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jverzani
2024-05-22 07:55:20 -04:00
parent f710cded15
commit 771bb06aa3
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6
quarto/integrals/volumes_slice.qmd
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@@ -242,17 +242,17 @@ cone = integrate(pi*R^2, (y, 0, h))
It is not unusual to parameterize a cone by the angle $\theta$ it makes and the height. Since $r/h=\tan\theta$, this gives the formula $V = \pi/3\cdot h^3\tan(\theta)^2$.
The frustum of a cone is simply viewed as a cone with its top cut off. If the original height would have been $h_0$ and the actual height $h_1$, then the volume remaining is just $\int_0^{h_1} \pi r(y)^2 dy$. We can see the formula for the frustrum of the right cone:
The frustum of a cone is simply viewed as a cone with its top cut off. If the original height would have been $h_0$ and the actual height $h_1$, then the volume remaining is just $\int_0^{h_1} \pi r(y)^2 dy$. We can see the formula for the frustum of the right cone:
```{julia}
@syms h0
frustrum = integrate(pi*R^2, (y, 0, h0))
frustum = integrate(pi*R^2, (y, 0, h0))
```
Simplifying, we can see this volume can be expressed using the ratio $(h_0/h_1)$:
```{julia}
frustrum - cone * ( 3h0/h - 3(h0/h)^2 + (h0/h)^3) |> simplify
frustum - cone * ( 3h0/h - 3(h0/h)^2 + (h0/h)^3) |> simplify
```