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I need to find a formula for the surface area of a solid of revolution rotated around the $y$-axis. The curve is $f(x)=x^2$ on $[0,1]$. However, my answer must be in terms of $f$, not $f^{-1}$.

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The only thing that changes when you revolve the curve about the $y$-axis instead of the $x$-axis is the expression for the radius of revolution. When you revolve it about the $x$-axis, the radius of revolution at a particular value of $x$ is $|f(x)|$, the distance from the curve to the $x$-axis; when you revolve it about the $y$-axis, the radius is $|x|$, the distance from the curve to the $y$-axis.

The element $dA$ of surface area at a given $x\in[0,1]$ is still $dA=2\pi x\,ds$, since the radius of revolution about the $y$-axis is simply $x$ in this case. From studying arc length you know that

$$ds=\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx=\sqrt{1+\big(f\,'(x)\big)^2}\;,$$ so

$$dA=2\pi x\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx=2\pi x\sqrt{1+\left(f\,'(x)\right)^2}\,dx\;;$$

now just integrate $dA$ over the appropriate range of values of $x$.

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