Volumes of Rotation: Where's my mistake?? I have tried to calculate the volume of of the solid bound by the function
$y = (-x^2+6x-5)^2$ and $y = x-1$ around the line $x =6$. Here is my work.  The answer should be $\frac{63\pi}{2}$ but I'm getting a different answer.
Have I made a mistake somewhere? Or is the answer key wrong.
I started by saying the functions intersect at $x = 1$ and $x = 4$, and since the functions are wrt $x$, I'm using shells, so I set up the integral:
$V =2\pi \int_1^4  x(6-(-x^2+6x-5))-x(6-(x-1))^2dx $
Which all simplifies to
$2\pi \int_1^4  2x^3-8x^2+6x dx$
and so the solution to the integral is:
$2\pi(3x^2+\frac{1}{2}x^4-\frac{8}{3}x^3)$
Plugging in the bounds gives me $9\pi u^3$.
I'm not sure if I set this up wrong or if I made some error, but it's been a while since I did volumes!  Any help is much appreciated
 A: You seem to have made a typo in writing the first function: $y = (-x^2 + 6x - 5)^2$, and I shall proceed by assuming that you meant $y = (-x^2 + 6x - 5)$.
For a rotation around the line $y=h$, the formula for the volume of rotation is:
$${\begin{cases}\displaystyle 2\pi \int _{a}^{b}(x-h)f(x)\,dx,&{\text{if}}\ h\leq a<b\\\displaystyle 2\pi \int _{a}^{b}(h-x)f(x)\,dx,&{\text{if}}\ a<b\leq h\end{cases}}$$
We break the bounded area into two, which you have done correctly. However, notice that you have:
$$2\pi\int_{1}^{4} x(6-(-x^{2}+6x-5))-x(6-(x-1))^2dx$$
According to the formula, what you should have done instead is:
$$2\pi\int_{1}^{4} (6-x)(-x^{2}+6x-5)-(6-x)(x-1)dx $$
Which yields $\frac{63\pi}{2}$, the desired answer.
$$$$
I checked my work using the washer method which, albeit less elegant, also yields $\frac{63\pi}{2}$:
$$
\pi\int_{0}^{3} \left[ \left(6 - \left(3 - \sqrt{4-y}\right)\right)^2 - \left(6 - (y+1)\right)^2 \right] dy + \pi\int_{3}^{4} \left[ \left(6 - \left(3 - \sqrt{4-y}\right)\right)^2 - \left(6 - \left(3 + \sqrt{4-y}\right)\right)^2 \right] dy$$
