Using the definition of the mean value of a function as $$\frac{1}{b-a}\int_a^by \space dx$$ Why can this not be used to form a rectangle with equal area of that of a curved function and then use $\pi r^2 h$ substituting the mean value of a function in for $r$ to find the volume of revolution instead of the usual volume of revolution formula.
I tried this method but it didn't appear to work and I'm wondering why it doesn't.
I am not sure what deep reasons are you looking for the fact that a wrong method of calculation does not work. In a nutshell, the reason is that the square function is not linear. That is, the mean value of the square of a function is not equal to the square of its mean value, unless the function is constant. More concretely, since $h$ in your notation must equal $b-a$, what you are suggesting is this formula: $$\pi\left(\frac{1}{b-a}\int_a^b f(x)dx\right)^2(b-a)=\pi\int_a^bf^2(x)dx$$ or $$\left(\frac{1}{b-a}\int_a^b f(x)dx\right)^2=\frac{1}{b-a}\int_a^bf^2(x)dx$$ and as pointed out above, this says that the mean value of the square of a function is equal to the square of its mean value, which only holds if $f$ is a constant function.