Find the volume of revoulution of the region enclosed by $y = x^3, x = 0, x = 1$ and $y=-1$ around the axis $y = -1$.
This question is most easily done via the disk method, however, I tried cylindrical shells and I have an incorrect answer.
The region is composed of two sections: A square with side 1 in the fourth quadrant and a region enclosed by $y=x^3$ in the first quadrant. So using shells, I believe I need to split the integral up, since the shells will be varying in height:
For ths lower region, I have $2\pi \int_{-1}^0 y\,dy = \pi$ For the upper region, I have $2 \pi\int_{0}^1 (1+y)(1-y^{1/3})\,dy = \frac{9\pi}{14}$ Evaluating and adding together yields an incorrect answer.
Many thanks.
Note that you made a small typo; the lower region should be set up as: $$2\pi \int_{-1}^0 -y\,dy = \pi$$
Thus, your total volume would be: $\pi+\dfrac{9\pi}{14}=\dfrac{14\pi}{14}+\dfrac{9\pi}{14}=\boxed{\dfrac{23\pi}{14}}$
Otherwise, your method looks correct. The easier solution via the disk method yields the same answer: $$\pi \int_0^1 (x^3+1)^2dx = \boxed{\dfrac{23\pi}{14}}$$