Shell method for calculating volume of solid of revolution  general
OK, here's as far as I got. It didn't quite work out yet, I think I made a calculation error somewhere, but I'm posting it as an answer because I'm confident that you'll be able to work out the details.
Start by letting $x = g(y)$, i.e. $y = f(x)$. Then $dy = f'(x) dx$ by the chain rule, and the integration interval changes to $[0, c] \mapsto g([0, c]) = [a, b]$, so we have
$$V = \pi \int_0^c (b^2  g(y)^2) \, dy = \pi \int_a^b (b^2  x^2) \cdot f'(x) \, dx$$
The first part is easy: $$\int_a^b b^2 f'(x) \, dx = \left. b^2 f(x) \right_a^b = b^2(f(b)  f(a)).$$
For the second part, use integration by parts: $$\int_a^b x^2 \cdot f'(x) \, dx = \left. x^2 f(x) \right_a^b  \int_a^b 2x \cdot f(x) \, dx $$ where $ \left. x^2 f(x) \right_a^b = b^2 f(b)  a^2 f(a)$.
Putting it together, you find $$\pi \int_0^c (b^2  g(y)^2) \, dy = \pi\left( \left[ b^2 f(b)  b^2 f(a) \right]  \left[ b^2 f(b)  a^2 f(a)  2 \int_a^b x f(x) \, dx \right] \right) $$
Now instead of all the constant terms cancelling, I am left with $C = \pi(a^2  b^2) f(a)$ and a final answer of $$ V = C + 2\pi \int_a^b x f(x) \, dx $$ I suspect an $a$ should really be a $b$ somewhere and you should be able to get this with $C = 0$.
[EDIT] I just realized that $f(a) = 0$ so indeed $C = 0$ as requested.
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LinAlgMan
Updated on January 10, 2020Comments

LinAlgMan almost 4 years
Let us have an injective continuous function $f : [a,b] \to [0,c]$ (such that $f(a)=0$ and $f(b)=c$). I want to calculate the volume of solid revolution of $f$ around the $y$ axis.
The first method is to write $g = f^{1}$ and then $$ V = \pi \int_0^c (b^2  g(y)^2) dy \ . $$
The second method is by using cyclindrical shells, which gives the formula $$ V = 2 \pi \int_a^b x f(x) dx \ . $$
The two methods are equivalent (I understanding it geometrically) but how do I prove this equivalence by algebraic or differential means? (That is, go from the 1st formula to the 2nd by algebraic and calculus and symbols manipulation.)

CompuChip almost 10 yearsI haven't gone through the exercise, but an obvious first step seems to substitute $$x = f^{1}(y)$$.

LinAlgMan almost 10 yearsAnd also use the chain rule $$ dy = \frac{dy}{dx}{dx} = y'(x)dx$$ to change integration variable but I can't simplify the resilting expression.


LinAlgMan almost 10 yearsYou have a mistake in your calculations, I put it into Wolfram Alpha and got that both methods give the same result.