Trouble finding the volume using the shell method A region is bounded by the line $y=3x+4$ and the parabola $y=x^2$ and is rotated about the line $x=4$.
First I found the limits of integration by finding the points of intersection. They are $(-1,1)$ and $(4,16)$. I then found the radius from the axis of rotation to be $r=4-x$ and the height to be $h=3x+4-x^2$.
I used the Volume formula $\int_{a}^{b}2\pi (radius)(height)dx$.
$$V=\int_{-1}^{4}2\pi(4-x)(3x+4-x^2)dx$$
$$2\pi \int_{-1}^{4}(x^3-7x^2+8x+16)dx$$
Evaluating this integral I get the volume to be $280\pi-\dfrac{1055}{6}$ but this is no where close to the actual volume. I believe that both my radius and height are right and that the formula is right. I've redid the problem a couple of times and ended up with the same answer, what am I missing here?

Everything is fine in what you have; the only thing missing is the actual evaluation of the definite integral.  I get $$\displaystyle 2\pi\left[\frac{x^4}{4}-\frac{7x^3}{3}+\frac{8x^2}{2}+16x\right]_{-1}^{4}$$
$$\displaystyle 2\pi \left[ \left(64-\frac{7(64)}{3}+64+64\right)- \left(\frac{1}{4}+\frac{7}{3}+\frac{8}{2}-16\right)\right]$$
From there, it's just a matter of simple algebra.  I get $\displaystyle \frac{625\pi}{6}$.
