A volume of frustum of right circular cone is calculated as follows. With known h, R & r of a container with the shape shown below, how to find out the rise in height for each time $7m^3$ of water is added in the container? Suppose there is no water in the container initially.
r is the radius at the topmost level. The rise in height of water should be different for each time. Please help. Thanks.
HINTS: At any instant, the radius $R_x$ at a vertical height $x$ from the bottom is given as $$R_x=R-\left(\frac{R-r}{h}\right)x$$ Now, consider an elementary slice of thickness $dx$ & radius $R_x$ at a a normal height $x$ from the bottom of frustum. Te volume of elementary slice $$dV=\pi R_x^2\ dx=\pi \left(R-\left(\frac{R-r}{h}\right)x\right)^2\ dx$$ Now, let the rise be $x_1$ in first time to fill $7\ m^3$ of water then using integration with proper limits $$\int_0^7\ dV=\int_0^{x_1}\pi \left(R-\left(\frac{R-r}{h}\right)x\right)^2\ dx$$ $$7=\frac{\pi h}{3(R-r)}\left(\left(\frac{R-r}{h}\right)^3x_1^3-3R\left(\frac{R-r}{h}\right)^2x_1^2+3R^2\left(\frac{R-r}{h}\right)x_1\right)$$ $$\frac\pi 3\left(\frac{R-r}{h}\right)^3x_1^3-\pi R\left(\frac{R-r}{h}\right)x_1^2+\pi R^2 x_1-7=0$$ Each time, one need to solve cubic equation, as obtained above, to find rises in height $x_1, x_2, ....$