Here is the question:
For the area between $x=g(y)$, $y-$axis, and $y=a$ and $y=2a$ lines, the volume is known to be $a^2$ when the area is rotated around the $y$-axis.
Similarly, for the are between $x=g(-y/2)$, $y$ axis and $y=-2a$ and $y=4a$ lines, the volume is known to be $a^3+3a$ when the area is rotated around the $y$-axis.
edit: $a>0$
Given these informations, find $g(y)$.
So in my understanding, the problem is: $$π\int_a^{2a} (g(y))^2 \, dy = a^2$$ and the other volume is this: $$π\int_{-2a}^{4a} (g(-y/2))^2 \, dy = a^3+3a$$
I thought maybe since the given volumes are polynomial, the function would also be polynomial. So I said if $g(y)$ is a $"n"$ degree polynomial, $g(y)^2$ would be $2n$ degree polynomial and if we integrate that, we would end up with a $2n+1$ degree polynomial. And because $a^2$ is a second degree polynomial, $2n+1$ would be $2$, and $n=1/2$. However, this doesn't make sense since square roots are not polynomials. On top of that the second integral results with a 3rd degree polynomial so it wouldn't make sense that way either. ı don't know how else I can solve this.
I'm really stuck on this and I have a big test tomorrow. Please help...
My illustrions of the areas created in MS Paint, with made up functions for $g(y)$ and $g(-y/2)$
