method of cylindrical shells

**FLTR** · August 29th 2006, 04:00 PM

Use the method of cylindrical shells to find the volume of the solid rotated about the line $x=-1$ given the conditions: $y=x^3-x^2; y=0;x =0$

**ThePerfectHacker** · August 29th 2006, 04:10 PM

Look at the picture.
Let us keep things positive.
$-(x^3-x^2)=x^2-x^3$
Now you are rotating about $x=-1$
Let us shift the coordinate axes to the right 1 unit.
That means replace $x$ for $x+1$ .
Thus, your new curve is, $(x+1)^2-(x+1)^3$
And you are rotating this time about the y-axis becase we shifted the curve.
Thus by cylindrical shells,
$2\pi \int_1^2 x[(x+1)^2-(x+1)^3]dx$
Note the new limits of integration because we shifted the curve.

**FLTR** · August 29th 2006, 05:04 PM

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.

$ 2\pi \int_1^2 x[(x+1)^2-(x+1)^3]dx$
$=2\pi \left[-\frac{1}{3}x(x(x+3)+3)\right}_{1}^{2}\approx-19.8968 $

**ThePerfectHacker** · August 29th 2006, 06:01 PM

I made a mistake. When you shift the coordinate axes to the right instead of $x+1$ it is $x-1$ .

**FLTR** · August 29th 2006, 07:06 PM

ok, based on that I have revised my anwser to this. Thoughts?

$ 2\pi \int_1^2 x[(x-1)^2-(x-1)^3]dx $
$ =2\pi \left[\frac{1}{3}x((x-3)+3)\right}_{1}^{2}\approx 2.0944 $

**galactus** · August 30th 2006, 04:22 AM

Don't get disks mixed up with shells.

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

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