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I know of the following formulas for calculating surface areas:

$\displaystyle A_S = 2\pi\int_{a}^{b}f(x)\sqrt{1+f'(x)^2}{\ dx}$ for the surface area ($A_S$) of the solid formed by revolving $f(x) = y$ about the line $y=0$, $a \leq x \leq b$.

$\displaystyle A_S = 2\pi\int_{a}^{b}f(y)\sqrt{1+f'(y)^2}{\ dy}$ for the surface area ($A_S$) of the solid formed by revolving $f(y) = x$ about the line $x=0$, $a \leq y \leq b$.

But, I'm uncertain of how to calculate the surface area of a solid of revolution revolved about some atypical line greater or lesser than $0$, or a surface area for an object formed about $y=0$ where $f(y)=x$, or an object formed about $x=0$ where $f(x)=y$. I have been unable to find any literature on the matter, could someone provide some insight.

I'm presently trying to calculate the surface area of the solid formed by revolution of $x=\frac{1}{3}(y^2+2)^{3/2} ,\ 1 \leq y \leq 2$ about $y=0$.

The predicament comes from the axis of rotation, as I'd said.

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You can use $\displaystyle S=\int_1^2 2\pi R(y)\sqrt{1+(f^{\prime}(y))^2}dy$ where $R(y)=y$

is the distance from a typical point $(x,y)$ on the curve to the axis of rotation $y=0$,

so this gives $\displaystyle S=\int_1^2 2\pi y\sqrt{1+y^2(y^2+2)}dy=2\pi\int_1^2y\sqrt{(y^2+1)^2}dy=2\pi\int_1^2y(y^2+1)dy$

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  • $\begingroup$ Why, and does this work inversely (were it to be the case where the curve were a function of $x$)? $\endgroup$ – alxmke Jul 5 '15 at 23:40
  • $\begingroup$ You can use $S=\int_a^b 2\pi R ds$ to find the area of a surface of revolution where $s$ represents arclength, and you can express this in terms of x if y is given as a function of x. $\endgroup$ – user84413 Jul 5 '15 at 23:46

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