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0:00

Felicia S.

If $f(x)=x+\sqrt{2-x}$ and $g(u)=u+\sqrt{2-u},$ is it true that $f=g ?$

00:56

Greninjack D.

01:09

01:59

Nutan C.

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So the last two examples took a line integral either a circulation integral or a flux integral and made it simpler by converting it to a double inter. Okay, but there are actually examples where they're double integral that are easy to evaluate, easier to evaluate. If you convert them to a line undergrowth or a circulation integral, that's going to be the case here. So we want to find the area bounded by this curve parameter rised by this factor value function. And so I recommend that you actually plot this yourself. But what this curve looks like, is it sort of like a? Well, it kind of like that. So it's a little bit hard to describe. So if you tried to do, even if you try to use polar coordinates or something like that, it's not even clear. You know what this is? You know, it's not a why is not a function of X here. It doesn't really even look like a polar function, so it looks a little bit difficult. Okay, but there's the curve. It's premature ized here and the inside region, which we call our That's what we want. The area of so recall that we had this nice way to kind of mix and match. And, you know, what we do is we introduced this kind of auxiliary vector valued function minus y X. And what we're able to do is say that the integral through the area of the region is really one half the integral over the boundary. So the circulation integral of this particular vector field minus why x dot with the unit tended direction. Okay, so let's see what happens when you plug all that in. So remember that we're going to do a line integral. So we need the parameter ization. We're going to need to take the derivative so the derivative is too consigned to t because 17 okay. And so this is gonna be one half. And then we have the inner girl from zero to pi. So we have We already have the privatization of the curve, which is great, and then minus why is going to be tu minus? Why is minus sign t and then we dot with the X component of the derivative. So this is gonna be minus two scientific because signs of duty plus X, which is signed to t dot with white components. So we'll have cousin t signed to t DT. Okay. And so this integral isn't super super easy to dio, but it's certainly easier than the alternative. And we can do this. Okay, So you can either just plug this in her computer or yeah, I mean, I'll show you how to do it without the thing to notice is well, the coastline to tea and assigned to t. I don't really like that. Okay? And so what I want to do is I actually want to convert those back to just sign of tea and society. So recall that cosign of two t is cosine squared. Um, t minus sign script t. So this is one half zero to pi and then we'll have minus two sign t and then we have cosign squared t minus science, great t and then we have plus sign tooty is to sign ti ko 70. So this is to coast sine squared t scientific, Super cool. Okay, D t. Now look what happens here. So the minus two scientist t times coastline square T plus to cosign square t sign t will cancel on what we're left with is actually one half and a girl from zero to pi of to sign cube tape. And it's even better than that because the one the one half cancels with the to and I just want to integrate teared up. I sign Cube T, and this is one of our classic, uh, you substitution. So we actually wanna let you, because I m t so that minus do you will be signed t e t do u d t. And then what do we get? Well, we're going to get so one of the scientists gets absorbed with the d t and we get a minus, d you? So we're going to get a minus and then let's see if cosine is, uh, let's see t zero cosign. Um, zero is one, and then co sign of pie is negative one and then we're left with the sine squared. But that's one minus cosine squared, which is one minus. You squared. Damn. And so we can flip these upsets minus 11 We'll have one minus you squared, Do you? And this is U minus. You cubed over three. Evaluated from minus 1 to 1. Cool. You evaluated from one to minus one is just going to be, too. And then we'll have minus. This is going to be, Let's see minus. Let's see you cute. That's gonna be one and then minus one of minus two thirds. So it's six thirds minus two thirds, which is four thirds. So that's the area of this crazy region. And we did it without really being able to describe the region. It all all we knew was this parameter ization of the surface. So again, you're starting to get a glimpse of how cool Green's theorem is were able to find areas of regions just by integrating along the boundary this kind of special vector fit, so it's really cool stuff.

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