# What's wrong with my Surface Area of a solid of revolution formula?

When I learnt about the derivation for the formula $$V=\pi\int_{x_1}^{x_2} y^2~dx$$ where $$V$$ is volume of the solid generated when $$y=f(x)$$ is rotated about the $$x$$ axis by $$2\pi$$ radians between $$x_1$$ and $$x_2$$, my teacher showed that this works by approximating the volume by splitting up the solid generated into very thin cylinders. Each cylinder would have width $$\delta x$$ and radius $$y$$, so we'd have $$V=\lim_{\delta x\to0}\sum_{x=x_1}^{x=x_2}\pi y^2\delta x=\pi\int_{x_1}^{x_2} y^2~dx$$

I thought of doing the same thing to find the surface area of thus solid that we have produced. Once more, consider the very thin cylinders that we used before. The surface area of the outer strip of each cylinder is $$2\pi y\delta x$$. Hence, I thought that the surface area should be equal to $$SA=\lim_{\delta x\to0}\sum_{x=x_1}^{x=x_2}2\pi y\delta x=2\pi\int_{x_1}^{x_2}y~dx$$ But this is wrong! Why is that? What is wrong my reasoning?

Thank you for your help.

• Kind of like considering $dL = dx$ in the formula for arc length is wrong, you have to use $dL = \sqrt{dx^{2}+dy^{2}}$ – Joshua Wang Mar 4 at 22:40
• See this answer. – Paramanand Singh Mar 7 at 16:24