What is the volume of the solid that results when the region under $y = 1 + x^3$ between x = 1 and x = 2 is revolved about the x- and y-axis?
For the x-axis: $$ \int_1^2 \pi (1+x^3)^2 \,dx = \frac{373\pi}{14} $$
For the y-axis, I solved the equation in terms of y and I computed for the y-coordinates, so the volume is: $$ \int_2^9 \pi [(y-1)^{\frac13}]^2 \,dx = \frac{93\pi}{5} $$
I only got 5/10 points in this problem, but I do not know which one is correct or wrong. Can you check these two answers?
It's your second one that is wrong: you need to integrate the region under $y=1+x^3$ so you'd be best served using the shells formula in that case. But as you mentioned in the comments you can't use that formula. That's okay, but now you'd need two integrals to get the answer. First do the lower portion which is just the region $1 \leq x \leq 2$ and $0 \leq y \leq 2$: $$ \int_0^2 \pi(2^2-1^2)dy = 6\pi. $$ Your second region is between $y=1+x^3, y=2$, and $x=2$ so use washer formula: $$ \int_2^9 (2^2 - (y-1)^{2/3})dy = \frac{47\pi}{5}. $$
That then adds up to the same answer you'd get from shells (much easier): $77\pi/5$.