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