What is the height of a rotating function when the volume is given

Question

Let the function $y = x^2$ rotate around the $y$-axis. If you would pour in $18$ units of fluid, how high would the fluid reach?

Answer 1

To calculate the volume of this "bowl" generated on rotating the parabola $y=x^2$, you would need an expression for its volume for any height (taken as the measurement along $Y$-axis) $y$.

Let $(x,y)$ be any point on the parabola. To find the volume of a "bowl" of height $y$, you need to integrate: $$A(y) = \int_{0}^{y}\pi x^2dy $$ $$ = \int_{0}^{y}\pi y dy$$ $$ = \pi \left[\frac{y^2}{2}\right]_{0}^{y} $$ $$ = \pi y^2/2 $$ To give the intuition behind the above integration: in this "bowl", all the points which have the same $y$ co-ordinate form a disk of radius $x$ and area $\pi x^2$. Volume of this infinitesimally thin disk of thickness $dy$ is $\pi x^2 dy$. Integrating this volume from $0$ to $y$ gives the volume of the entire "bowl", till the height $y$. This is only an intuition behind the procedure and must not be taken as a rigorous proof that this formula gives what we want.

Okay, now that we have the volume as an expression in $y$, we need to find the height (or in this case the value of $y$) for which volume = 18 cube units. $$\pi y^2/2 = 18$$ $$y=\frac{6}{\sqrt{\pi}} $$ (Note that $y=-\frac{6}{\sqrt{\pi}}$ is not a solution as $y=x^2$, so it is always positive.)

Answer 2

From the start i told myself that cylindrical shells was the way to go but it always turnet out wrong. After days of struggling with this i finally tried plane slices and of course it was a lot easier and i got it right :)

The answer is that the cone that is created must have a height of 6/(sqrt(pi)) units.