Volume of solid of revolution by revolving the region $y=x^2$,$x=0$,$y=9$

Question

Find the volume generated by revolving the region shown below:

I have done as follows: Revolve the rectangle formed by the vertices $(0,0),(3,0),(3,9),(0,9)$ about $X$ axis. The required volume generated is the volume of cylinder of radius $9$ and height $3$ which is $$V_1=\pi (9)^2(3)=243\pi$$

Now revolve the region formed by $y=x^2$,$x=3$ and $X$ axis about the $X$ axis. By Disk method the volume is: $$V_2=\pi \int_{0}^{3}(x^2)^2dx=\frac{243\pi}{5}$$ So the required volume is $$V_1-V_2=\frac{972\pi}{5}$$ Is there any other approach?

Answer 1

We could set it up by integrating volume of cylindrical shells as well (shell method). As we are rotating the region around x-axis,

At distance $y$ from x-axis, the shell width = $ \sqrt y \ $ and $0 \leq y \leq 9$.

So $ \ V = \displaystyle \int_0^9 2 \pi y \ \sqrt y \ dy = \frac{972 \pi}{5}$

We could also set up using the disk method as (similar to what you did),

$\displaystyle \int_0^3 \int_{x^2}^{9} 2 \pi y \ dy \ dx = \frac{972 \pi}{5}$

Answer 2

Calculating surface or volumes of revolution can often be done using Guldin's theorems (or Pappus centroid I think in english).

$V=2\pi\times S\times \operatorname{dist}(G,axis)$

Where $S$ is the area of the surface to revolve and $G$ its centroid.

Though using the moments $\begin{cases}M_x=\iint y\mathop{dx}\mathop{dy}\\M_y=\iint x\mathop{dx}\mathop{dy}\end{cases}\quad$ we can get rid of $S$ since $\begin{cases}x_G=\frac 1SM_y\\y_G=\frac 1SM_x\end{cases}$

In the present case, the axis of revolution is $(Ox)$ thus $\operatorname{dist}(G,axis)=y_G$ and the volume is given by $$V=2\pi\times S\times y_G=2\pi\times M_x=2\pi\int_0^3 \int_{x^2}^9 y\mathop{dy}\mathop{dx}=\frac{972\pi}{5}$$

Notice this is the "disk method" referenced in Math Lover's answer.