Consider the function $f(x) = \sqrt{x}\,$ where $\,0<x<1$
1- This function is rotated about the $x-axis$ by an angle $\theta = \frac{\pi}{2}$
2- Normally the volume of this rotation can be found by $$\frac{\pi}{4}\int_{0}^{1} \left(\sqrt{x}\right)^2 dx $$
3- However, the points are not connected by an arc, but rather by a straight line forming triangles rather than sections of a circle as shown below.
4- I must instead use the method of cross sections to find the volume between the two functions where the area of a right triangle is $\frac{1}{2}bh$ which results in
$$ volume = \frac{1}{2}\int_{0}^{1}bh \,dx \,= \,\frac{1}{2}\int_{0}^{1} (\sqrt{x})(\sqrt{x})\,dx = \frac{1}{4} $$
Q - Am I correct in my reasoning and result?
Thank you for your time and help.