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05:28

Felicia S.

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

00:06

Jeffery W.

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00:56

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alrighty. So here we're going to do a surface integral of a vector field over a curve. And as you know, the most important thing to do here is to parameter rise the curve. So actually, here it's relatively easy because I can write Z as a function of X and y. What I can do is I can just let X beat s and y b t. And so my premature ization, it's just going to be S t and then one minus s minus team. Okay, so thistle is something. It's important to remember that a lot of times when you're parameter izing a surface. If that surface is actually coming from a function of X and Y, it's really easy to promote dress. Same thing for a curve. If the curve actually is just a function of X, y is a function of X. It's pretty straightforward to parameter rise. So that's what we have here and now what we do need to be careful of so recall that are region looks something like this. So it's in the first Occident and we have this plane. It passes through one on all of the coordinate axes and So if I look down in the X Y plane, the projection down there is actually a triangle. And so if I think about fixing an X value or an s value between zero and one in that triangle down in the X Y plane so I could say s is between zero and one. But now for a fixed s why? Or tea is just going to go from zero up to this line, which now remember Zia zero. So the line is actually just one minus s. Because if I plug in zero and saw for why I get one minus X and s is equal to X. Okay, so here's my domain of my premature ization. It's actually a triangle. And so one thing that's interesting here is that we are specifying an upward orientation, and so we're gonna have to make sure that our premature ization actually has an upward orientation. But that's not so hard to dio. Okay, so let's actually find that the direction of the orientation. So in other words, we want to find a normal vector and show that it's actually pointing up away from the origin and so recalled that that normal vector is given by the cross product. So R s, which is 10 negative one in our tea, which is 01 negative one and said the normal vector is going to be the cross product, which is okay, we're going to have zero minus negative one. That's gonna be one that we have negative one minus zero. But it's negative because it's the Y component. So one, one and then for the Z component we have one is well, so the normal vector is actually just 111 And now it's a little confusing here because in general, when we do this cross product, this will actually be a function of S and T. But because we're dealing with the plane, actually, the normal vector is constant. It's just constantly 111 in this vector points up out of the plane. So it's not towards the origin. That would be a negative orientation. It's actually pointing up out of the origin. So that's a upward orientation upward. Just meaning sort of in the general positive Z direction. Okay, so now we just want to do our surface integral of the Specter field, but notice what's gonna happen when I do this dot product. So here's my surface s here. I have my region are in the S t plane, so f got with the normal direction. Since the X and Y coordinates of my vector field are the same and I do the dot product, I'm just gonna end up adding them together. So I'm going to get zero, which is actually really helpful because thes functions air nasty. And so I could just go ahead and do the stock product here and notice that well, I'm going to plug in the parameter ization and again when I take the dog product here, even if after I plug in X s t y S t, that should actually be of S t, of course. So if I plug in X S t y best easy of S t here. This plus, this is gonna be zero. So I actually only have the Z component when I do the dot product, which is gonna be one times the Z component but evaluated at the premature ization. It's a one minus s minus t. So I'm just integrating one minus s minus t. And now s is going from 0 to 1. T is going from 01 minus s. This is D T D s. Okay, so this is I have a T. So that's a one minus s. And then here, I'm gonna have a T s. So that's minus s times one minus s. And then I have a t squared over two that's gonna be minus one minus s squared over to Okay, so we're going to continue evaluating the center girl. Of course I need a d s here. But the first thing I'm going to do is I'm actually going to expand what we have here. So this is the inner girls. You're the 11 minus s minus s plus s squared, minus one half and then pull us two s over two. So s. And then minus a squared over two, which is so minus s plus s one minus a half is one half. We have minus s in the s squared minus s squared over two. It's gonna be plus escort over to Yes, and then we'll take any anti derivative you have s over to minus escort over two plus s cubed over six. Evaluated from 0 to 1, just one half minus a half plus the sixth, which is 1/6. Just a reminder that our region was a triangle and we had our vector field. And what this 16 represents is the flux of the field across this surface here. And so, if it was water, you could just think about this, the amount of water passing through the region or if the vector field represented an electric field. That's the electric flux over the region, etcetera. Uh, yeah, so really cool has a really nice physical interpretation very much related to the line. Integral flux. Integral that we did just It's the amount of the field passing out of the service.

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