Volume of revolution of $y=\sqrt {x+2},y=x,y=0$ about $x$-axis Q: Volume of revolution of $y=\sqrt {x+2},y=x,y=0$ about $x$-axis using the shell method.My Approach: I know how to use shell method. For shell method height and radius are the main thing, because $V=2\pi\int R(y)H(y)dy$  That's why I found height $$H(y)=y-(y^2-2).$$ But I cannot write $R(y)$, because for $x\in[-2,0]$ the lower curve is $y=0$ and for $x\in[0,2]$ the lower curve is $y=x$. Then how could i find the volume using shell method. Any hints or solution will be appreciated.Thanks in advance.
 A: The axis of rotation is the x axis.
Since you are using the shells method, then the shells shall be the horizontal hollow cylinders made by the 
rotation of the rectangles having a side on the line at constant $y$, and height $dy$  . The lenght of the horizontal side is 
$\Delta x = x_2(y)-x_1(y)=y-(y^2-2)$.
Now, $\Delta x$ remains non-negative on the range $0 \le y \le 2$, which is the range over which you shall integrate.
The segment $(-2,0),(0,0)$ is horizontal, and does not represent an obstacle using this method.

So the volume is
$$
\eqalign{
  & V = 2\pi \int_0^2 {y\left( {y - \left( {y^{\,2}  - 2} \right)} \right)dy}  =   \cr 
  &  = 2\pi \int_0^2 {\left( { - y^{\,3}  + y^{\,2}  + 2y} \right)dy}  = 2\pi \left( { - {{16} \over 4} + {8 \over 3} + 2{4 \over 2}} \right) =   \cr 
  &  = {{16} \over 3}\pi  \cr} 
$$
which is the answer you are given.
You can countercheck this by performing the integration by "disks", i.e. vertical circular cross-sections.
These are solid cylinders, with a circulat basis centered at $(x,0)$, having radius $y(x)$, and thickness of $dx$.
In this method, the discontinuity between the line $y=0$ and $y=x$ requires some attention.
We just have to split the integral into two parts :
$$
\eqalign{
  & V = \int_{x =  - 2}^0 {\pi y_{\,1} (x)^{\,2} dx}  + \int_{x = 0}^2 {\left( {\pi y_{\,1} (x)^{\,2}  - \pi y_{\,2} (x)^{\,2} } \right)dx}  =   \cr 
  &  = \pi \left( {\int_{x =  - 2}^0 {\left( {x + 2} \right)dx}  + \int_{x = 0}^2 {\left( {x + 2 - x^{\,2} } \right)dx} } \right) =   \cr 
  &  = \pi \left( {{4 \over 2} + \left( {2 \cdot 2 + {4 \over 2} - {8 \over 3}} \right)} \right) =   \cr 
  &  = {{16} \over 3}\pi  \cr} 
$$
