I've been faced with this problem originally:
Use the shell method to find the volume of the solid generated by revolving the region bounded by the line $y=3x+4$ and the parabola $y=x^2$ about the following lines.
... c) The x-axis.
My problem: According to my math lab, the answer is $250\pi$, however my answer is slightly above that and I can't explain why.
I got the inverse of both functions such that $x=\frac{(y-4)}3$ and $x=\sqrt{y}$. Looking at the graph of these functions, I know that the radius of the shell is $y$, and the height must be the right-most function minus the left-most function.
Knowing this along with the formula for the shell method, I get $$2\pi\int_0^{16}y(\sqrt{y}-\frac{(y-4)}3)$$ $$2\pi\int_0^{16}y^\frac{3}2-\frac{1}3y^2+\frac{4}3y$$ $$2\pi[\frac{2}5y^\frac{5}2-\frac{1}9y^3+\frac{2}3y^2]$$
Now here I would plug in 16 (I didn't bother with 0 since all of the terms have a y in it) and would get $\frac{11264\pi}{45}$, aka $250.311\pi$. My answer is so close, but I don't see what I could've done wrong here. Any help is appreciated.
Between $y = 0$ and $y = 1$, your cylindrical shell doesn't go from $x = \frac{y-4}3$ to $x = \sqrt y$, it goes from $y = -\sqrt y$ to $x = \sqrt y$. That small "triangle" between $(0,0)$, $(-1, 1)$ and $(-\frac43, 0)$ shouldn't be included in the calculations, but you have included it. See the below image where the "height" of your cylindrical shell is marked in red, and the actual shell in black:
So you need to do this as two separate integrals, one from $y = 0$ to $1$, and one from $y = 1$ to $16$.