Integrate $f(x, y, z)=x+\sqrt{y}-z^{2}$ over the path $C_{1}$ followed by $C_{2}$ followed by $C_{3}$ from (0,0,0) to (1,1,1) (see accompanying figure) given by

$$C_{1}: \quad \mathbf{r}(t)=t \mathbf{k}, \quad 0 \leq t \leq 1

$$$C_{2}: \quad \mathbf{r}(t)=t \mathbf{j}+\mathbf{k}, \quad 0 \leq t \leq 1$$$C_{3}: \quad \mathbf{r}(t)=t \mathbf{i}+\mathbf{j}+\mathbf{k}, \quad 0 \leq t \leq 1$$

(GRAPH CANT COPY)

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Johns Hopkins University

Harvey Mudd College

University of Michigan - Ann Arbor

Boston College

Okay, folks. So in this video, we're gonna take a look at this this line into growth. This is problem number 16 on your book. We had this function, which is a function of position is given by this F is equal to X plus route. Why minus disease squared. And we're given three separate parents that we would like to integrate over and some of them all up, Um, she wanted C two and C three. So the way we're gonna do this is just the usual. The usual way of doing lined girls, which is by, uh, by writing. Well, okay, so that we want to do this is by doing it separately. Uh, were first gonna be integrating every c one. That's that's one of the inter girls. And then we're gonna do see, too. So that's another integral. And then and we're gonna do see three. So that's our last underground. Then we're gonna add them all up, so that's gonna be are finally answer. Okay, What about? So So let's do see one first. What? What is the first integral? Well, the first integral is gonna be f uh, multiplied by. Yes, of course. I'm sure you all know that DS is the infinite testament line segment, um, and say messy too, and same as C three. So I was gonna write them all out F d s. OK, so that's our, uh, total expression. That's Ah, let's treat them separately. So I'm going to write the first integral as f ah ah, square root of the D X day t squared plus de y e t squared plus d z d t squared. Okay, I'm sure you all know that. And then, of course, when a multiply by D T and same as C two and C three. So when apply this thing or in every we're gonna redo this thing three times, basically for C one and C two and C three. So for each of the of each of the paths, we're gonna figure out what the X DT is and what he why did he is and what the DDT is. And then we're just gonna plug and chug back them back into this expression, and then we can evaluate and crank out the answer stats. That's our That's our strategy. Okay, so let's, um let's do it. So that's Ah that's first. Take a look at the first path and figure out what the what these little derivatives are. Well, if you look at the first path right here, you see that the X and Y are both zero. Them is the derivatives with respect to T or both zero. Um, and the derivative of the Z variable with respect to T is simply just a d t d t which is one so so for C one, we have ah, integral. See, one of f multiplied by the square root of of DC DT is one. So one squared is still one right as times one times d t So now we can do it. We can We can evaluate this expression, which is Ah, uh f is X plus route by minus B squared. But X and Y are both zero summonses. Ignore those two terms. Um um And so now I only have minus Z squared, but Z issue with a TV, you could look here and see. It's just equal t. So minus Z squared is the same thing as minus t squared. So I'm gonna write minus t squared right here, multiplied by one. I'm not. I'm not gonna keep writing that one because his pointless multiplied by DT. Okay, um, and then the limits of integration is from 0 to 1. Okay, so that's see one. That's very easy. Um, what about C two c to do the same thing over and over again? See, to is, um Ah. Well, if you look here, X zero and Z is one. And why is square root from skirt of why? But why is t so square? Good tea. Okay. So I'm going to simplify the expression for the function f as I'm the simplify it as zero, which I'm going to Ah, ignore plus plus route. Why? But why is t so I'm plus root tea minus z square. But Z is one so minus one. So that's our function f and then multiplied by the square root the extra t zero d Y d t is one thes e d t zero That simple. I think I'm gonna let me You have. You have to allow me to raise this thing because I'm not gonna keep writing one's everywhere. It's pointless. A multiplied by D t. And then this is gonna be simplified into from 0 to 1. Rooty minus one D t. Don't worry. We're gonna crank out all of the final numbers later on before now, let's Ah, let's right out into girls first for C three. We're gonna do the same thing again. See three ah, along C three X's t so t plus route. Why? Why is one route why is one minus z squared? But see is one So Z squared is still one and then multiplied by the square root The X DT is one and then d y t t and d zd tear both zero. Right, um, and then multiplied by DT. So that's simplified this a little bit more. So we're integrating from 0 to 1 of t plus one minus +10 So we have T d t. OK, so we have three of these things. That's very nice. Um, we're gonna crank out the answers, and they want to add them all up. That's gonna be our final answer. And let's see what that is. We have minus t cubed over three, uh, from 0 to 1, which is gonna give you minus 1/3. Okay. Um, what about this one? This one is a little bit trickier, because rooty is simply just want t to the power of 1/2. So I'm gonna do the usual integral t of you know, what's 1/2? Plus one is 3/2 over 3/2 and then minus T from 0 to 1 would just give you, uh, 1/3, half minus one, which is just gonna be two or three in minus one, which is minus 1/3. Okay, that's the second, Inderal. What about the 3rd 1? While the 3rd 1 is very simple, it looks like a so we have t squared over two from 0 to 1 of, uh, that's just gonna be 1/2. Okay, lets Adam all up. We have minus 1/3, minus 1/3. Plus one has. That's very simple. That's just minus 23 plus one tap, which is ah, six to the power of three minus four. So that's minus 1/6. All right, so that's our Ah, that is our final answer of how exciting. All right, that's it for this video. Thank you very much. And, um, let's see, that's all there is about this video. That's Oh, yeah, that's it. That's it for this video. Thank you very much. See the next video, But

University of California, Berkeley