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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.