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00:33

Dungarsinh P.

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

0:00

Felicia S.

Jsdfio K.

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All right, so we want to find the mass of the hemispherical shell. A radius one. Yeah. Okay. So what is the mass? Well, let's write our favorite equation. So the mass, it's just going to be the sum over this surface, the hemispherical shell of all the mass elements. But what is a mass element? Well, really, I'm doing a double integral over the surface of the density times. A small surface element two sigma. All right, so now all this look to do most of the work is in the parameter ization of the surface. But we already know the parameter ization of the sphere, so I just need to take half of it so I can use this premature ization. So I think about S is being fee. So that will be sign of s Kosan T. And then sign of s sign of tea and co sign of s frizzy. Okay. And now let's take some derivatives. It will take the derivative with respect to s. And that will be co sign of s go sign of tea and then will have co sign a mess again. Now sign of tea and then negative sign. Uh, s. And then our tea is going to be Well, we're going to get negative Sign yes, scientific, and then sign of s co sign of tea and then finally zero because the Z component doesn't depend on t. All right, so what is the cross product? Well, let's see. Let's find the cross product first. So our S cross our t. So we'll ignore these first, and we'll do this minus that this time to this. And so that's going to be sine squared s CO santi. Okay. And then the y component is going to be this minus this minus this minus this times this, but it's gonna be negative. So looks like we have three negatives, so it's gonna be negative because they're subtracting, But then it's going to be positive. So this should be sine squared s sign t. Then what about the Z coordinate we're gonna have? It's like coziness, Sinus cosign t and then a cosign t squared, actually. Okay, so we'll have co sign s sign us co sign square and then minus, but it's gonna be plus co sign s Okay, Sign as the same thing. But with sine squared t so that's just gonna get. Actually, give me a one here. So we're just off the coast minus. Sign us and getting us because we get the same thing, but with a coastline squared and then the same thing Plus with sine squared and there we go. So what's the magnitude here? Well, I'm gonna square both of these and Adam and I'm gonna have coastline square t Science Square t So it looks like I have just sign now to the fourth. Should be square root. It's gonna be a square root of signed to the fourth s. Okay, And then plus this squared, which is going to be cosine squared us times sine squared us. But look at this. This is sine squared as times sine squared us and this is cosine squared s times sine squared us So I have a sine squared s posco sine squared s, which gives me a one. And so I'm just left with the square root of sine squared s. Which v recall? What's my rain for Ed s? It's between zero and pie. So it's actually going to be Actually, it's between, um, zero empire for two. So I didn't actually write the range here, but let's go ahead. And because this is gonna be important because we're taking the square root. So we need If we know that this is always positive, we can just remove the square root. So this is really living in s is going from zero to pi over two because it's the hemispherical shell and then cross 0 to 2 pi. So I really can say that this is just a sign of s like that. Okay, well, that was ah lot, but that's most of the work. So what this ends up being is the integral So t goes from 0 to 2 pi s goes from zero to Piper to of okay, we have a signed s that's coming from the D Sigma, and then this is one minus z. But see is just coziness. So this is one minus coziness times sign of s. And this is gonna be d a, but D A it's just ds d t. Yeah. Okay, well, this is just a nice substitution problem. I could just let you be co sign and then I just get minus sign ds, So I'm going to get a negative sign 0 to 2 pi. I'm letting you because Sinus. So that's gonna be 1 to 0. And then this is just gonna be one minus you. Do you doing the use substitution, detainee. Okay. And so what do we get? This looks like when we get a two pi. And then this is just going to be Aiken. Switch the limit. The order of the limits here with the negative. And so I just skip a few steps. I'll have U minus. You squared over two. Evaluated from 0 to 1. That's just gonna be one half. And so the final answer is Pop, I'll let you work through the details. I know I did that kind of fast, so that's the mass of this hemispherical shell.

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