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This is what I've done so far:

$V_1 = \pi\int_0^af(x)^2dx = -\pi\int_0^by^2 (1/f'(x))\ dy = -\pi\int_0^by^2 (1/f'(g(y)))\ dy = -\pi\int_0^by^2 g'(y)\ dy$

Integrating by parts: $u=y^2,\ du=2y\ dy, \ v=g(y), \ dv = g'(y)\ dy$

$y^2g(y)|_0^b -2\pi\int_0^by\ g(y)\ dy = -ab^2 -2\pi\int_0^by\ g(y)\ dy = -ab^2-V_2$

so it looks like $V_1 = -ab^2-V_2$ which looks wrong.

I've also tried other ways of integrating, substituting and I always end up with a proof that $V_1 \ne V_2$. Can someone point me to the right direction?

Charles Khunt
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2 Answers2

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Starting from the formula for $V_2=2\pi\int_0^b yg(y)dy$, make the substitution $y=f(x)$, notice that $g(f(x))=x$, $dy=f'(x)dx$, we get $$V_2=2\pi\int_a^0 f(x)f'(x)xdx=-\pi\int_0^a [f(x)^2]'xdx$$ Then integrate by parts, $$V_2=\pi f(x)^2x]_0^a+\pi\int_0^a f(x)^2dx=V_1$$

Kle
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You've made two mistakes. $y^2g(y)|_0^b=b^2\cdot0-0^2\cdot a=0$, not $b^2a$. So that leaves you with $V_1=-V_2$ after your integration by parts. But it also seem you're neglecting the minus sign before the integrating by parts. Actually, you are neglecting the $\pi$ too in part as the expression should be $-\pi\, y^2g(y)|_0^b+2\pi\int_0^by\,g(y)\,dy$.

Make these two corrections and you have it.

2'5 9'2
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