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# Find the line integral of $f(x, y, z)=x+y+z$ over the straightline segment from (1,2,3) to (0,-1,1).

## $3 \sqrt{14}$

Integrals

Vectors

Vector Functions

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##### Lily A.

Johns Hopkins University

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Michael J.

Idaho State University

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### Video Transcript

Okay, folks. So in this video, we're going to take a look at this. Line it to grow. This is problem number 13. We have this function F which is a function of X, y and Z, and it's equal to the sum of the three variables X, y and Z, and we're integrating from 1 to 3. Um, via a straight line segment 20 negative one and one. So, in other words, we want the line didn't grow from 1 to 3, 20 negative 11 of dysfunction of this function X, y and Z multiplied by the usual D s, which is an infinite testament line segment. So unfortunately, we're not given the privatization, so we're gonna have to find the parameters Asian theirselves. So the way we're going to find the privatisation if x y and Z with respect to t is by using the condition um, I hope you remember this condition x minus x zero of ex fine on minus X zero equals why minus y zero of why final minus y zero is equal to Z E minus Z zero z final minus Z zero. So we have this condition. We have this condition right here that every point on a line segment needs to satisfy. So this is this is this is ah, set of three expressions that are all equal to each other, and we're gonna use them to obtain our parameter ization. By that, I mean, I'm gonna set all of them t equals t. So, um so from this, we can get three separate equations for X and Y and Z each as a function of time. So let's do it. So, for for X we're gonna use this is equal to T. We're gonna use this to get what's x zero. While x zero is just this number right here That's actually wrote X minus one. So we have What about X Final while X finally zero and ex My ex zero is one again. That is equal to t. So, as you can see, X minus one must be equal to negative t. Therefore, we have X equals one minus t. So we have obtained our first function, which is a which is X, which is a function of time. OK, so that's X. Let's do the same thing for why we have why mine is Why not? Why not is to a zoo you can see here. So why minus two over. Why? Find on which is negative. One negative one minus y two is just Excuse me. Negative one minus two. That's equal to T. So we get why minus two over mine of three equal t. So that means that we have Why equals minus three t plus two. Okay, um so we have obtained our function for Why? What about Z while Z we're gonna do the same thing again. So Z minus the 00 is three, as you can see above, Um, the final is one and then one minus three diocese with a T. So from this, we're gonna get Z minus three over minus two. Is you with a T? Okay, so we have this disease looking like a two here. Okay, so we have Z equal three minus two t. So, as you can see, we have obtained all three of our functions. Now, what we can do is we can just plug it in using the usual, You know, the way you the way you do a line into grows by first parameter rising it and then rewriting the d s as a function of from of tea. The way you write that is, by first of all, you say that DS is equal to ah ex prime squared Plus why Prime Squared pleasant Z primes squared the whole thing multiplied by t. Um, and then ex prime is just the driven at the time derivative of the function X. So, as you can see here, um, the time and derivative of this function, it's just a negative one. So negative one square to just one. So we have one. Plus, we're gonna do the same thing for why you take the derivative of why, with respect to you get negative three. So So you square that you get nine plus plus four okay, multiplied by D t. And then when you add these these three things up 10 plus four, just 14 d t. Which is, um, two times seven. You can't do anything about three times seven inside of the square root. So I'm gonna I'm gonna leave this right here, okay? I'm not I'm not gonna do anything about it. Um, so we have DS as an expression of DT, and then now What we're gonna do is we're gonna plug it back into this line integral here. So we have what's African Office X plus Y plus Z, But now we're not gonna write X wines. You were going to write all of them as a function of t. So we have one minus t. That's for X Plus wise in the to minus three t. Okay, um, and Z is gonna be three minus two t. So we have these three guys multiplied by the Yes, but the S is just a constant times d t square root of 14 times DT. Let me copy this down. Should I have more room running on a room right now? Okay, so we have we have this thing. Let's simplify this a little bit. So we have one plus two plus three is to six. Um, and then minus T minus three t in the minus two TSX minus 60 Route 14 d t. So we have six. Route 14 one minus t. Did he? I'm just pulling all the count, all of the constants out of the integral, because I could do that. And then now what we're gonna do is we're gonna just get rid of it. Every the whole integral. Now we have t minus one. Have to You squared. Um, what about the limits of integration while we can calculate the limits of integration by by the usual trick Where you, uh, where you were? You look at this. 133 And you and you try to figure out what tea is. Well, what is t t is even what? Well, well, as you know, X first of all, we find X as a function of T excess. One minus t x is one minus t Well, when x is one at the origin, that means in T is equal to two. Actually, there's no true. That's not true. That means tease you with zero. Okay, so that's the beginning. Point the point where we start. And then as for the end limit, which is right here, so we have one minus t, which is x x zero. That means that t is one. So in other words, were integrating against of the variable t from 0 to 1. And that's a very simple ah integral, weaken, weaken. Do so we have from 0 to 1 0 to 1 six, route 14 one minus 1/2 and then minus zero. I'm not gonna write the zero terms out. Um, so now we have this thing is just 1/2 1 half time, six of three. So three, Route 14 is the answer to this interval. And that's it for this video. It went a little bit longer than I expected, but that's fine. Um, thank you.

University of California, Berkeley

#### Topics

Integrals

Vectors

Vector Functions

##### Lily A.

Johns Hopkins University

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Michael J.

Idaho State University

Lectures

Join Bootcamp