Find the volume generated between two revoluting two arcs

Question

I am looking forward to solve numerically the volume generated between between two arcs of circles undering revolution around y-axis of given radii and centers that can change dynamically depending on their opening angles. So here AB is tangent to x-axis while BC is not. Here I should use use the washer method by writing: $$V=\pi \int x(y)^2dy$$ here i got stuck. about writing $x(y)$. what i thought is to take for each arc as a portion of ellipse, for example AB of center $(x_A, y_B)$ and BC of center $(x_C,y_B)$, then use the equation of ellipse to write $x(y)$. I watched a youtube series here, it should not be $x(y)$ but rather like $x_B^2 - x(y)^2$ where $x(y)$ is: $$x(y) = \begin{cases} x_A + \dfrac{x_B-x_A}{y_B}\sqrt{1-(y-y_B)^2} \ \ \ if x \in arc AB \\ x_C - \dfrac{x_C-x_B}{y_B}\sqrt{1-(y-y_B)^2} \ \ \ if x \in arc BC \end{cases}$$ These two eqautions correspond to equation of an ellipse. so the volume should be written as:: $$V = \pi \int_A^B x_{B}^2 - x(y)^2 dy + \pi \int_B^C x_B^2+x(y)^2 dy$$ so I am not sure if this way of thinking is correct, because I am not able to know what is the expression of the volume. This way is giving me some negative values, especially i am not sure what kind of shape this will up with. Any suggestions?

Answer 1

We need the equations for the two arcs $x_{AB}(y)$ and $x_{CB}(y)$ between $y=0$ and $y=y_B$ such that

$$V=\pi \int_0^{y_B} (x_{CB}(y))^2-(x_{AB}(y))^2 \,dy$$

which is equivalent to

$$V=\pi \int_0^{y_B} (x_{CB}(y))^2-x_B^2 \,dy+\pi \int_0^{y_B} x_B^2-(x_{AB}(y))^2 \,dy$$

In your expression I can't completely understand the derivation you have used for $x(y)$ and why you are using ellipse if the arcs are circular. In any case the correct set up for the integral you are considering should be

$$V = \pi \int_A^B x_{B}^2 - x(y)^2 dy + \pi \int_B^C x_B^2-x(y)^2 dy=$$

$$= \pi \int_A^B x_{B}^2 - x(y)^2 dy + \pi \int_C^B x(y)^2-x_B^2 dy$$