use quarto, not Pluto to render pages
This commit is contained in:
jverzani committed
1 parent
93c993206a
commit
7b37ca828c
879 files changed
+793311
-2678
No files matched your search
@@ -0,0 +1,15 @@
|
||||
{
|
||||
"hash": "f41415b8d9e2ee6ebd4bde5e3b3947cf",
|
||||
"result": {
|
||||
"markdown": "# Volumes by slicing\n\n\n\nThis section uses these add-on packages:\n\n``` {.julia .cell-code}\nusing CalculusWithJulia\nusing Plots\nusing QuadGK\nusing Unitful, UnitfulUS\nusing Roots\nusing SymPy\n```\n\n\n\n\n---\n\n\n\n\n\nAn ad for a summer job says work as the Michelin Man! Sounds promising, but how much will that costume weigh? A very hot summer may make walking around in a heavy costume quite uncomfortable.\n\n\nA back-of-the envelope calculation would start by\n\n\n * Mentally separating out each \"tire\" and lining them up one by one.\n * Counting the number of \"tires\" (or rings), say $n$.\n * Estimating the radius for each tire, say $r_i$ for $1 \\leq i \\leq n$.\n * Estimating the height for each tire, say $h_i$ for $1 \\leq i \\leq n$\n\n\nThen the volume would be found by adding:\n\n\n\n$$\nV = \\pi \\cdot r_1^2 \\cdot h_1 + \\pi \\cdot r_2^2 \\cdot h_2 + \\cdot + \\pi \\cdot r_n^2 \\cdot h_n.\n$$\n\n\nThe weight would come by multiplying the volume by some appropriate density.\n\n\nLooking at the sum though, we see the makings of an approximate integral. If the heights were to get infinitely small, we might expect this to approach something like $V=\\int_a^b \\pi r(h)^2 dh$.\n\n\nIn fact, we have in general:\n\n\n> **Volume of a figure with a known cross section**: The volume of a solid with known cross-sectional area $A_{xc}(x)$ from $x=a$ to $x=b$ is given by\n>\n> $V = \\int_a^b A_{xc}(x) dx.$\n>\n> This assumes $A_{xc}(x)$ is integrable.\n\n\n\nThis formula is derived by approximating the volume by \"slabs\" with volume $A_{xc}(x) \\Delta x$ and using the Riemann integral's definition to pass to the limit. The discs of the Michelin man are an example, where the cross-sectional area is just that of a circle, or $\\pi r^2$.\n\n\n## Solids of revolution\n\n\nWe begin with some examples of a special class of solids - solids of revolution. These have an axis of symmetry from which the slabs are then just circular disks.\n\n\nConsider the volume contained in this glass, it will depend on the radius at different values of $x$:\n\n\n\n\n\nIf $r(x)$ is the radius as a function of $x$, then the cross sectional area is $\\pi r(x)^2$ so the volume is given by:\n\n\n\n$$\nV = \\int_a^b \\pi r(x)^2 dx.\n$$\n\n\n:::{.callout-note}\n## Note\nThe formula is for a rotation around the $x$-axis, but can easily be generalized to rotating around any line (say the $y$-axis or $y=x$, ...) just by adjusting what $r(x)$ is taken to be.\n\n:::\n\nFor a numeric example, we consider the original Red [Solo](http://en.wikipedia.org/wiki/Red_Solo_Cup) Cup. The dimensions of the cup were basically: a top diameter of $d_1 = 3~ \\frac{3}{4}$ inches, a bottom diameter of $d_0 = 2~ \\frac{1}{2}$ inches and a height of $h = 4~ \\frac{3}{4}$ inches.\n\n\nThe central axis is straight down. If we rotate the cup so this is the $x$-axis, then we can get\n\n\n\n$$\nr(x) = \\frac{d_0}{2} + \\frac{d_1/2 - d_0/2}{h}x = \\frac{5}{4} + \\frac{5}{38}x\n$$\n\n\nThe volume in cubic inches will be:\n\n\n\n$$\nV = \\int_0^h \\pi r(x)^2 dx\n$$\n\n\nThis is\n\n::: {.cell execution_count=6}\n``` {.julia .cell-code}\nd0, d1, h = 2.5, 3.75, 4.75\nrad(x) = d0/2 + (d1/2 - d0/2)/h * x\nvol, _ = quadgk(x -> pi * rad(x)^2, 0, h)\n```\n\n::: {.cell-output .cell-output-display execution_count=7}\n```\n(36.917804295114436, 7.105427357601002e-15)\n```\n:::\n:::\n\n\nSo $36.9 \\text{in}^3$. How many ounces is that? It is useful to know that 1 [gallon](http://en.wikipedia.org/wiki/Gallon) of water is defined as $231$ cubic inches, contains $128$ ounces, and weighs $8.34$ pounds.\n\n\nSo our cup holds this many ounces:\n\n::: {.cell execution_count=7}\n``` {.julia .cell-code}\nozs = vol / 231 * 128\n```\n\n::: {.cell-output .cell-output-display execution_count=8}\n```\n20.456618830193282\n```\n:::\n:::\n\n\nFull it is about $20$ ounces, though this doesn't really account for the volume taken up by the bottom of the cup, etc.\n\n\nIf you are poor with units, `Julia` can provide some help through the `Unitful` package. Here the additional `UnitfulUS` package must also be included, as was done above, to access fluid ounces:\n\n::: {.cell execution_count=8}\n``` {.julia .cell-code}\nvol * u\"inch\"^3 |> us\"floz\"\n```\n\n::: {.cell-output .cell-output-display execution_count=9}\n```\n20.456618830193282 fl ozᵘˢ\n```\n:::\n:::\n\n\nBefore Solo \"squared\" the cup, the Solo cup had markings that - [some thought](http://www.snopes.com/food/prepare/solocups.asp) - indicated certain volume amounts.\n\n\n\n\n\nWhat is the Line truncated
|
||||
"supporting": [
|
||||
"volumes_slice_files"
|
||||
],
|
||||
"filters": [],
|
||||
"includes": {
|
||||
"include-in-header": [
|
||||
"<script src=\"https://cdnjs.cloudflare.com/ajax/libs/require.js/2.3.6/require.min.js\" integrity=\"sha512-c3Nl8+7g4LMSTdrm621y7kf9v3SDPnhxLNhcjFJbKECVnmZHTdo+IRO05sNLTH/D3vA6u1X32ehoLC7WFVdheg==\" crossorigin=\"anonymous\"></script>\n<script src=\"https://cdnjs.cloudflare.com/ajax/libs/jquery/3.5.1/jquery.min.js\" integrity=\"sha512-bLT0Qm9VnAYZDflyKcBaQ2gg0hSYNQrJ8RilYldYQ1FxQYoCLtUjuuRuZo+fjqhx/qtq/1itJ0C2ejDxltZVFg==\" crossorigin=\"anonymous\"></script>\n<script type=\"application/javascript\">define('jquery', [],function() {return window.jQuery;})</script>\n"
|
||||
]
|
||||
}
|
||||
}
|
||||
}
|
||||
Reference in new issue
Block a user