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# Evaluate $\int_{C}(x-y+z-2) d s,$ where $C$ is the straight-line segment $x=t, y=(1-t), z=1,$ from (0,1,1) to (1,0,1).

Integrals

Vectors

Vector Functions

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### Video Transcript

Okay, folks. So in this video, we have this line into grow that we need to evaluate. We have this integral along see of X minus y plus Z minus two. And where we're doing the integral with respect to s the variable s. And, um so the way we're gonna do this is we're gonna as usual when you convert this DS into something with respect to T. If you remember, the US is really just the x d t squared plus d y d t squared plus d c g t squared multiplied by d t outside of the square root. And so now we're gonna evaluate X and Y and Z evaluate the derivative with respect to time, So excessive function of tea is just tea. So so ex prime of tea is one and ex prime. I mean, why prime the tea is minus one, because why is one minus? T and Z is a constant. So when you take the derivative of the of a constant with respect to a variable, we get zero z prime of t zero. So we're gonna cross this out because it's zero. Now we put this in here. We get the s equals. Um, square root of one squared, which is one and a negative. One squared is also one multiplied by DT. So we have route to DT. So we're gonna plug this back into the integral, which is right here, and so are integral becomes X as a function of time is t minus. Why? Why is one minus t plus Zied C is a constant C is one and then minus two. So this is the Inter Grand. The DS becomes now routes to DT. All right, that's good. So we're gonna evaluate what this whole thing is. So it's, um t minus one plus T plus one minus two route to DT. Okay, so now I'm going to do, um What's this? Minus one plus one. Okay, so we have to t minus two. So two of T minus one, right to t minus two easters, two of T minus one times route to. So we have to root two t minus one. DT. Okay, so this is a constant. We're gonna pull the constant out. So we have to root two of of the inter grow of T minus one D t, which is gonna be to root two of what this t squared over two minus t. Um, evaluated are the two limits. What's the first limit? Well, the first limit is when x y Z is 011 And the second limit is when x y Z is 101 Well, when x at the lower limit, X zero when x zero, t zero. Okay, so we have t equals zero. What about the second? The upper limit? Well, the upper limit says that. Why is zero right? What? When? Why is zero that means one minus t. Because why equals one minus t zero? That means tea is one. Okay, so let's plug in the two limits of integration. We have to root two of 1/2 minus one. That's the 1st 1 minus zero minus zero. Okay, so we have this minus 1/2. This is gone because zero, um, this, too. And this to cancel out minus route to is where we have left. All right. I think that's it for this video. Thank you very much. Let me raise this arrow for you. I wanted to draw a narrow, but I failed. Okay, so we have negative route to That's it for this video. Thank you.

University of California, Berkeley

Integrals

Vectors

Vector Functions

Lectures

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