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Problem 14 Medium Difficulty

Find the line integral of $f(x, y, z)=\sqrt{3} /\left(x^{2}+y^{2}+z^{2}\right)$ over the curve $\mathbf{r}(t)=t \mathbf{i}+t \mathbf{j}+t \mathbf{k}, 1 \leq t \leq \infty$.

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Okay, folks. So in this video, we're gonna take a look at this. Um, this integration here, we have a line integral. This is problem number 15. We have aligned to grow that we need to do. We have. First of all, we have the function f that we want to integrate against X y Z. Um, so that's equal to X plus route. Why? Mine is Z squared. So this is our function and were given to space cars. The 1st 1 is see, one in the 2nd 1 is C two, Um, And because of the fact that as you can probably see from the graph C one and C two, the the point where the joint is not to continue is not differential. There's a the derivative. The has kind of a jump, this continuity there at at the 0.110 which is where C one C two meets. So we can't just do one thing to grow. We have to do to winter growth, and then we're gonna add them to get together. So we're gonna integrate the function f on the curve C one first, and then we're gonna add that to the integral of the function F on along the curve. C two. Okay, so we're gonna have 200 girls to do instead of just one so that to do it. So for the first, see one, we have the space curve. R is equal to t i, which is I'm gonna write I as x had, because that's the same thing. Plus t squared in the white hat direction and then t is from 0 to 1. Okay, so that's for C one. What about C two Will see two is just one ex head plus one J hat. Plus, let's see what this is. Oh, plus t. Um does he had okay, And then the the limits of integration is also from zero all the way upto one. Okay, I realized I missed something here. I missed the tea. All right, so now we have given we have been given the to space curves that we won't integrate along. So that so the first thing you brought us do. Uh um, that's do the integral for C one. And then we have C one F as a function of x, y and Z multiply by the S um, plus C two of f a za function of x y and Z multiplied by the Yes, we're going to calculate the the these two integral separately and then we're gonna add them together. So that's our strategy for over this one we have now I'm gonna rewrite. What I'm gonna do is I'm gonna rewrite X, y and Z all of him as a function of team. So as you can see alone on the curve, C one x as a function of T If you look here is just tea. Okay, so we have tea plus the square root of T squared, which is going to give you, you know, which is going to give you the absolute value of tea. But because of the fact that we're restricting ourselves, Teoh only the points where t is bigger than zero. So the actual value of tea in this range is really just t itself. That's kind of a triviality that you don't need to pay too much attention to. So we have square root of T square. It is going to give you tea and then Z is zero. So I'm gonna leave this term alone. Um, multiply it against a D s. But what's DS? Well, the s is really just a, you know, ex prime squared. Plus why Prime Squared plus z prime squared multiplied by D t. And then what this is going to give you is, um that's calculated Prime. I mean, ex prime Well, excess teeny. So ex prime is 11 square. Desist one. Uh, plus Well, why is a function of teas two square? So, Teoh, the derivative of that is gonna give you to tea squared. Plus, there's no Z. So So the last party zero multiplied by DT. Okay, so this is our first curve. Um, and what we're gonna do, let's figure out what the limits of integration is. Well, we're going from zero 21 OK, so that's for the 1st 1 Let's see what this is. Um, so we're givens. We have this two t here multiplied by eternity squared plus one. Um, d t from zero 21 I'm going to write the to t squared as as four t squared. And then if we evaluate this integral here, we're gonna do Ah, we're gonna do a U substitution. We're going to say that that 40 squared plus one. I'm going to find that as a variable. You therefore d'you is he would 80 duty. Um so now we have this integral as, um what's to t? Well, to t is just If you divide both sides by four, you get to t d t. Which is what we have here. So to tt tt you over for multiplied by the square root of you but the square root of you I'm going to write that as u to the power of 1/2. And that is going to give you, um, you 3/2 over three. Half evaluated, um, on the two. Ah, the two limits. The first limit is when t zero, what went t zero use one, and the second limit is went to use one. Well, when tea is one, plug this into here. You get four times one plus one, which is just five. Okay, there was my dog. Okay, so now we have Let's do Ah, algebra. Here. We have 1/4 times. 2/3. Um, that's our constant. Five to the power three. Half minus one. Okay, so we had This is this is just 1/6 5 3/2 minus one. OK, but that, as you remember, is really just the first part too far. Ah, of our journey. The second part is this thing. Which, which, which I'm going to treat in a very similar manner. Tom. So we have this thing is equal to, um it's it's still t plus tty, but and it's still, um the S. But But D s here is a little bit different. Well, actually, excuse me, this is not true, because for for this, integral were integrating along a different different curve than the C one. So our from our parameter ization is gonna be different. Um, we're gonna focus on C two here, but for C two, they're all constants for C two. X and Y are all constants. They're all one, okay. And the only thing that's not a constant Z component that Z z as a function of t is gonna be t and X as a function of T is, he would've why the function of T is equal to one. So let's plug these functions back into F and see what we get. Well, f now becomes X, which is one plus the square root of why? Which is one and then minus Z squared. But Zia's team So we have minus t squared. Okay. And what about um what about the Yes? Well, d s, as you remember, is just ex prime squared. Plus why Prime Squared plus z prime squared multiplied by D t. But because X wire all constant, these two terms become zero. We're left with the last term, which is see Prime Square. But Z is t so the prime is one. So we have this whole thing. Square root becomes one. That's very nice. Multiplied by DT No, let's do this. Let's have What's this? This is two minus two squared. Do t um, from 0 to 1. So now we can evaluate this thing. We have to t minus t a cubed over three from zero the ones when we get to minus 1/3 minus zero. And that's really just Ah, 5/3. Because 6/3 U minus 1/3 is five. Number three. Now we have successfully evaluated both of our inter girls. We're gonna add them up. We have 1/6 of 5 3/2 minus one close I have over three. And this is gonna be the result of our integration. And we're done for this video. It's Ah, it's much longer than I expected Purple, because this problem is a little bit more complicated than it usually is. But yeah, that's it for this video. Thank you. Bye bye.