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# Find $I_{x}$ for the arch in Example 4.

## $I_{x}=2 \pi-2$

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### Video Transcript

Okay, folks. So in this video, we're gonna talk about problem number 41. So we're given Arch, and we're gonna be looking for it The x component of the moment of inertia for this particular shape. So, as you recall, we have an expression for for the X component of the moment of inertia and that expression can be written in this way. We have i X we have I actually quality integral of delta, where delta is the density function which may or may not be constant, depending on the problem. Um, multiplied by r squared minus x squared, multiplied by D. S. So this is Thea expression for the X component of the moment of inertia matrix for for for any, any shape. And in this problem, it zone, arch. So, um, and what we're given in this problem is we're given a density function which varies with position. So let's write that down. We have density being equal two to minus Z multiplied by r squared minus X squared. All right, R squared minus X squared gives you wise court plus disease Course. Let me write this down. We have art y squared plus z squared um because as you remember, R squared is just X squared plus y squared plus z squared, right. That's the square. That's the square of the of the distance from the origin to the any point on the shape that we're looking at. Anyway, Why Square plus Z squared happens to be a very nice number, and it's given in the problem. That number is one. So we're just going to ignore that number. We're going to ignore one. We're not going to run it out every time. Okay, so that leaves us with DS. However, there's another way to write DS Azi remembered. The S is just an infinite decimal line segment. So we can write that as square root of d y squared plus Z squared. Okay, Um, and the picture, the picture that I want you to have in your mind is this This is Thea. I'm not drawing it very well with this city. The shape that we're looking at and this is why axis and this the Z axis. So, um, any infinitives malign sackman, for example, this one could be you could be written as, um, the square root of D y score, plus easy scored investors Pythagorean theorem. Anyway, um, Z tu minus lee this Z, we can write it in another way. We could write it as the square root of one minus y squared because they, as you remember Z squared plus y squared gives you one. So that means Z squared is equal to one minus y squared. And because because of the fact that that in this particular problems, he's always positive we could ignore the negative sign in front of the square root. And just so we just have the square root of one minus y squared for Z, um, and this thing, we're going to write it as square root of one plus, Do you see, Do you? Why squared? Multiply it by the wise, um and this is the usual manipulation that I'm sure all of you guys have seen. Um, and this we're going to rewrite as, um, to minus squared of one minus y squared, multiplied by the square root of one plus well, z prime it. We have to first figure out what It's easy. D why? Well, we take this. We take this equation and we do kind of an implicit differentiation. we have to see d c equals minus two y the y. And so we have easy d y is simply equal to minus Why over easy. But I'm not going to write Z. I'm going to write square root of one minus y squared. So this is thes e d. Y. Now, if you were to square it to get y squared over one minus y squared multiplied, by the way, okay. And so this we're going to split. We're going to Ah, multiply out the parentheses. Here we have twice of one minus y squared one d Why minus, um square one minus y squared, multiplied by one minus y squared one de why and from negative Oneto one same as this one. And this thing is simply ageist. Um, I'm gonna write. Okay, First of all, I want you to take a look at this thing. What is the integral of one over square root of one minus y squared. Right. Um and that integral is simply sine inverse of why, Okay, because the derivative assign numbers gives you one over a square root of one minus westward. Um, from negative, 1 to 1. All right. So there's that minus. And if you know this here we get one. OK? Because that's one over the skirt of one minus y squared, multiplied by square root of one minus y squared and they cancel. So you get one. So you have de y from negative 1 to 1. And this thing is obviously too, Um and let's see what we have for this thing. We have to times sine inverse of one give you pie half, obviously minus minus pi half minus two. So this thing is pie, right? And twice pie gets you to pry minus two. And that is gonna be the answer for problem number 41. And we're done for this video. Thank you. Bye.

University of California, Berkeley

Integrals

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Vector Functions

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