Flächeninhalt des Umdrehungsrechners

Your input: find the area of the surface of revolution of $f\left(x\right)=x^{2}$ rotated about the x-axis on $\left[0,1\right]$

The surface area of the curve is given by $S = 2\pi \int_a^b f \left(x\right) \sqrt{\left(f'\left(x\right)\right)^2+1}d x$

First, find the derivative: $f '\left(x\right)=\left(x^{2}\right)'=2 x$ (steps can be seen here)

Finally, calculate the integral $S = \int_{0}^{1} 2 \pi x^{2} \sqrt{\left(2 x\right)^{2} + 1} d x=\int_{0}^{1} 2 \pi x^{2} \sqrt{4 x^{2} + 1} d x$

The calculations and the answer for the integral can be seen here.