Kivindu's question via email about volume by revolution

[Prove It]

View attachment 5626

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

View attachment 5627

Clearly the x-intercept of $\displaystyle \begin{align*} y = 3 - 3\,\sqrt{x} \end{align*}$ is (1, 0) so the terminals of the integral will be $\displaystyle \begin{align*} 0 \leq x \leq 1 \end{align*}$.

We should note that the volume will be exactly the same if everything is moved up by 2 units, with the advantage of being rotated around the x-axis.

So we want to get the volume of the solid formed by rotating $\displaystyle \begin{align*} y = 5 - 3\,\sqrt{x} \end{align*}$ around the x-axis and then subtract the volume of the solid formed by rotating $\displaystyle \begin{align*} y = 2 \end{align*}$ around the x-axis. So the volume we want is

$\displaystyle \begin{align*} V &= \int_0^1{ \pi\,\left( 5 - 3\,\sqrt{x} \right) ^2 \,\mathrm{d}x } - \int_0^1{ \pi\,\left( 2 \right) ^2 \,\mathrm{d}x } \\ &= \pi\int_0^1{ \left[ \left( 5 - 3\,\sqrt{x} \right) ^2 - 2^2 \right] \,\mathrm{d}x } \\ &= \pi\int_0^1{ \left( 25 - 30\,\sqrt{x} + 9\,x - 4 \right) \,\mathrm{d}x } \\ &= \pi\int_0^1{ \left( 21 - 30\,x^{\frac{1}{2}} + 9\,x \right) \,\mathrm{d}x } \\ &= \pi\,\left[ 21\,x - \frac{30\,x^{\frac{3}{2}}}{\frac{3}{2}} + \frac{9\,x^2}{2} \right] _0^1 \\ &= \pi\,\left[ 21\,x - 20\,x^{\frac{3}{2}} + \frac{9\,x^2}{2} \right] _0^1 \\ &= \pi\,\left\{ \left[ 21 \,\left( 1 \right) - 20\,\left( 1 \right) ^{ \frac{3}{2}} + \frac{9\,\left( 1 \right) ^2}{2} \right] - \left[ 21\,\left( 0 \right) - 20 \,\left( 0 \right) ^{\frac{3}{2}} + \frac{9\,\left( 0 \right) ^2 }{2} \right] \right\} \\ &= \pi \,\left( 21 - 20 + \frac{9}{2} - 0 \right) \\ &= \frac{11\,\pi}{2}\,\textrm{units}^3 \end{align*}$

 

[jvicens]

Thank you for sharing your sketch and problem. I find this type of problem interesting and I would like to offer my thoughts on it.

Firstly, your approach to finding the volume of the solid is correct. However, I would like to point out that the line $y=2$ does not need to be rotated around the x-axis, as it is already parallel to it. Therefore, we can simply subtract the volume of the solid formed by rotating $y=2$ around the x-axis from the volume of the solid formed by rotating $y=5-3\sqrt{x}$ around the x-axis.

Additionally, I would like to mention that when rotating a function around an axis, it is important to consider the orientation of the function. In this case, the function $y=5-3\sqrt{x}$ is below the x-axis, so we would need to use the formula $\pi\int_a^b (f(x))^2 dx$ instead of $\pi\int_a^b (f(x))^2 dx$.

Finally, I would like to suggest using the disk method instead of the shell method to find the volume in this problem. The disk method would involve slicing the region into thin disks perpendicular to the x-axis and summing their volumes, while the shell method would involve slicing the region into thin shells parallel to the x-axis and summing their volumes. The disk method may be easier to visualize and calculate in this case.

I hope my insights are helpful to you. Good luck with your problem!