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Lecture Comments (4)

1 answer

Last reply by: Professor Hovasapian
Wed May 11, 2016 3:35 AM

Post by Tram T on May 7 at 09:18:10 PM

In example 2 last minute, for j^ term, i think it should be -yz^4 + x^2y.

1 answer

Last reply by: Professor Hovasapian
Sun Jan 20, 2013 10:19 PM

Post by mateusz marciniak on January 19, 2013

professor in example 1 why did you not distribute the (-) sign to e^x yz

Divergence & Curl in 3-Space

Find the divergence of the vector field F = (2x + y,x + y + z,x − 2y).
  • The divergence of a vector field F(x,y,z) = (f1(x,y,z),f2(x,y,z),f3(x,y,z)) is div(F) = [(df1)/dx] + [(df2)/dy] + [(df3)/dz]
  • For f1 = 2x + y, f2 = x + y + z and f3 = x − 2y we have [(df1)/dx] = 2, [(df2)/dy] = 1 and [(df3)/dz] = 0.
Hence div(F) = 2 + 1 + 0 = 3
Find the divergence of the vector field F = (cos(x),sin(x),tan(x + y + z)).
  • The divergence of a vector field F(x,y,z) = (f1(x,y,z),f2(x,y,z),f3(x,y,z)) is div(F) = [(df1)/dx] + [(df2)/dy] + [(df3)/dz]
  • For f1 = cos(x), f2 = sin(x) and f3 = tan(x + y + z) we have [(df1)/dx] = − sin(x), [(df2)/dy] = 0 and [(df3)/dz] = sec2(x + y + z).
Hence div(F) = − sin(x) + sec2(x + y + z)
Find the curl of the vector field F = ( − y2 + z,x − y + z, − x2 + z).
  • The curl of a vector field F(x,y,z) = (f1(x,y,z),f2(x,y,z),f3(x,y,z)) is curl(F) = ( [(df3)/dy] − [(df2)/dz],[(df1)/dz] − [(df3)/dx],[(df2)/dx] − [(df1)/dy] )
  • For f1 = − y2 + z,, f2 = x − y + z and f3 = − x2 + z we have curl(F) = ( − 2 − 1,0 − 1,1 − 1 ).
Simplifying yields div(F) = ( − 3, − 1,0)
Find the curl of the vector field F = (e − xyz,exyz,exy).
  • The curl of a vector field F(x,y,z) = (f1(x,y,z),f2(x,y,z),f3(x,y,z)) is curl(F) = ( [(df3)/dy] − [(df2)/dz],[(df1)/dz] − [(df3)/dx],[(df2)/dx] − [(df1)/dy] )
For f1 = e − xyz, f2 = exyz and f3 = exy we have curl(F) = ( xexy − xyexyz, − xye − xyz − yexy,yzexyz + xze − xyz ).
Find ∇×F and ∇×F of the vector field F = ( 1 − [(x2)/4],1 − [(y2)/4],z2 ).
  • Note that ∇ is the operator assigning ( [d/dx],[d/dy],[d/dz] ), so that ∇×F corresponds to the divergence and ∇×F corresponds to the curl of the vector field F.
  • Then ∇×F = ( [d/dx],[d/dy],[d/dz] ) ×( 1 − [(x2)/4],1 − [(y2)/4],z2 ) = [d/dx]( 1 − [(x2)/4] ) + [d/dy]( 1 − [(y2)/4] ) + [d/dz]( z2 ) = − [x/2] − [y/2] + 2z
And ∇×F = (
i
j
k
[d/dx]
[d/dy]
[d/dz]
1 − [(x2)/4]
1 − [(y2)/4]
z2
) = (0 − 0)i − (0 − 0)j + (0 − 0)k or (0,0,0)
Find ∇×F and ∇×F of the vector field F = ( √{xy} ,√{xz} ,√{xy} ).
  • Note that ∇ is the operator assigning ( [d/dx],[d/dy],[d/dz] ), so that ∇×F corresponds to the divergence and ∇×F corresponds to the curl of the vector field F.
  • Then ∇×F = ( [d/dx],[d/dy],[d/dz] ) ×( √{xy} ,√{xz} ,√{xy} ) = [d/dx]( √{xy} ) + [d/dy]( √{xz} ) + [d/dz]( √{xy} ) = [y/(2√{xy} )] + 0 + 0 = [y/(2√{xy} )]
And ∇×F = (
i
j
k
[d/dx]
[d/dy]
[d/dz]
√{xy}
√{xz}
√{xy}
) = ( [x/(2√{xy} )] − [x/(2√{xz} )] )i − ( [y/(2√{xy} )] − 0 )j + ( [z/(2√{xy} )] − [x/(2√{xz} )] )k or [1/(2√x )]( [x/(√y )] − [x/(√z )], − √y ,[z/(√y )] − [x/(√z )] )
Find ∇×F and ∇×F of the vector field F = ( [1/yz],[1/xz],[1/xy] ).
  • Note that ∇ is the operator assigning ( [d/dx],[d/dy],[d/dz] ), so that ∇×F corresponds to the divergence and ∇×F corresponds to the curl of the vector field F.
  • Then ∇×F = ( [d/dx],[d/dy],[d/dz] ) ×( [1/yz],[1/xz],[1/xy] ) = [d/dx]( [1/yz] ) + [d/dy]( [1/xz] ) + [d/dz]( [1/xy] ) = 0 + 0 + 0 = 0
And ∇×F = (
i
j
k
[d/dx]
[d/dy]
[d/dz]
[1/xy]
[1/xz]
[1/xy]
) = ( − [1/(xy2)] + [1/(xz2)] )i − ( − [1/(x2y)] − 0 )j + ( − [1/(x2z)] + [1/(xy2)] )k or [1/x]( − [1/(y2)] + [1/(z2)],[1/xy], − [1/xz] + [1/(y2)] )
Find ∇×F and ∇×F of the vector field F = ( xcos(y)ez,ycos(x)ez,z ).
  • Note that ∇ is the operator assigning ( [d/dx],[d/dy],[d/dz] ), so that ∇×F corresponds to the divergence and ∇×F corresponds to the curl of the vector field F.
  • Then ∇×F = ( [d/dx],[d/dy],[d/dz] ) ×( xcos(y)ez,ycos(x)ez,z ) = [d/dx]( xcos(y)ez ) + [d/dy]( ycos(x)ez ) + [d/dz]( z ) = cos(y)ez + cos(x)ez +
And ∇×F = (
i
j
k
[d/dx]
[d/dy]
[d/dz]
xcos(y)ez
ycos(x)ez
z
) = ( 0 + ycos(x)ez )i − ( 0 − xcos(y)ez )j + ( − ysin(x)ez − xsin(y)ez )k or ez( ycos(x),xcos(y), − ysin(x) − xsin(y) )
Let F = (x + y + z,x − y − z, − x + y − z) be a vector field.
i) Find ∇×F
We compute ∇×F = (
i
j
k
[d/dx]
[d/dy]
[d/dz]
x + y + z
x − y − z
− x + y − z
) = ( 1 + 1 )i − ( − 1 − 1 )j + ( 1 − 1 )k to obtain ( 2,2,0 )
Let F = (x + y + z,x − y − z, − x + y − z) be a vector field.
ii) Find ∇×(∇×F)
  • Since ∇ is an operator, we follow our usual order of operations. From our previous problem ∇×F = ( 2,2,0 ).
Then ∇×(∇×F) = ∇×( 2,2,0 ) = (
i
j
k
[d/dx]
[d/dy]
[d/dz]
2
2
0
) = 0i − 0j + 0k to obtain ( 0,0,0 )
Let F = (x + y + z,x − y − z, − x + y − z) be a vector field.
iii) Find ∇×(∇×F)
  • Since ∇ is an operator, we follow our usual order of operations. From our previous problem ∇×F = ( 2,2,0 ).
Then ∇×(∇×F) = ∇×( 2,2,0 ) = ( [d/dx],[d/dy],[d/dz] ) ×( 2,2,0 ) = 0 + 0 + 0 = 0
Let F = (x2,y2,z2) be a vector field and ∇f = (2x,2y,2z) be the gradient of F.
i) Find ∇×F
We compute ∇×( x2,y2,z2 ) = ( [d/dx],[d/dy],[d/dz] ) ×( x2,y2,z2 ) to obtain 2x + 2y + 2z
Let F = (x2,y2,z2) be a vector field and ∇f = (2x,2y,2z) be the gradient of F.
ii) Find ∇2f
  • Note that ∇2f = ∇×(∇f) which is equivalent to taking the divergence of the gradient of F.
Then ∇×(∇f) = ∇×(2x,2y,2z) = ( [d/dx],[d/dy],[d/dz] ) ×(2x,2y,2z) = 2 + 2 + 2 = 6
Let F = (x2,y2,z2) be a vector field and ∇f = (2x,2y,2z) be the gradient of F.
iii) Find ∇×[ ( ∇2f )F ]
  • Note that ∇2f = 6 which is a number. Then ( ∇2f )F is just a multiplication which yields 6(x2,y2,z2) = (6x2,6y2,6z2).
Then ∇×[ ( ∇2f )F ] = (
i
j
k
[d/dx]
[d/dy]
[d/dz]
6x2
6y2
6z2
) = ( 0 − 0 )i − ( 0 − 0 )j + ( 0 − 0 )k or ( 0,0,0 )

*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.

Answer

Divergence & Curl in 3-Space

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 and Curl in 3-Space 0:26
    • Introduction to Divergence and Curl in 3-Space
    • Define: Divergence of F
    • Define: Curl of F
    • The Del Operator
    • Symbolically: Div(F)
    • Symbolically: Curl(F)
    • Example 1
    • Example 2

Transcription: Divergence & Curl in 3-Space

Hello and welcome back to educator.com and multi variable calculus.0000

Today's lesson we are going to be talking about divergence and curl in 3-space.0004

So we have talked about divergence and curl before, we did it for 2-space when we discussed Green's Theorem.0008

Now we are going to talk about it in 3-space. We are going to introduce some new symbolism that will take care of all cases all at once.0016

Let us just go ahead and jump right on in. Okay.0024

So, we will let f(x,y,z) = f1(x,y,z), f2(x,y,z), and f3(x,y,z).0028

This is just a highly explicit way of representing a vector field. A vector field, these are the coordinate functions, the coordinate functions are functions of x,y,z themselves. I have just written everything out.0051

So, let this be a vector field on an open set s in R3, in 3-space, 3-dimensional space.0064

Okay. That is... and I know you know what this is but I am just going to repeat it for the sake of being complete.0084

For each point in s, there is a vector given by f pointing in some direction away from that point... pointing in some direction... starting at that point. That is it.0095

So, if I take some 3-space, so in this particular case let us just say that our open set s happens to be all of 3-dimensional space.0140

If I pick a point at random, well if I pick a point and I put that point in for f, I am going to get a vector going this way, maybe for this point I have a vector this way, maybe for this point a vector this way, that way, that way, that way, that way, could be any number of things. That is a vector field in 3-space. That is all it is.0150

Okay, let us define the divergence of f.0171

The divergence of f, which is symbolized as div(f) = df1 dx + df2 dy + df3 dz, or in terms of capital D notation, D1f1 + D2f2, + D3f3. Okay.0180

It is a scalar, it is a number. It is a scalar. A number at a particular point, Df dx + Df2 dy + Df3 dz at a particular point you put that particular point into this thing and it is just going to spit out a number.0218

Okay. That is the divergence in 3-space. It is just the analog of the divergence in 2-space.0246

So the curl is the... it is the analog of the curl in 2-space that we discussed previously, but it is a little bit more notationally complicated.0253

So, let us go ahead and write it out, and then we will go ahead and give you a symbolic way of representing it. 0262

So, curl of f... define the other definition, the curl of the vector field... is equal to... this one I am going to do the capital D notation first, and then I will do the regular partial derivative notation.0267

It is going to be D2f3 - D3f2, D3f1 - D1f3, D1f2 - D2f1. Notice this is not a scalar. This is a vector. The curl of a vector field gives you another vector field.0282

In terms of partial derivative notation, this looks like this. So, D2 of f3, this is Df3 dy - Df2 dz.0316

This is Df1 dz - Df3 dx.0335

This is Df2 dx - Df1 dy.0346

I hope to heaven I have got all of those correct. Okay. That is the curl. It is a vector.0356

So, when you are given a vector field f, when you take the divergence of it, you want to put a number at a given point.0367

When you take the vector field f, the curl of that vector field, what you end up with is another vector field. It is a vector at a given point.0374

Okay. Now we are going to introduce something called the Del operator.0386

The del operator is a symbolic way of simplifying the calculations. That is really what it is. And making things just look more elegant. The del operator, okay, it is usually an upside down triangle, or you can just write del.0392

So, the definition of the operator is the following. For the time being, the notion of an operator, an operator is just something that tells you to do something to something else.0413

In other words, we talk about the differential operator d dx. So, for example you knew what this is, d dx. If I apply this differential operator to some function f, d dx of f, well, that is just df dx.0423

That just means take the derivative of it. The integral operator. The integral operator says integrate on the function f, you are going to end up getting something else, so that is all an operator is.0440

It is a fancy term that says do something to this function. Well, the del operator is the same thing.0450

It is an operator that says do this to a given function, except it has multiple parts. These are individual operators. The differential, the integral operators. The del operator actually is written symbolically in the form of a vector.0456

So, but other than that you treat it exactly the same way as you do anything else. It says do something to something.0473

As it turns out, the del operator is a differential operator. It is a partial differential operator. You will see what we mean in just a minute.0480

So, let us go ahead and write out the symbol and then do some examples and of course everything will make sense.0487

Let us see, so this, or del, is equivalent to d dx, d dy, d dz, or D1, D2, D3, notice there is no function here because it is an operator. 0493

I have to choose a function and then say do this to it. That is the whole idea. We write it as a vector because it is going to operate as a vector on another vector. That is the whole idea.0518

So, now, let us go ahead and write what we mean by divergence and curl symbolically, using this del operator notation.0534

Okay. So, symbolically, and again, this is all symbolic. Symbolically, the divergence of f = the del operator dotted with the vector field, f.0544

Well, the del operator dotted with the vector field f, is equal to... well, the del operator is D1, D2, D3, that is my symbolic operator for del, dotted with well... f is f1, f2, f3.0568

Well, I know what a dot product is. It is just this × this, this × this, this × this, added together.0585

Except I am multiplying these two numbers, this is a symbolic vector operation. It says do d1 to f1, in other words, take the derivative of f1 with respect to x. Take the derivative of f2 with respect to y, take the derivative of f3 with respect to z. This is a symbolic notion.0591

We are symbolizing using the idea of a vector, that is what an operator is. So, this is equal to... so, it is d1f1 + d2f2 + d3f3, except this is not multiplication, this d1f1... this is a differential operator.0614

It says take the derivative of f1 with respect to the first variable, this says take the derivative of f2 with respect to the second variable, this says take the derivative of f3 with respect to the 3rd variable. I hope this makes sense.0637

Now the curl of f. Okay. This is going to be kind of interesting. Let me go to the next page.0651

So, the curl of f. Now you can memorize this any way you want to. If you want to just go back to what I initially wrote on the first page of this lesson, when I defined divergence and curl, you are more than welcome to remember curl in that way if you want to remember the indices -- 23 32, 13 31, 12 21.0657

That is fine, or you can remember it this way, symbolically. 0677

So, the curl of f, it is defined as del cross f, the del operator crossed with the vector field. Well we know what a cross product is. We have been dealing with it symbolically. It is the symbolic determinant i,j,k, and now del cross f... it is like this is a vector, this is a vector.0682

Well, the first vector is in the second row, so we will just write D1 D2 D3, and f is just f1 f2 f3.0705

We will symbolically take the determinant of that. When you do that, you end up with the following.0720

You end up with (D2f3 - D3f2)i - (D1f3 - D3f1)j + (D1f2 - D2f1)k. i... j... k... this is the first component function, second component function, third component function.0726

What you end up with is exactly what we had earlier. You get D2f3 - D3f2, this - turns this into a negative, turns this into a positive... what you end up with is D3f1 - D1f3 and you get D1f2 - D2f1.0784

This is the vector representation, this is the i,j,k, representation. This is the curl.0813

So, all I have done is I have taken this... I have created this thing called the del operator, given it a symbol like a vector, and I have been able to define the divergence and curl in terms of the two vector operations that I have.0819

The divergence of the vector field f is equal to del · f, and then curl of the vector field f is equal to del cross f.0835

This is the symbolic way of keeping things straight. So, let us just do some examples and I think it will make sense here. So, Example 1.0843

We will let our vector field f = the first component function is x2, the second component function is xy, and our third component function is going to be ex,y,z.0859

Okay. So, the divergence of f, divergence is always going to be easier because you are just sort of taking partial derivatives 1 by 1. So, the divergence, the partial of this with respect to x is going to be 2x, + the partial of this with respect to y, so this is going to be cos(xy) × x, so it is going to be + x × cos(xy), and the partial of this with respect to z is just going to be ecy.0875

There you go. That is the divergence. It is a scalar, not a vector. 0906

Now, let us go ahead and form the curl of f. The curl of f, well, we said that the curl of f = del cross f, so let us go ahead and form our symbolic determinant here... i, j, k.0915

We have d1, d2, d3, or if you want, let us go ahead this time for this particular example, let us use our partial differential and our d dx, d dy, it does not really matter.0934

So, let us do i, j, k, so we have d dx, d dy, an d dz, that is the del vector operator, and then we have f, which is x2, sin(xy), and exyz. We want to form this determinant.0950

Okay. So, let us see what we get. When we expand along the first row, we get the derivative of this with respect to y - the derivative of this with respect to z, so I end up getting the derivative of this with respect to y is exz, ex × z - 0 × i - ... because remember it is + - +, alternating sign.0978

The derivative of this with respect to x - the derivative of this with respect to z.1008

So the derivative of this with respect to x is exyz is e2 × yz - 0, and that is going to be j.1014

Of course our last one k is going to be the derivative of this with respect to x - the derivative of this with respect to y.1025

So, the derivative of this with respect to x is going to be y × cos(xy) - 0 × k, so is the... so I get this, this, and this... and so I end up with ex × z is my first component function of my curl, ex × yz is the second component function of my curl, and y × cos(xy) is the third component of my curl vector. There we go.1032

This is a scalar divergence... curl is a vector at a given point.1071

If I take all of the points, it gives me a vector field. That is it.1078

Okay. Let us do another example. So, example 2.1085

f(x,y,z) = x2yz, xy3z, and xyz4. Okay, well, let us see what we have got. 1095

The divergence of f, we said, is equal to del · f, well del · f is d1f1 + d2f2 + d3f3, but we are not multiplying the d and the f, this means take the derivative of f.1122

Well, the derivative of this with respect to x is going to be 2xyz + this one, the derivative with respect to y is going to be + 3xy2z, and the derivative with respect to z of this one is going to be +4xyz3. This is our divergence of the particular vector field.1141

Okay. So, now let us go ahead and do the curl of the vector field. The curl of this vector field, well it is equal to del cross f, and del cross f, well, is symbolic.1169

It is going to be i, j, k... let us do capital D notation here... D1, D2, D3, and we have x2yz, we have xy3z, and we have xyz4.1187

Okay. So, now let us go ahead and expand along the first row. It is going to be the derivative with respect to y of this - the derivative with respect to z of this. The derivative with respect to y of this is xz4.1208

The derivative with respect to z of this is xy3. This is the i component... - , now we go to the next one.1227

The derivative with respect to x of this - the derivative with respect to z of this, derivative with respect to the first variable which is x, derivative with respect to the third variable which is z. That is what is going on here.1241

This × this, this × this, except it is not times, it is symbolic. It means operate on this. 1253

Okay. So, the derivative with respect to x of this is yz4 - the derivative with respect to z of this, x2y. This is the j component.1262

Okay, we are almost done. Now, the derivative with respect to x of this - the derivative with respect to y of this. 1273

So, the derivative with respect to x of this is going to be y3z - the derivative with respect to y of this is going to be x2z.1282

Again, I hope that you are confirming this for me. There are lots of x's, y's, z's, i's, j's, k's, floating around. y3z, there you go.1294

I will go ahead and actually leave it in this form. That is the first coordinate function of the curl, that is the second coordinate function of the curl, and that is the third coordinate function of the curl.1304

Actually, you know what, let me go ahead and write it out. It is not a problem.1319

So, I will go ahead and erase these stray lines here.1320

So, we have got xz4 - xy3, that is the first component function, we have yz4 -x2y, and then we have y3z - x2z, notice the divergence is a scalar, the curl is a vector. 1329

It has three component functions... an x, a y and a z. At a given point (x,y,z), there is some vector pointing in some direction and away from that point. That is the whole idea.1359

Again, it is all based on this notion of what we call an operator. It is just a symbolic way of telling you what to do to a given function.1370

It is a unifying scheme, so you had this thing. You can take the divergence and curl of a vector field. We want to be able to express that in terms that we know.1380

Well, if we gave this... we called it a del operator, we have it a symbolic representation s D1, D2, D3, as a symbolic vector... If we do del · f, we get divergence of f, if we get del cross f, we get the curl of f. That is it. It is just a unifying scheme. 1390

Okay, thank you for joining us here at educator.com for divergence and curl. We will see for a discussion of the divergence theorem in 3-space. Take care, bye-bye.1410