Jesse's question via email about volume by revolution

[Prove It]

View attachment 5633

Here is a sketch of the region to be rotated and the line to be rotated around.

View attachment 5634

Notice that the volume will be exactly the same if we were to move everything up by 3 units, but with the advantage of rotating around the x axis. So we want to find the volume of the region under $\displaystyle \begin{align*} y = 12 - x^2 \end{align*}$ between $\displaystyle \begin{align*} x= 0 \end{align*}$ and $\displaystyle \begin{align*} x = 3 \end{align*}$ rotated around the x axis, and then subtract the volume of the region under $\displaystyle \begin{align*} y = 3 \end{align*}$ between $\displaystyle \begin{align*} x = 0 \end{align*}$ and $\displaystyle \begin{align*} x = 3 \end{align*}$ rotated around the x axis. Thus the volume we want is

$\displaystyle \begin{align*} V &= \int_0^3{ \pi\,\left( 12 - x^2 \right) ^2\,\mathrm{d}x } - \int_0^3{ \pi\,\left( 3 \right) ^2\,\mathrm{d}x } \\ &= \pi\int_0^3{ \left[ \left( 12 - x^2 \right) ^2 - 3^2 \right]\,\mathrm{d}x } \\ &= \pi\int_0^3{ \left( 144 - 24\,x^2 + x^4 - 9 \right) \,\mathrm{d}x } \\ &= \pi\int_0^3{ \left( 135 - 24\,x^2 + x^4 \right) \,\mathrm{d}x } \\ &= \pi\,\left[ 135\,x - 8\,x^3 + \frac{x^5}{5} \right] _0^3 \\ &= \pi\,\left\{ \left[ 135\,\left( 3 \right) - 8 \,\left( 3 \right) ^3 + \frac{3^5}{5} \right] - \left[ 135\,\left( 0 \right) - 8 \,\left( 0 \right) ^3 + \frac{0^5}{5} \right] \right\} \\ &= \pi\,\left( 405 - 216 + \frac{243}{5} - 0 \right) \\ &= \frac{1188\,\pi}{5}\,\textrm{units}^3 \end{align*}$

 

[jvicens]

Hello,

Thank you for sharing your calculation of the volume of this rotated region. I would like to point out that your approach is correct and follows the principles of integral calculus. However, I would also like to suggest an alternative method that may be more efficient and intuitive.

Instead of subtracting the volume of the region under $y=3$, we can simply add the volume of the region between $y=3$ and $y=12-x^2$. This is because when we rotate the region around the x-axis, the volume under $y=3$ will be included in the total volume.

Using this method, we can rewrite the volume as:

$\displaystyle \begin{align*} V &= \int_0^3{ \pi\,\left( 12 - x^2 \right) ^2\,\mathrm{d}x } + \int_0^3{ \pi\,\left( 3 - x^2 \right) ^2\,\mathrm{d}x } \\ &= \pi\int_0^3{ \left[ \left( 12 - x^2 \right) ^2 + \left( 3 - x^2 \right) ^2 \right]\,\mathrm{d}x } \\ &= \pi\int_0^3{ \left( 144 - 24\,x^2 + x^4 + 9 - 6x^2 + x^4 \right) \,\mathrm{d}x } \\ &= \pi\int_0^3{ \left( 153 - 30\,x^2 + 2x^4 \right) \,\mathrm{d}x } \\ &= \pi\,\left[ 153\,x - 10\,x^3 + \frac{2x^5}{5} \right] _0^3 \\ &= \pi\,\left\{ \left[ 153\,\left( 3 \right) - 10 \,\left( 3 \right) ^3 + \frac{2 \cdot 3^5}{5} \right] - \left[ 153\,\left( 0 \right) - 10 \,\left( 0 \right) ^3 + \frac{2 \cdot 0^5}{5} \right] \