Need some calculus help...

Question

anonymous · Answer

what is q?

rogue · Answer

The problem is in the attachment. This is what is given$$\int\limits_{0}^{1} f(x) dx = 1$$And I have to find the maximum possible value for the volume. So I tried playing around with this$$V = \pi \int\limits_{0}^{1} f^2 (x) dx$$but I got nowhere :(

anonymous · Answer

Try taking a simple case. If f(x) is such that the area under it from [0,1] is 1, then maybe f(x)=1? That will make a 1 by 1 square which has an area of 1. I think a right circular cylinder in that situation would be as much volume as you could get.

anonymous · Answer

f(x) can be different from the constant 1.
Take f(x) =2 x then $ \int_0^1 2x dx= 1$

anonymous · Answer

Take
$$
f(x) = n x^{n-1}\
\int_0^1 f(x) dx =1\
\pi \int_0^1 (f(x))^2 dx =\frac{n^2\pi}{2 n-1}\
\lim_{n\to \infty}\frac{n^2\pi}{2 n-1}=\infty

$$
So if you do not know f, the maximum can be as big as you want.
@mukushla @experimentX  @satellite73

anonymous · Answer

If you consider pi times the integral from 0 to 1 of f(x)^2 dx, to be a single spike that extends really high, with width dx, then the area is pi/dx^2.

So, if this equals n * pi, then dx = sqrt{n}/n, which is infinite as n -> infinity.
This is possible because the volume varies with x^2 dx, and the area with x dx.

rogue · Answer

Thank you very much for the help; I really appreciate it :)