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update volumes_slice.qmd

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some typos.
This commit is contained in:
Fang Liu
2023-05-26 09:51:48 +08:00
parent 768c649d76
commit bfa8ffd767
2 changed files with 9 additions and 9 deletions
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2
quarto/integrals/mean_value_theorem.qmd
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@@ -200,7 +200,7 @@ What integral will show the intuition of the Merton College scholars that the di
#| hold: true #| hold: true
#| echo: false #| echo: false
choices = [ choices = [
"``\\int_0^t (v_0 + au) du = (v_0 t + a\cdot u^2/2)\\big|_0^t``", "``\\int_0^t (v_0 + au) du = (v_0 t + a\\cdot u^2/2)\\big|_0^t``",
"``\\int_0^t (v(0) + v(u))/2 du = v(0)/2\\cdot t + x(u)/2\\ \\big|_0^t``", "``\\int_0^t (v(0) + v(u))/2 du = v(0)/2\\cdot t + x(u)/2\\ \\big|_0^t``",
"``(v(0) + v(t))/2 \\cdot \\int_0^t du = (v(0) + v(t))/2 \\cdot t``" "``(v(0) + v(t))/2 \\cdot \\int_0^t du = (v(0) + v(t))/2 \\cdot t``"
] ]

16
quarto/integrals/volumes_slice.qmd
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@@ -56,7 +56,7 @@ Then the volume would be found by adding:
$$ $$
V = \pi \cdot r_1^2 \cdot h_1 + \pi \cdot r_2^2 \cdot h_2 + \cdot + \pi \cdot r_n^2 \cdot h_n. V = \pi \cdot r_1^2 \cdot h_1 + \pi \cdot r_2^2 \cdot h_2 + \cdots + \pi \cdot r_n^2 \cdot h_n.
$$ $$
The weight would come by multiplying the volume by some appropriate density. The weight would come by multiplying the volume by some appropriate density.
@@ -299,7 +299,7 @@ Vₜ, _ = quadgk(h -> pi * rad(h)^2, 0, 50)
For the actual shoot, the vase is to be filled with ash, to simulate a funeral urn. (It will then be knocked over in a humorous manner, of course.) How much ash is needed if the vase has walls that are 1/2 centimeter thick For the actual shoot, the vase is to be filled with ash, to simulate a funeral urn. (It will then be knocked over in a humorous manner, of course.) How much ash is needed if the vase has walls that are 1/2 centimeter thick
We need to subtract $0.5$ from the radius and `a` then recompute: We need to subtract $0.5$ from the radius and `h` then recompute:
```{julia} ```{julia}
@@ -432,7 +432,7 @@ So the integral becomes:
$$ $$
V = \int_0^h A_{xc}(u) du = A_{xc}(0) \int_0^h (1 - \frac{u}{h})^2 du = A_{xc}(0) \int_0^1 v^2 \frac{1}{h} dv = A_{xc}(0) \frac{h}{3}. V = \int_0^h A_{xc}(u) du = A_{xc}(0) \int_0^h (1 - \frac{u}{h})^2 du = A_{xc}(0) \int_0^1 v^2 h dv = A_{xc}(0) \frac{h}{3}.
$$ $$
This gives a general formula for the volume of such cones. This gives a general formula for the volume of such cones.
@@ -618,12 +618,12 @@ h225 = find_zero(h -> r_vol(h) - 225, 10)
numericq(h225) numericq(h225)
``` ```
* For the straight-sided glass, what is the percentage of the total height when the glass is half full. (For a cylinder it would just be 50. * For the straight-sided glass, what is the percentage of the total height when the glass is half full. (For a cylinder it would just be 50.)
```{julia} ```{julia}
#| echo: false #| echo: false
numericq(h225/450 * 100, 2, units="percent") numericq(h225/h450 * 100, 2, units="percent")
``` ```
* People often confuse the half-way by height amount for the half way by volume, as it is for the cylinder. Take the height for the straight-sided glass filled with $450$ mm, divide it by $2$, then compute the percentage of volume at the half way height to the original. * People often confuse the half-way by height amount for the half way by volume, as it is for the cylinder. Take the height for the straight-sided glass filled with $450$ mm, divide it by $2$, then compute the percentage of volume at the half way height to the original.
@@ -655,12 +655,12 @@ h_225 = find_zero(h -> s_vol(h) - 225, 10)
numericq(h_225) numericq(h_225)
``` ```
* For the curved-sided glass, what is the percentage of the total height when the glass is half full. (For a cylinder it would just be 50. * For the curved-sided glass, what is the percentage of the total height when the glass is half full. (For a cylinder it would just be 50.)
```{julia} ```{julia}
#| echo: false #| echo: false
numericq(h_225/450 * 100, 2, units="percent") numericq(h_225/h_450 * 100, 2, units="percent")
``` ```
* People often confuse the half-way by height amount for the half way by volume, as it is for the cylinder. Take the height for the curved-sided glass filled with $450$ mm, divide it by $2$, then compute the percentage of volume at the half way height to the original. * People often confuse the half-way by height amount for the half way by volume, as it is for the cylinder. Take the height for the curved-sided glass filled with $450$ mm, divide it by $2$, then compute the percentage of volume at the half way height to the original.
@@ -771,7 +771,7 @@ plot!(x-1)
#| echo: false #| echo: false
Ra(y) = cbrt(y+1) Ra(y) = cbrt(y+1)
ra(y) = y + 1 ra(y) = y + 1
a,b = 0, 1 a,b = -1, 0
val, _ = quadgk(x -> pi * (Ra(x)^2 - ra(x)^2), a,b) val, _ = quadgk(x -> pi * (Ra(x)^2 - ra(x)^2), a,b)
numericq(val) numericq(val)
``` ```