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

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

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

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

## Discussion

## Video Transcript

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

## Recommended Questions

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

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

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

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

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)

Integrate $f ( x , y , z ) = ( x + y + z ) / \left( x ^ { 2 } + y ^ { 2 } + z ^ { 2 } \right)$ over the path $\mathbf { r } ( t ) = t \mathbf { i } + t \mathbf { j } + t \mathbf { k } , 0 < a \leq t \leq b$

Integrate $f(x, y, z)=(x+y+z) /\left(x^{2}+y^{2}+z^{2}\right)$ over the path $\mathbf{r}(t)=t \mathbf{i}+t \mathbf{j}+t \mathbf{k}, 0<a \leq t \leq b$.

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

$$\begin{array}{ll}C_{1}: & \mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{j}, \quad 0 \leq t \leq 1 \\C_{2}: & \mathbf{r}(t)=\mathbf{i}+\mathbf{j}+t \mathbf{k}, \quad 0 \leq t \leq 1\end{array}$$

(GRAPH CANT COPY)

The paths of integration for Exercises 15 and 16

The paths of integration for Exercises 15 and 16

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

$$

\begin{array} { l l } { C _ { 1 } : } & { \mathbf { r } ( t ) = t \mathbf { i } + t ^ { 2 } \mathbf { j } , \quad 0 \leq t \leq 1 } \\ { C _ { 2 } : } & { \mathbf { r } ( t ) = \mathbf { i } + \mathbf { j } + t \mathbf { k } , \quad 0 \leq t \leq 1 } \end{array}

$$

The paths of integration for Exercises 15 and 16

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

$$

\begin{array} { l l } { C _ { 1 } : } & { \mathbf { r } ( t ) = t \mathbf { k } , \quad 0 \leq t \leq 1 } \\ { C _ { 2 } : } & { \mathbf { r } ( t ) = t \mathbf { j } + \mathbf { k } , \quad 0 \leq t \leq 1 } \\ { C _ { 3 } : } & { \mathbf { r } ( t ) = t \mathbf { i } + \mathbf { j } + \mathbf { k } , \quad 0 \leq t \leq 1 } \end{array}

$$

Integrate $f(x, y, z)=-\sqrt{x^{2}+z^{2}}$ over the circle

$$\mathbf{r}(t)=(a \cos t) \mathbf{j}+(a \sin t) \mathbf{k}, \quad 0 \leq t \leq 2 \pi$$

Integrate $f ( x , y , z ) = - \sqrt { x ^ { 2 } + z ^ { 2 } }$ over the circle

$$

\mathbf { r } ( t ) = ( a \cos t ) \mathbf { j } + ( a \sin t ) \mathbf { k } , \quad 0 \leq t \leq 2 \pi

$$

Integrate $f(x, y, z)=-\sqrt{x^{2}+z^{2}}$ over the circle

$$\mathbf{r}(t)=(a \cos t) \mathbf{j}+(a \sin t) \mathbf{k}, \quad 0 \leq t \leq 2 \pi$$.