\displaystyle 2{\pi}\int_{4}^9y\left(\frac{6-y}{2}+\frac{\sqrt{y}}{2}\right)dy={\pi}\int_{4}^9\ left(6y-y^2+y^{\frac{3}{2}}\right)dy$
Evaluating this yields about $\displaystyle 57.73{\pi}$
The sum of the 2 volumes of revolution is $\displaystyle (57.73+25.6){\pi}=83.33{\pi}$