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Given the equation $y = x + \sin(x)$, as far as I am aware, an explicit equation $x = f(y)$ can not be found. Is there still a way to compute the volume of the solid of revolution of this function rotated about the y-axis, because the volume of the solid of revolution about the x-axis can be found ($V = \pi \int (x + \sin(x))^2 dx$). Considering $x = f(y)$ this "function" appears to have points with infinite slope on integer multiples of the point $(\pi,\pi)$ which may also complicate the computation of the solid of revolution.

Graph of $y = x + \sin(x)$

N. F. Taussig
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unnamed
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  • You could express y as a Taylor series in x and invert it to obtain x as a function of y. The "infinite slope" should not present a problem, as the volume (between finite limits) is finite. – Phil Freedenberg May 26 '23 at 22:19
  • See here for inverses of $x+\sin(x)$. You could also try an inverse substitution like $x\to f^{-1}(x)$ to remove the inverse function from the integrand – Тyma Gaidash May 26 '23 at 22:28
  • You can actually slice the region in a particular way to avoid using inverses. – Sean Roberson May 26 '23 at 22:33
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    One could try $\int\pi x^2,\mathrm{d}y=\int\pi x^2(1+\cos(x)),\mathrm{d}x$ – robjohn May 26 '23 at 23:10
  • Using the substitution dy = (1 + cos(x))dx in the integral produces an anti-derivative that shares many similar terms with the cylindrical shell method and substituting f(x) with f(b) - f(x). However it is not exactly the same and does not produce the same answer, except in the case when the lower bound a is set equal to zero. – unnamed May 28 '23 at 15:41

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You've not specified a bounded region or interval over which to compute the volume of revolution, but in general, we can use the method of cylindrical shells to compute the volume when the cross-section is rotated about the $y$-axis. So if $f$ does not have a closed-form inverse, we can still compute the volume as

$$V_1(a,b) = \int_{x=a}^b 2\pi x f(x) \, dx,$$ for some suitable interval satisfying $0 \le a < b$, whereas the volume of the region rotated about the $x$-axis is given by the method of disks/washers; e.g.,

$$V_2(a,b) = \int_{x=a}^b \pi f(x)^2 \, dx$$ where $f \ge 0$ on $[a,b]$.

heropup
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  • This cylindrical shell method finds the volume of the area between the x-axis and f(x) rotated about the y-axis which is close to what I was looking for. I found that by substituting in the integral f(x) with f(b) - f(x) (where b is the upper bound of integration) the volume between the y-axis and f(x) rotated about the y-axis is found which is what I was looking to compute. – unnamed May 28 '23 at 15:35