Is the integral in the meme picture correct when it comes to calculating the efficiency of the glass? Assuming the cup is a perfect frustum of a cone.
2026-03-28 15:35:56.1774712156
Calculating the efficiency of glass
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The two point formula for a line gives us $$\frac{r-r_1}{h-0}=\frac{r_2-r_1}{h_2-0}$$ or $$r=r_1+\frac{r_2-r_1}{h_2}h$$ Which can be checked by substituting in $h=0$ and $h=h_2$. Then the volume up to $h_1$ is $$V_1=\int_0^{h_1}\pi\left[r_1+\frac{r_2-r_1}{h_2}h\right]^2dh$$ And up to $h_2$ is $$V_2=\int_0^{h_2}\pi\left[r_1+\frac{r_2-r_1}{h_2}h\right]^2dh$$ This is not equivalent to the formula in the picture. Note that in that case the integrands are constant, so the volume would be proportional to $h$ which only works for a cylinder $r_2=r_1$ in which case the expressions contain division by zero.