Volume of revolution of $y=\sqrt {x+2},y=x,y=0$ about $x$-axis

Question

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.

Answer 1

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 : $$ \eqalign{ & V = \int_{x = - 2}^0 {\pi y_{\,1} (x)^{\,2} dx} + \int_{x = 0}^2 {\left( {\pi y_{\,1} (x)^{\,2} - \pi y_{\,2} (x)^{\,2} } \right)dx} = \cr & = \pi \left( {\int_{x = - 2}^0 {\left( {x + 2} \right)dx} + \int_{x = 0}^2 {\left( {x + 2 - x^{\,2} } \right)dx} } \right) = \cr & = \pi \left( {{4 \over 2} + \left( {2 \cdot 2 + {4 \over 2} - {8 \over 3}} \right)} \right) = \cr & = {{16} \over 3}\pi \cr} $$