Tyler Clarke
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48 lines
4.1 KiB
HTML
[!]
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[=post-]
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<p>
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Welcome once more to another thrilling textbook section review. This one covers Thomas 15.6, which is a truly strange section-
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called simply "Applications", it doesn't cover any new material as much as demonstrate some interesting possibilities with triple integrals.
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</p>
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<p>
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The first thing to consider is a geometric interpretation of the integral of `F` over a volume - if `F` is a function for density,
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the result of the integration is mass! This has some surprising consequences. For instance: with triple integrals, you can find the first moment about
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any given axis. If your mass integration is `int int_D int F(x, y, z) dV`, then the first moment about the x-axis is `int int_D int xF(x, y, z) dV`,
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the moment about the y axis `int int_D int yF(x, y, z) dV`, etc. These are handy for physical reasons, but the immediate useful point is
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finding the center of mass: the center of mass about an axis is the first moment about that axis divided by total mass, so the center
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of mass x-component is `frac {int int_D int xF(x, y, z) dV} {int int_D int F(x, y, z) dV}`, etc.
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</p>
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<p>
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Let's do an example. Given a constant density function `F(x, y, z) = 2`, what is the center of mass of a shape bounded below by the xy plane and
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above by the function `z = 4 - x^2 - y^2`?
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</p>
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<p>
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To find this, we first need to find the bounds of integration. The z bounds are obvious, but what about x and y?
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Fortunately, fate smiles upon us, and we can just take the intersection of the z bounds `z=0` and `z=4-x^2-y^2`. A quick substitution and some algebra
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gives us `x^2 + y^2 = 4`, which is just a circle with radius 2! Assuming we want to integrate in `dz dy dx` (`dz dx dy` would also be valid), this means
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the x-bounds are from -2 to 2, and the y bounds are from `-sqrt(4 - x^2)` to `sqrt(4 - x^2)`.
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Our mass is thus given by `int_{-2}^{2} int_{-sqrt(4 - x^2)}^{sqrt(4 - x^2)} int_{0}^{4-x^2-y^2} 2 dz dy dx`.
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The inner integral in `dz` evaluates nicely to `8 - 2x^2 - 2y^2`. This leaves us with `int_{-2}^{2} int_{-sqrt(4 - x^2)}^{sqrt(4 - x^2)} 8 - 2x^2 - 2y^2 dy dx`. Ew.
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It's useful to learn how to do these thorny ones properly, but without the much-desired help of cylindrical coordinates it's also <i>very annoying</i>, and out of the scope
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of this post. Wolfram|Alpha to the rescue! Our mass is 50.2655 (note that this is unitless).
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</p>
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<p>
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Because this shape is symmetrical in the x and y directions and the density is constant, the first moment and thus center of mass in both directions should be zero.
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This leaves the z direction. To find that, we first need to find the first moment in z: `int_{-2}^{2} int_{-sqrt(4 - x^2)}^{sqrt(4 - x^2)} int_{0}^{4-x^2-y^2} 2z dz dy dx`.
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I categorically refuse to do this in Cartesian. For those who are interested, the integral in cylindrical coordinates is just `int_{0}^{2pi} int_{0}^{2} int_{0}^{4-r^2} 2zr dz dr d theta`,
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which is a lot easier to evaluate. We'll see more discussion of this problem in the post on 15.7, but for now, just believe me when I say that it evaluates to 67.0206. Dividing
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`frac {67.0206} {50.2655}` gives us a fairly nice number: `frac {4} {3}`. So our center-of-mass is `\[0, 0, frac {4} {3}]`. Nice!
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</p>
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<p>
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The next thing to talk about is moment of inertia. This is actually a physics concept, but it's an interesting illustration of another property of triple integrals.
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Moment of inertia can be thought of as <i>rotational mass</i> - it's the property that resists torque, much like how mass is the property that resists force.
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The moment of inertia is also known as the <i>second moment</i>. You can find the second moment about any given line or axis just by multiplying the density by the
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function for squared distance from said line or axis - `int int_D int r^2(x, y, z)F dV`. For instance, the moment of inertia about the x axis is
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`int int_D int (y^2 + z^2)F dV`. The process here is otherwise exactly the same as finding the first moment.
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</p>
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[/]
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[=title "Multivariable Exam 3 Review: Thomas 15.6"]
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[=author "Tyler Clarke"]
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[=date "2025-3-30"]
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[=subject "Calculus"]
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[#post.html] |