Volume of this solid of revolution? I'm currently taking Calculus II as an undergrad, and I am having a lot of issues understanding solids of revolution. Unfortunately neither my professor or my textbook has been much help and I'm feeling very "black boxed"--that is, when the formula doesn't work I am completely lost as to how to fix it.
The current problem I am having difficulty with is the following:
The area bounded by the y-axis and the curves $x=1+(y-2)^2$ and $x=2$ is rotated about the x-axis. Find the volume of the solid created by using the method of cylindrical shells.
I have attempted this problem and got the following integral: $ \int_1^2 (2 \pi)(y)(1+(y-2)^2dx $, which evaluates to $\frac{23 \pi}{6}$. The textbook gives the answer as $\frac{16 \pi }{3} $ with no explanation for this answer. I am completely lost and could really use an explanation of why my integral isn't working and what will work.
 A: For this problem, the region is in the first quadrant, and it is bounded on the left by part of a parabola opening to the right, and on the right by a vertical line. Its bottom tip is the point $(2,1)$, its top tip is the point $(2,3)$, and its left-most point is $(1,2)$.
Since it's being revolved around the $x$-axis, we could either use a $dx$ integral with the "washers" method, or else a $dy$ integral with the "shells" method. The first option would have us cut up the region into vertical slices, expressing the edges as functions of $x$, which seems annoying. The second option allows us to use the fact that we already have the left and right edges expressed as functions of $y$.
We need to integrate from bottom to top, i.e., from $y=1$ to $y=3$. The integrand, with the shell method, is given by the cylindrical lateral area formula: $A=2\pi r h$. We have $r=y$, and $h=2-(1+(y-2)^2)$, because that's the right edge, minus the left edge. Putting that together, we get:
$$\begin{align}
\int_1^3 2\pi\cdot y\cdot(2-(1+(y-2)^2) \,dy &= 2\pi\int_1^3 y(2-(1+y^2-4y+4))\,dy\\
&=2\pi\int_1^3 y(-3+4y-y^2)\,dy\\
&=-2\pi\int_1^3 y^3-4y^2+3y\,dy\\
&=-2\pi\left[\frac{y^4}{4}-\frac{4y^3}{3}+\frac{3y^2}{2}\right]_1^3\\
&=\frac{16\pi}{3}
\end{align}$$
Does this help?
A: The first thing you want to do is to plot the function and evaluate what you have (see the figure below). The washer method tells us that for rotation about the $x$-axis,
$$V=\pi\int_1^2(y_1^2-y_2^2)~dx$$
where
$$
y_1,y_2=2\pm\sqrt{x-1}\\
\begin{align}
y_1^2-y_2^2
&=(2+\sqrt{x-1})^2-(2-\sqrt{x-1})^2\\
&=8\sqrt{x-1}
\end{align}
$$
Then,
$$V=\pi\int_1^2(y_1^2-y_2^2)~dx=\pi\int_1^2 8\sqrt{x-1}~dx=\frac{16\pi}{3}$$

A: Looking at the graph we can see that the "top function" is $x=2$ and the bottom function is $x=1+(2-y)^2$ in your integral you forgot to account for the 2 functions. These intersect when $y=1$ and when $y=3$ so our shell formula is $2\pi\int_{a}^{b}y(f(y)-g(y))\,\mathrm{d}y$ plugging in all our values we get$2\pi\int_{1}^{3}y(2-(1+(2-y)^2)\,\mathrm{d}y$ Also in the future, a valuable strategy that I learned when functions are given in terms of $x=f(y)$ as opposed to the typical $y=f(x)$ as far as the integration and volume goes you can just swap the $x$'s and $y$'s in order for you to get a better mental picture of the region, all the bounds will be the same and if done correctly you'll get the same answer 
