# Shell method for calculating volume of solid of revolution - general

1,189

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.

Share:
1,189

Author by

### LinAlgMan

Updated on January 10, 2020

• 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 years
I haven't gone through the exercise, but an obvious first step seems to substitute $$x = f^{-1}(y)$$.
• LinAlgMan almost 10 years
And 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 years
You have a mistake in your calculations, I put it into Wolfram Alpha and got that both methods give the same result.