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### Divergence & Curl, Continued

Find the divergence of the vector field F(x,y) = (ax,by) for constants a > 0 and b > 0.
• We define the divergence of a vector field F(x,y) = (f1(x,y),f2(x,y)) as div(F) = [(df1)/dx] + [(df2)/dy].
• Since f1(x,y) = ax then [(df1)/dx] = a. Similarly, f2(x,y) = by and so [(df2)/dy] = b.
Hence the div(F) = [(df1)/dx] + [(df2)/dy] = a + b.
Find the divergence of the vector field F(x,y) = ( √{ax + by} ,0 ) for constants a > 0 and b > 0.
• We define the divergence of a vector field F(x,y) = (f1(x,y),f2(x,y)) as div(F) = [(df1)/dx] + [(df2)/dy].
• Since f1(x,y) = √{ax + by} then [(df1)/dx] = [a/(2√{ax + by} )]. Similarly, f2(x,y) = 0 and so [(df2)/dy] = 0.
Hence the div(F) = [(df1)/dx] + [(df2)/dy] = [a/(2√{ax + by} )].
Find the divergence of the vector field F(x,y) = ( [b/(a − x2)],[a/(b − y2)] ) for constants a > 0 and b > 0.
• We define the divergence of a vector field F(x,y) = (f1(x,y),f2(x,y)) as div(F) = [(df1)/dx] + [(df2)/dy].
• Since f1(x,y) = [b/(a − x2)] then [(df1)/dx] = [2bx/(( a − x2 )2)]. Similarly, f2(x,y) = [a/(b − y2)] and so [(df2)/dy] = [2ay/(( b − y2 )2)].
Hence the div(F) = [(df1)/dx] + [(df2)/dy] = [2bx/(( a − x2 )2)] + [2ay/(( b − y2 )2)].
Find the divergence of the vector field F(x,y) = ( asin(x),bsin(y) ) for constants a > 0 and b > 0.
• We define the divergence of a vector field F(x,y) = (f1(x,y),f2(x,y)) as div(F) = [(df1)/dx] + [(df2)/dy].
• Since f1(x,y) = asin(x) then [(df1)/dx] = acos(x). Similarly, f2(x,y) = bsin(y) and so [(df2)/dy] = bcos(y).
Hence the div(F) = [(df1)/dx] + [(df2)/dy] = acos(x) + bcos(y).
Find the divergence of the vector field F(x,y) = (aexy,bexy) for constants a > 0 and b > 0.
• We define the divergence of a vector field F(x,y) = (f1(x,y),f2(x,y)) as div(F) = [(df1)/dx] + [(df2)/dy].
• Since f1(x,y) = aexy then [(df1)/dx] = ayexy. Similarly, f2(x,y) = bexy and we obtain [(df2)/dy] = bxexy.
Hence the div(F) = [(df2)/dx] − [(df1)/dy] = ayexy + bxexy = exy(ay + bx).
Evaluate the divergence of the vector field F(x,y) = ( − y2 + √{x + y} , − x2 + √{x + y} ) at (1,1).
• First we compute div(F) = [(df1)/dx] + [(df2)/dy].
• Since f1(x,y) = − y2 + √{x + y} then [(df1)/dx] = [1/(2√{x + y} )]. Similarly, f2(x,y) = − x2 + √{x + y} and we obtain [(df2)/dy] = [1/(2√{x + y} )].
• Hence the div(F) = [(df1)/dx] + [(df2)/dy] = [1/(2√{x + y} )] + [1/(2√{x + y} )] = [1/(√{x + y} )].
Thus the div(F(1,1)) = [1/(√{1 + 1} )] = [1/(√2 )].
Evaluate the divergence of the vector field F(x,y) = ( sin(x)cos(y),tan(xy) ) at ( [p/4], − [p/4] ).
• First we compute div(F) = [(df1)/dx] + [(df2)/dy].
• Since f1(x,y) = sin(x)cos(y) then [(df1)/dx] = cos(x)cos(y). Similarly, f2(x,y) = tan(xy) and we obtain [(df2)/dy] = xsec2(xy).
• Hence the div(F) = [(df1)/dx] + [(df2)/dy] = cos(x)cos(y) + xsec2(xy).
Thus the div( F( [p/4], − [p/4] ) ) = cos( [p/4] )cos( − [p/4] ) + ( [p/4] )sec2( − [(p2)/16] ) = − [1/2] + [p/4]sec2( [(p2)/16] ).
Evaluate the divergence of the vector field F(x,y) = ( [1/xy],[1/(x2y2)] ) at ( − [1/2],[1/4] ).
• First we compute div(F) = [(df1)/dx] + [(df2)/dy].
• Since f1(x,y) = [1/xy] then [(df1)/dx] = − [1/(x2y)]. Similarly, f2(x,y) = [1/(x2y2)] and we obtain [(df2)/dy] = − [2/(x2y3)].
• Hence the div(F) = [(df1)/dx] + [(df2)/dy] = − [1/(x2y)] + ( − [2/(x2y3)] ) = − [1/(x2y)] − [2/(x2y3)] = [( − y2 − 2)/(x2y3)].
Thus the div( F( − [1/2],[1/4] ) ) = [( − ( 1 \mathord/ \protect phantom 1 4 4 )2 − 2)/(( − 1 \mathord/ \protect phantom 1 2 2 )2( 1 \mathord/ \protect phantom 1 4 4 )3)] = [( − 33 \mathord/ \protect phantom 33 16 16)/(1 \mathord/ \protect phantom 1 256 256)] = − 528.
Find the net flow of the vector field F(x,y) = ( − x2 + y2,xy) over the curve C:x2 + y2 = 9 and determine if F is expanding or contracting.
• The net flow of a vector field F over the curve C is defined as ∫C F = where A is the region contained in C.
• Computing div(F) yields div(F) = [(df1)/dx] + [(df2)/dy] = − 2x + x = − x.
• Since our curve is a circle, we can compute using polar coordinates with θ ∈ [0,2π] and r ∈ [0,3]. Also − x = − rcosθ.
• Hence = ∫003− rcosθ rdrdθ = ∫003− r2cosθ drdθ .
• Integrating yields ∫003− r2cosθ drdθ = ∫0 ( − [(r3)/3]cosθ ) |03dθ = ∫0− 9cosθdθ = − 9sinθ |0 = 0.
Thus the net flow equals zero and F is neither expanding or contracting.
Find the net flow of the vector field F(x,y) = ( − √{x2 + y2} ,√{x2 + y2} ) over the curve C: r = cos2θ, − [(π)/4] ≤ θ ≤ [(π)/4] . Do not integrate.
• The net flow of a vector field F over the curve C is defined as ∫C F = where A is the region contained in C.
• Computing div(F) yields div(F) = [(df1)/dx] + [(df2)/dy] = − [x/(√{x2 + y2} )] + ( [y/(√{x2 + y2} )] ) = [( − x + y)/(√{x2 + y2} )].
• Since our curve is defined in polar coordinates, we can compute with θ ∈ [ − [(π)/4],[(π)/4] ] and r ∈ [0,cos2θ]. Also [( − x + y)/(√{x2 + y2} )] = − cosθ+ sinθ.
Hence = ∫ − π \mathord/ \protect phantom − π 4 4π\mathord/ \protect phantom π4 40cos2θ ( − cosθ+ sinθ ) rdrdθ = ∫ − π \mathord/ \protect phantom − π 4 4π\mathord/ \protect phantom π4 40cos2θ r( − cosθ+ sinθ ) drdθ .

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

### Divergence & Curl, Continued

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

• Intro 0:00
• Divergence & Curl, Continued 0:24
• Divergence Part 1
• Divergence Part 2: Right Normal Vector and Left Normal Vector
• Divergence Part 3
• Divergence Part 4
• Divergence Part 5
• Example 1

### Transcription: Divergence & Curl, Continued

Hello and welcome back to educator.com and multivariable calculus.0000

In the previous lesson, we introduced these type of derivatives of a vector field called the divergence and the curl.0004

We concentrated mostly on the curl, the circulation curl form of Green's theorem.0011

Today we are going to talk about the flux divergence, and we are going to see that it is completely analogous, it just measures something else, so let us jump on in and get started.0015

Again, given our vector field... so given f = (f1,f2), I am going to leave off the x's and y's, so again we are talking about the plane so it is x and y.0025

We said that the divergence of f is equal to df1 dx + df2 dy.0043

Now, recall that we said the divergence of f at a point is a measure of the extent to which the vector field at that point is moving away from or towards the point.0058

So, the divergence at a specific point, it measures the extent to which the vector field is going away or it is coming in... diverging, contracting... or we could say diverging, antidiverging.0109

Now, we introduced the flux divergence form of Green's theorem.0127

It looked like this... of f of c · n dt is equal to the double integral over a of the divergence of f × dy dx.0157

So, let me go ahead and... you are going to see the closed curve, however I am going to start using this notation, this Da, looks like a derivative with a capital letter -- I used it in the last lesson.0176

If this is a region a, this derivative a is actually a symbol, it is just a symbol for the boundary of a.0197

The reason we use that symbol is to establish a connection between these two... I mean we know there is a connection, but we actually want to show it symbolically.0205

So, the integral of a given vector field over the boundary of the region, is equal to the integral of the divergence of that vector field over the area enclosed by that region. That is it.0216

This is just the closed curve, the symbol for the closed curve, it is not an actual derivative. Here is where the derivative takes place.0225

Okay. Now let us go ahead and see what these integrals mean, and what these integrands mean. So, let me write that down.0234

Now, let us see -- you know what, I do not have to write that down.0241

So, now we want to talk about what the integrands mean, and what the actual integrals mean. Okay.0250

So let us go ahead and draw our curve again... like that.0256

Okay. Let me go ahead and do it over here in fact. Let me do it over here, to the left. So these is just some closed curve, this is the region a, this is the curve c, we are traversing it in the counter clockwise direction, and we said that we had a point, there is of course this vector field... this curve is passing through this vector field, we do not know which way it is.0267

Some of the points on that curve, the vector field can actually be evaluated per point on that curve, so let us say this is f, we ended up actually forming... so this is c(t), this is f(c) and the previously lesson we ended up forming c', the tangent vector.0304

Now, we found c'. Now we are going to define... we can also form 2 vectors orthogonal to c'. There are 2 vectors orthogonal to c'.0327

One of them is this direction, it points away from the region, the other one points into the region.0355

As you are traversing the curve, the tangent vector is going to be in the direction of traversal.0364

If you go to the right of it, we call that the right normal vector. This one. Okay? The right normal vector nr.0372

If you go to the left of the, as you are moving, if you go to the right so right normal vector, if you go to the left you are looking in the direction of the tangent, it is called the left normal vector.0381

So, orthogonal to c... so, let us say given c(t) = c1(t), c2(t) -- these are just the two coordinate functions of the curve in 2-space -- well, c'(t) = c1',c2'. You just take the derivative of each component function.0395

Now let us go ahead and define our left and right normal vectors.0428

Define nr(t), it is equal to c2' - c1'. This is the right normal vector.0433

I am given a particular curve, when I take the derivative of it, that is my tangent vector, that is my c'.0455

Now, I have c1, c2, if I switch the order of c1 and c2, make c2', c1, and make c1', -- let me write this down... so c' = c1' c2'.0460

If I switch the places of c1 and c2, make c2 x and c1 y, and if I negate the y, I get the right normal vector. That is how I do it. That is how I construct the vector.0482

The norm of this vector is the same as the norm of the tangent vector, because all I have done is switch places. I have literally created a vector that is orthogonal to the tangent vector.0493

Now, the left normal of t, well, again we switch c1 and c2, but this time we negate the first one, the first coordinate.0506

This is the left normal vector. Okay. We are only going to be concerned with the right normal vector, so we do not even have to worry about this, but I wanted you to see that you can define both.0521

You are talking about something orthogonal to the right, something orthogonal to the left. We are going to be concerned with what is orthogonal to the right, so it is the right normal vector that we are going to concern ourselves with.0536

Therefore, we are no longer going to be using this nr, we are just going to call it n.0544

So, we are only concerned with the right normal vector, so we will just call it n... n(t).0551

Okay. It is the vector to the right of c'(t) and perpendicular to it as you traverse c in the counterclockwise direction, keeping the region contained, keeping the region to your left.0583

Remember that was one of the things about Green's theorem. When we traverse a curve, the region that is contained by the curve has to be to the left. We are moving in the counter clockwise direction.0641

It always points away from the region... from the region enclosed.0655

So, given a particular closed curve, this is f, this is the tangent vector, that is the normal vector, this is c', this is f(c), it always points away.0670

We are traversing in this direction. That is it. Okay. So, actually let me draw it a little bit better here, because I am going to use this picture somewhat anyway.0683

So, I have got... that is f... that is c'... that is n, so this is c', this is n, this is f(c)... and this point of c(t), and we are traversing in this direction and this is the region a to our left.0700

f(c), the integrand of the line integral f(c) · n is the, again, component of f along n and again we said that f(c) · c', when we were talking about curl in some sense is the component of this vector along this.0728

You can think of it as the projection, and we say component because this c' is not a unit vector. Well, it is the same thing. I can project this vector onto this vector, the normal vector, and this f(c) · n is the component of f along the normal vector.0761

That is what it is. That is how you want to think about it. Okay. So, think of f(c) · n as the extent to which f is moving away from the point, but I am going to say moving across -- oops, we definitely do not want to do that... you know what, I am going to start on another page here, because I do not want these crazy random lines all over the place.0780

So, let us go ahead and draw our picture again, very, very important to have our pictures.0833

SO this is our f, that is our c', that is our n, this is our f(c)... c'.0839

This is c(t), we are travelling this way, so think of f(c) · n as a measure of the extent to which f is moving across the curve and away from the point and away from the point.0850

So, again, f is this f at this point, but it is also moving in that direction so it is a measure of the extent to which f is actually flowing, passing, moving away from that point.0900

I think this word "across" will make a little bit more sense when we actually integrate everything.0917

But for right now, think of it as just the extent to which it is actually moving in the direction of the normal vector.0922

That is the best way of thinking about it. Now, when we form, when we take the integral... when we add up the f(c) · n for every single point around the curve, so when we take the integral of f(c) · n dt, we get the net flux or flow.0929

That is what flux means -- it just means flow in Latin -- of f across the entire curve.0967

That is what we are doing, we are measuring this... at a certain point, the vector field is moving that way, moving that way, moving this way, maybe here it is moving that way, maybe here it is moving out, here it is moving in, here it is moving out, here it is moving in, here it is moving in, in, in, out, out.0980

When I add up all of these things, the ins the outs, I get the net flow, the net out or in across that curve. That is what this measure is, that is what this integral f(c) · n measures.1002

That is why we call it the flux integral, flux means flow, it is a measure of the net flow of the vector field going out of the region or coming into the region, or in terms of the curve passing the curve. That is it.1018

If I have some water flowing in, let us say at a bunch of points and water flowing out at a bunch of points, if I add all of those, there is going to be some measure.1032

It is going to be either positive or negative, there is going to be a net flow of water into the region, a net flow of water out of the region, or there is going to be a net flow across the boundary. There is going to be a net flow across the boundary out, that is all this is measuring.1041

Now, what about the double integral of a of the divergence of f. Okay.1059

Well, we said that the divergence of f at a point is a measure of the extent to which the vector field is expanding or contracting at that point.1073

Okay. Now, if we integrate, if we actually take... so now we are forming the divergence, we are actually forming, we are taking the divergence of all the points this time inside the closed curve, in the region, not along the curve itself.1117

Well, when we add up all of the extents to which -- you know -- the points, the vector field is either contracting or expanding, we get the net contraction or expanding into the region or away from the region.1132

That is what is going on. Let me write that down. Let me see... if we integrate, you know this one I am actually going to move to another page here, because I know that the crazy lines are going to start again.1148

If we integrate, that is... add up -- that is what integration is, it is always good to remember what you are doing when you are integrating, you are adding things up, you are making an infinite addition -- add up all the divergences for every point in the region enclosed by the curve, we get the net expansion or contraction of the vector field over the region.1162

Green's theorem says that these two things are equivalent. Okay. So if I take the line integral, if I do the integral of f(c) · n, I am measuring... let me go ahead and draw my quick region here, remember?1239

If I have a bunch of outs and ins, I am measuring the extent... the net flow of the vector field across the boundary of that region.1258

Well, instead of doing the line integral, instead of measuring that along the line integral... I can go ahead and calculate the divergence of all of the points inside the region that is enclosed by that curve.1267

I can add all of those up and because the divergence is a measure of the extent to which its vector field is moving away or coming into a point, when I add all of those up, it is going to give me the net measure of the extent to which the vector field is moving away from the region or moving into the region.1280

Green's theorem tells me that those 2 numbers are equal.1300

So, Green's theorem, the flux divergence form, Green's theorem tells me these two values are equal. Hence, flux divergence.1306

The flux integral is a line integral, the divergence integral is an area integral. Green's theorem tells me that these things are the same.1337

So, the integral of a vector field, the integral of -- well, let me write it out first and then I will say it out loud -- so, the divergence of f dy dx, the flux of a vector field across a closed curve is equal to the integral of the divergence of that vector field over the area enclosed by that curve.1345

Let me say that again. The flux of a vector field across a closed curve is equal to the integral of the divergence of that vector field over the area enclosed by that curve. This is called the divergence theorem.1376

You will see it referred to as that. It is just the flux divergence of Green's theorem. Okay. It is not a different theorem, it is the same thing. It is equivalent, but it just measures something else.1393

You will see this referred to as the divergence theorem, but it is still just Green's theorem.1405

So, you will have a circulation curl, we have the flux divergence, they measure different things.1416

Okay. Let us go ahead and do an example. We will let f equal to... we will use the same vector field as before... so we have cos(xy), and we have sin(x2y), actually kind of a complicated vector field.1421

We have c(t) which is as before, we will right the parameterized version which is 4cos(t), and 2sin(t).1450

Or we will write it in Cartesian, x2 + 4y2 = 16 is the ellipse centered at the origin, major focal radius 4, minor focal radius 2.1458

So, we are going to go ahead and calculate both integrals. We are going to calculate the line integral, we are going to calculate the double integral.1473

Let me go ahead and start this over here. Let me just draw out our particular region, so we remember what we are talking about here.1482

So, we have this ellipse, this is -4, this is 4, this is 2, this is -2, so we are going to go ahead and calculate the line integral and we are going to integrate in the counterclockwise direction, keeping the region to our left, and then we will go ahead and calculate the double integral.1490

Okay. So, we want to do our first one. We want to calculate the integral over the boundary of f(c) · n dt.1507

Well, f(c), that is going to equal -- let me do this a little bit below so that we have as much room as possible -- so, let us see... I have got the f(c) part, I am just going to put c into f.1520

It is going to equal cos(8) cos(t) sin(t), and I am going to get sin(32), cos2(t), sin(t).1540

Again, I hope that you will confirm this for me, and if I make a mistake it should not be a problem. Just make sure -- you know, go with what you think it is if in fact I have made a mistake, and just put it into your math software and you will get the particular answer.1557

Okay, well we need c'(t), right? f(c), we need c'(t), so that we can form n, and then we can take the dot product.1572

So, c'(t), that is equal to -4sin(t), and 2cos(t), right? So now we form n.1582

Well, n(t), we switch these two and we negate the second. So, this equals 2cos(t) and this becomes -4sin(t), - (-4sin(t)), so it becomes 4sin(t).1592

Now let us go to blue, now I have my f(c), that is this one. I have my n(t), that is this one. Now I take the dot product of those 2.1610

So, f(c), that is it, that is all I am doing. Just plugging things in. f(c) · n = ... it is a little long... 2cos(t)cos(t)cos(t)sin(t) + 4sin(t) × sin(32)cos2(t)sin(t).1621

It is a little long. i do not want to keep writing this. I am just going to keep referring to this as z.1657

Therefore, the integral around the closed curve of f(c) · n dt, in other words, the flux of this vector field across this closed curve is going to equal the integral from 0 to 2pi of this z dt.1662

When I put it into my math software, I get the number 0.5466. There we go, it is just a number.1683

It is a negative number. Negative divergence... negative flux means that the vector field is actually not moving away from, it is not moving across the curve outward, it is moving inward.1693

In other words, it is this way. The vector field is actually contracting, fluid is flowing into the region. Fluid is not flowing out of the region. That is what this is telling me, for this particular vector field.1708

If this particular vector field happened to represent fluid flow across this ellipse.1725

So, now let us go ahead and do the double integral just to make sure that you do actually end up getting the same number and that Green's theorem is correct... it is, but you know it is good practice because we need to actually solve the double integral.1731

So, now the double integral -- I really like this blue ink, it is nice -- double integral, okay.1743

So, we want to solve a divergence of f dy dx.1757

Well, we calculated the divergence of f, so let us just recall what that was and if we do not recall, let us just go ahead and do it again. We have our vector field.1767

It is -y × sin(xy) + x2 × cos(x2y). Okay.1777

Now, let us go ahead and put it in. So, the double integral over a of the divergence of f dy dx.1789

Our x is going from -4 to +4.1808

Our y is moving from -16 - x2 under the radical, divided by 2 all the way up to sqrt(16) - x2/2, the positive. The lower... that is the upper. That is what y is doing.1816

Our divergence is -y × sin(xy) + x2 × cos(x2y) dy dx.1835

When I go ahead and put this into my math software, guess what number I get? 0.5466, negative 0.5466.1856

This confirms that the flux integral is equal to the divergence integral. It confirms the fact that this particular vector field is actually... it has a negative divergence. It is contracting. Fluid is flowing, there is a net flow. This is a net, remember? We are integrating.1866

At a different point, at different points inside there, you might have a positive divergence but when you add up everything, the net flow is into the region.1885

Okay. Let us go ahead and write this out. Since -0.5466 is negative, this means that there is a net flow of the vector field into the region.1896

In other words, if we have our... the vector field as a whole is contracting. Okay. Now, in the previous lesson, we found that the curl of this particular vector field, the curl of f was equal to 0.1934

So, this particular vector field is contracting into the region but it is not rotating left or right.1964

This is really, really powerful, that we can make this kind of statement about a vector field given a particular closed region. Really, really extraordinary.1972

Thank you for joining us here at educator.com for divergence and curl. We will see you next time. Bye-bye.1982