79 lines
6.4 KiB
HTML
79 lines
6.4 KiB
HTML
[!]
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[=post-]
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<p>
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Hey there, everyone! I come to you after a long and boring Spring break with unholy tidings of exams approaching.
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My first exam in multivariable will be covering everything in Thomas' Calculus (really good book, much recommend) between 15.5 and 16.8. To do those very complex sections
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justice, I'm writing a separate piece for each of them. It's dense stuff!
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</p>
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<h2>Basics of Triple Integrals</h2>
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<p>
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15.5 starts with a discussion of triple integrals. These really aren't all that complicated - if you know how to do basic integration, they shouldn't pose a challenge.
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Usually they're written as `int int_D int f dV`. In this case, `D` is the region, and `dV` is the <i>area element</i>
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- the `1 dx dy dz` part, if this was in terms of cartesian coordinates, but as we'll see there are many options.
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</p>
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<p>
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The hard part of triple integration is determining bounds. Usually speaking, you'll have to either figure out the bounds yourself from a set of boundary functions, OR convert the bounds to a different
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form. There's a knack to it that you can pick up with practice, but the process amounts to picking an order of integration and finding bounds that don't leave dangling variables
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(for instance, if you try to integrate `int_0^y int_0^z int_1^3 sin(x) dz dy dx`, you'll end up with a function rather than a value). You also want to pick bounds of integration that
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arne't too complicated - in many, many cases, it will actually be easier to change variables to spherical or cylindrical coordinates rather than toughing it out in cartesian with nasty bounds.
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</p>
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<p>
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Let's go through a few examples. A classic is the triangular area one: we have a region in the first quadrant below the plane x + y + z = 4
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over which to integrate the function `f(x, y, z) = x`. This area is convenient because we don't have to think too much about it: it's obviously the lower triangular half of the (4, 4, 4) cube.
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We need to find the bounds of the triple integral that describes this area. We don't need to do anything special with the coordinate system here, so our `dV = dz dy dx`. How did I pick that order?
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This shape is symmetrical, so I picked at random - it'll be equally easy to produce bounds for any given order.
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</p>
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<p>
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Because the x-bounds are on the outside, we want them to be constant (if they contain variables, our final result will contain variables). Because this is a region in the
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first quadrant, x must be no larger than 0; the largest possible value of x is the x-intercept, `x=4`. Thus we have our first constraint: `0 <= x <= 4`.
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Onto `y`! `y` is a bit more complex. We can't just say `0 <= y < = 4`, because for z=0 that would produce a square rather than a triangle on the xy plane.
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If you can visualize the intercept of the region and the xy plane, it's clearly a simple line `y = 4 - x`: we can use this! `0 <= y <= 4 - x` is valid.
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</p>
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<p>
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Finally, the z-bounds. These are a bit tricky to intuit, but fortunately we've a nice equation! `x + y + z = 4` can be easily reduced to `z = 4 - x - y`, and this is
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in fact the upper bound of z over this entire region. `0 <= z <= 4 - x - y`!
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</p>
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<p>
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We can insert those bounds and our element of integration directly into the integral, to get `int_0^4 int_0^{4-x} int_0^{4 - x - y} f(x, y, z) dz dy dx`. Beautiful!
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Because `f(x, y, z) = x`, this is equivalent to `int_0^4 int_0^{4-x} int_0^{4 - x - y} x dz dy dx`. Let's integrate! The innermost integral, `int_0^{4 - x - y} x dz`,
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evaluates obviously to `4x - x^2 - xy` (if you can't see how I got this, practice integration). This means our integral is now `int_0^4 int_0^{4-x} 4x - x^2 - xy dy dx`.
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The innermost integral is a bit trickier for this one, but works out to `4x(4 - x) - x^2(4 - x) - x(4 - x)^2 / 2`.
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</p>
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<p>
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Yikes.
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</p>
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<p>
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Before we can finish the integration, we have to simplify. A binomial expansion and some distribution leaves us with
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`16x - 4x^2 - 4x^2 + x^3 - x^3/2 + 4x^2 - 8x`, which simplifies to `8x - 4x^2 + x^3/2`. Not terrible! Integrating `int_0^4 8x - 4x^2 + x^3/2 dx` leaves
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us with a simple `|_0^4 4x^2 - 4/3x^3 + x^4/8`, which evaluates to `frac{32}{3}`. Not too hard!
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</p>
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<p>
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An example of a harder problem, ripped straight from the textbook, is to find the volume of a region bounded below by `z = 3` and bounded above
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by a sphere of radius 5 (centered on the origin). I'll leave solving this one as an exercise to the reader.
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</p>
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<p>
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A particularly common problem is finding the volume of a tetrahedron given its corner points. There's no real trick here - we just have to find
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a set of lines describing the extrema, and solve them properly. For instance (yoinked from the textbook): given a tetrahedron described by the points (0, 0, 0), (0, 1, 0), (1, 1, 0), and (0, 1, 1),
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what are the bounds?
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</p>
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<p>
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Like the last one, it doesn't matter too much which dimension we pick; I'm reusing `dV = dz dy dx`. The bounds for x are obvious enough: the lowest x-value is 0, and the highest is 1,
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so `0 <= x <= 1`. The bounds for y are conveniently described by our points on the z=0 plane: (0, 0, 0), (0, 1, 0), and (1, 1, 0). These form a right triangle; the line we're concerned
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with is (coincidentally) the hypotenuse, described by `y = x`. In this case, `y` is <i>above</i> the line and below 1, so our y-bounds are `x <= y <= 1`.
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Finally, onto the z-bounds! To find these, consider that z increases with y and decreases with x, hinting not-so-subtly
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at a function quite like the one from before: `z = y - x`. `z` is below the function, here, so `0 <= z <= y - x`.
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</p>
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<p>
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These are very difficult to derive without graphics. While sketching is out of the scope of this website (seeing as I'm so bad at it), I'd recommend anyone learning multivariable
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practice sketching complex functions in a way that conveys the area reasonably well. Oftentimes, the functions you need are blisteringly obvious once you can look at them drawn out.
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</p>
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<p>
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Other basic calculus ideas like the average value of a function in a region apply here as well. The average value of `F` over a region is simply
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`frac {int int_D int F dV} {int int_D int 1 dV}` - the integral of `F` over the region, divided by the volume of the region. Easy!
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</p>
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[/]
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[=title "Multivariable Exam 3 Review: Thomas 15.5"]
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[=author "Tyler Clarke"]
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[=date "2025-3-27"]
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[=subject "Calculus"]
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[#post.html] |