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I am using Stewart Calculus and trying to understand one of the formulas for the surface area of revolution generated by a curve about an axis on an interval.

The standard formula for the surface area generated by a curve $y = f(x)$, $a \leq x \leq b$, about the $x$ axis is given by:

$$\int_a^b 2 \pi y \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\; dx.$$

If instead the curve is given as a function of $y$, i.e. $x = g(y)$, $c \leq y \leq d$, but we are still revolving it about the $x$ axis, then Stewart gives the formula as:

$$\int_c^d 2 \pi y \sqrt{1 + \left(\frac{dx}{dy}\right)^2}\; dy.$$

I am struggling to grasp how this second formula is derived. To me it seems that if we are revolving about the $x$ axis we would need to still integrate with respect to $x$. Would anyone be able to clarify or give an example?

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Denote by $dSA$ the surface area of the sliced element of the arc. Then $$dSA=2\pi y\cdot ds$$ where $(x,y)$ is the center of the sliced element. If $x=f(y)$ and $\frac{dx}{dy}$ exists, then one can write $ds$ as $$ds=\sqrt{1+\bigg(\frac{dx}{dy}\bigg)^2}dy.$$ Thus, $$SA=\int_c^ddSA.$$

This shows that even if the axis of rotation is the $x$-axis, it does not imply that the variable of integration is $dx$.

Try the following example.

The arc of the line $y=\sqrt{3}x$ from $(0,0)$ to $(1,\sqrt{3})$ is rotated about the $x$-axis. Find the area of the resulting surface.

Soln 1. $\frac{dy}{dx}=\sqrt{3}$. Thus, $ds=\sqrt{1+(\sqrt{3})^2}dx=2dx$ and so $$SA=\int_0^12\pi y\cdot ds=\int_0^12\pi\sqrt{3}x\cdot2dx=2\pi\sqrt{3}$$

Soln 2. $\frac{dx}{dy}=\frac{1}{\sqrt{3}}$ and so, $ds=\sqrt{1+\bigg(\frac{1}{\sqrt{3}}\bigg)^2}dy=\frac{2}{\sqrt{3}}dy$. Thus, $$SA=\int_0^{\sqrt{3}}2\pi y\cdot ds=\int_0^{\sqrt{3}}2\pi y\cdot\frac{2}{\sqrt{3}}dy=2\pi\sqrt{3}$$

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