# How to find the volume of a revolution by splitting into two revolutions

I have a region that is bounded by $y=\sqrt{x}$, $y=7$, $x=0$, $x=4$ and I need to find the volume of this region when revolved around the y-axis

Now, the easiest way of doing this I believe would be using the shell method, but I was thinking of how to achieve this using the washer method, so I decided to divide the region into two sections:

So to find the volumes:

$$V_1 = \int_2^7{4^2 - 0^2}dy$$ $$V_2 = \pi\int_0^2{(y^2)^2 - 0^2}dy$$

The answer I get adding the volumes ($V_1 + V_2$) is 100.106, but I know the correct answer is $\frac{432 * \pi}{5}$

Why does the washer method not work here?

## 2 Answers

Note that

$$V_1 = \color{red}{\pi}\int_2^7{4^2 - 0^2}dy=80\pi$$

$$V_2 = \pi\int_0^2{(y^2)^2 - 0^2}dy=\frac{32}5\pi$$

thus

$$V_1+V_2=\frac{432}5\pi$$

...or you could write all with $\;x\;$ as a function of $\;y\;$ and doing it "the usual" way (and splitting, certainly):

$$\begin{cases}V_1:\;\;&x=y^2\;,\;\;y=0\;,\;\;y=2\\{}\\V_2:\;\;&2\le x\le7\;,\;\;y=4\end{cases}$$

and thus we get:

\begin{align*}&V_1:\;\;\pi\int_0^2y^4\;dx=\left.\frac\pi5y^5\right|_0^2=\frac{32\pi}5\\{}\\ &V_2:\;\;\pi\int_2^716\;dx=\pi\cdot16\cdot(7-2)=80\pi\end{align*}

And thus the wanted volume is

$$V_1+V_2=\frac{32\pi}5+80\pi=\frac{432\pi}5$$

• Precisely! I think the OP get wrong forgetting the $\pi$ term in front of the integral for $V_1$. – user Apr 2 '18 at 0:07