WEBVTT
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so you may be asked sometimes to integrate a line
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in a girl. And a lot times you'll see
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that written as even three variables and then potentially be
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given like line segment that you're going to find that
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medical over some sort of bound for the problem so
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you could consider a question and think about. Maybe
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the integral itself starts with three variables with X,
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Y and Z. Maybe it's written like X plus
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, the square root of why minus z squared you'll
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be given then abound on it, like from the
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origin to some sort of point. Remember, the
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origin is 000 and then let's say we're going to
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111 so x Y Z coordinates there. So now
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let's consider some line segments we can integrate this across
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. So let's consider the first line segment. Let's
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make it our F T is T K. Let's
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do another line segment. R T is T J
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plus K, and then let's do 1/3 line segment
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like T I Class J Plus K. Now you
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would want to make sure these were bounded on the
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same interval, so like we would want t to
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be bounded, going from 0 to 1 and then
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I actually have enough information now to try and set
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up some integration. So to integrate, I will
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create each of these as a parameter for the function
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at I'm gonna change it into terms of tea and
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then that gives me really just one variable to focus
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on for integration instead of the three variables that I
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originally started with. So to make that change,
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remember that you're going to start with some sort of
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in a girl of X y Z, and you
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can change that parameter into a new integral in terms
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of hopefully some sort of function of t so that
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they're all the same function. They will have the
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same variable that is, they don't see the same
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function. And then when you're doing that, you
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need to use a velocity vector. So you wanna
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also have a V of tea in here, and
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that's gonna help you change your variable for reference.
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V F. T comes from the derivatives of each
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of your terms. So as we go through,
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we're going to see this played out in this problem
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. But you're using the derivative of acts with respect
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to T. You're using the derivative of why with
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respect to t. And then you're using the derivative
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Z with respect to teams. So those three things
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based on your parameters are going to help us define
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that velocity vector that we need as well. So
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let's go ahead and fill in our line segments.
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So I'm going to add together are three line segments
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. So our three line segments we have are of
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t is T. K. Remember that if you're
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thinking about the vector you're thinking about I j k
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set up here So when it says T k,
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it's actually telling you're just filling in t for the
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last term for that K term. In this case
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, that's dizzy. So I really just want to
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replace Z. Since I don't see anything with an
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eye on that 1st 1 I don't see anything with
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a J. Then I'm not filling in anything.
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I have zero for both X and y, and
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then I need to figure out my velocity vector so
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that velocity vector comes from again the derivatives. So
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the derivatives here would be, well, the derivative
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of zero for the first term. Well, derivative
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of zero is zero derivative of the Y term zero
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and then the derivative of the Z term is one
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. So that velocity vector, I hope we can
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all agree becomes just one under the square root There
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it's a just a one. And then I know
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I'm gonna go from 0 to 1 for my bounds
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cause now I've put it in terms of tea is
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my variable. So I know the limits of integration
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Here are is your own one. Plus, I
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want to do the next line segment, so I'm
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changing the parameters in terms of tea. Now I
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have a term for J in cases still no X
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term, right? No, I part of that
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vector. But I have tea for the y.
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And I have one for the easy because there would
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be a one in front of this K term for
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that second line segment. If I were to do
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that Velocity vector just to save a moment of time
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again, actually get ah one because your constant goes
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to zero. Do you have the square root of
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zero squared plus one squared plus zero squared. Since
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this Z term would have a derivative of C grow
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so still velocity Vector one. And then let's do
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our last line segment. We have I, J
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K all included, and to fill it in and
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order, I have tea for the X one in
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front of the J one in front of the K
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as I do that velocity vector for derivatives derivatives to
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be 10 and zero. So one more time I
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have a square root of one on this question and
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then DT Now that we've written out the function for
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in terms of the t parameter, we're going to
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actually fill these into the equation. So for this
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1st 1 as in the example as I fill in
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00 t into the equation, that would mean I
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have zero plus the square root of zero minus t's
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squared. I'm filling in the X and the Y
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in the sea. But really, I don't need
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to write those zero terms. So we take those
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away and just call it minus T squared and then
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DT plus on the 2nd 1 If I fill in
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zero as X t is the y and then one
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as the Z, I would have a square root
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of T minus one squared coming from that X Y
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and Z term. And then on the last integral
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. If I fill in t one and one as
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X, y and Z, I'm gonna have t
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plus the square root of one minus one squared and
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one minus one that's gonna cancel out for us there
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. So, really, this last integral is just
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the integral of tea with respect to teams. So
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now we can take some anti derivatives and evaluate and
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we're nearly at that final answer. So the anti
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derivatives become negative 1/3 t cubed, going from 0
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to 1. Plus, this is 2/3 t to
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the three halfs minus t going from 0 to 1
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. And then lastly, I have 1/2 a T
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squared, going from 0 to 1. If we
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evaluate these limits of the integration plugging in an upper
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limit minus lower limit, I get negative 1/3 for
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the first actually get negative 1/3 again for the second
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. And then on the last one, I get
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a 1/2 adding those altogether. My final answer is
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a negative 16 So that would be the integration of
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f of X, y and Z over these particular
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three line segments, given