# method of cylindrical shells

• Aug 29th 2006, 04:00 PM
FLTR
method of cylindrical shells
Use the method of cylindrical shells to find the volume of the solid rotated about the line $\displaystyle x=-1$ given the conditions: $\displaystyle y=x^3-x^2; y=0;x =0$
• Aug 29th 2006, 04:10 PM
ThePerfectHacker
Look at the picture.
Let us keep things positive.
$\displaystyle -(x^3-x^2)=x^2-x^3$
Now you are rotating about $\displaystyle x=-1$
Let us shift the coordinate axes to the right 1 unit.
That means replace $\displaystyle x$ for $\displaystyle x+1$.
Thus, your new curve is, $\displaystyle (x+1)^2-(x+1)^3$
And you are rotating this time about the y-axis becase we shifted the curve.
Thus by cylindrical shells,
$\displaystyle 2\pi \int_1^2 x[(x+1)^2-(x+1)^3]dx$
Note the new limits of integration because we shifted the curve.
• Aug 29th 2006, 05:04 PM
FLTR
once again i thank you for your paitence. To finish the problem I have came up with the following, however I feel something is wrong.

$\displaystyle 2\pi \int_1^2 x[(x+1)^2-(x+1)^3]dx$
$\displaystyle =2\pi \left[-\frac{1}{3}x(x(x+3)+3)\right}_{1}^{2}\approx-19.8968$
• Aug 29th 2006, 06:01 PM
ThePerfectHacker
I made a mistake. When you shift the coordinate axes to the right instead of $\displaystyle x+1$ it is $\displaystyle x-1$.
• Aug 29th 2006, 07:06 PM
FLTR
ok, based on that I have revised my anwser to this. Thoughts?

$\displaystyle 2\pi \int_1^2 x[(x-1)^2-(x-1)^3]dx$
$\displaystyle =2\pi \left[\frac{1}{3}x((x-3)+3)\right}_{1}^{2}\approx 2.0944$
• Aug 30th 2006, 04:22 AM
galactus
Don't get disks mixed up with shells.

$\displaystyle \int_{0}^{1}\overbrace{(-1-x)}^{\text{radius}}\overbrace{(x^{3}-x^{2})}^{\text{height}}\overbrace{dx}^{\text{thick ness}}=\frac{4{\pi}}{15}$