I am trying to find the volume of revolution of solid formed by $ y = x^2 $ and $ y=2x $ about the $ y $-axis where both graphs lie in the first quadrant only.
The graph is shown below:
Since it is a rotation about the $y$-axis, let's express $x$ in terms of y instead.
Hence, $ y = x^2 $ becomes $ x = \sqrt{y} $ (first quadrant only) and $ y = 2x $ becomes $ x = \frac{y}{2} $
Now, we will integrate with respect to y to obtain the volume of revolution:
- Find points of intersection:
\begin{align} \text{ Let } x = \sqrt{y} = \frac{y}{2} \end{align} \begin{align} \sqrt{y} = \frac{y}{2} \end{align} \begin{align} y = \frac{y^2}{4} \end{align} \begin{align} y^2 - 4y = 0 \end{align} \begin{align} \therefore \ y = 4 \text{ or } y = 0 \end{align}
- Integrate with respect to y
\begin{align} \text{Volume of revolution} &= \int_{0}^{4}{\pi\left(\sqrt{y}-\frac{y}{2}\right)^2}dy \\ &= \int_{0}^{4}{\pi\left[y-y\sqrt{y}+\frac{y^2}{4}\right]}dy \\ &= \pi\left[\frac{y^2}{2}-\frac{2y^\frac{5}{2}}{5}+\frac{y^3}{12}\right]_0^4 \\ &= \pi\left[\frac{16}{2}-\frac{64}{5}+\frac{64}{12}\right] \\ &= \pi\left[\frac{160}{12}-\frac{64}{5}\right] \\ &= \pi\left[\frac{40}{3}-\frac{64}{5}\right] \\ &= \frac{8}{15}\pi \\ \end{align}
Somehow, the answer appears to be $ \frac{8}{3}\pi $ which was supposedly obtained by integrating with respect to $x$. Now, this leaves me puzzled. Shouldn't the volume produced be the same since the area bounded between the graphs are the same with respect to $x$-axis and $y$-axis?
Could someone please advise me on what went wrong here?
volume of revolution=
$$\int_0^4 \pi ( (\sqrt{y})^2-(\frac{y}{2})^2)dy$$
= difference between the two volumes.
you will find
$\pi \frac{8}{3}$.