The last video was very

abstract in general, and I used you know, f of x

and g of t, and h of t. What I want to do in this video

is do an actual example. Let’s say I have f of xy. Let’s say that f of

xy is equal to xy. And let’s say that we have

a path in the xy plane, or a curve in the xy plane. And I’m going to define my

curve by x being equal to cosine of t, and y being

equal to sine of t. And we’re going to go from–

you know, we have to define what are our boundaries in our

t –and we’re going to go from t is equal to 0– or t is going

to be greater than or equal to 0 –and then less

than or equal to. We’re going to deal in

radians, pi over 2. If this was degrees, that

would be 90 degrees. So that’s our curve. And immediately you might

already know what this type of a curve looks like. And I’m going to draw that

really fast right here in and we’ll try to visualize this. I’ve actually drafted

ahead of time so that we can visualize this. So this curve right here, if I

were to just draw it in the standard xy plane– do that in

a different color so we can make the curve green; let’s say

that is y, and this is right here x –so when t is equal

to 0, x is going to be equal to cosine of 0. Cosine of 0 is 1, y is going to

be equal to sine of 0, which is 0, so t is equal to 0. We’re going to be at x equal to

1, that’s cosine of 0, and y is sine of 0, or y is going to

be 0, so we’re going to be right there. That’s what t is equal

to; t is equal to 0. When t is equal to pi of 2,

what’s going to happen? Cosine of pie over 2–

that’s the angle; cosine of pi over 2 –is 0. Sine of pi over 2 is 1. We’re going to be

at the point 0, 1. So this is when we’re at

t is equal to pi over 2. You might recognize what we’re

going to draw is actually the first quadrant of the unit

circle; when t is equal to pi over 4, or 45 degrees, we’re

going to be at square root of 2, square root of 2. You can try it out for

yourself, but we’re just going to have a curve

that looks like this. It’s going to be the

top right of a circle, of the units circle. It’s going to have radius 1. And we’re going to go in that

direction, from t is equal to 0, to t is equal to pi over 2. That’s what this

curve looks like. But our goal isn’t here just to

graph a parametric equation. What we want to do is raise a

fence out of this kind of base and rise it to this surface. So let’s see if we can do that

or at least visualize it first, and then we’ll use the tools

we used in the last video. So right here I’ve graphed this

function, and I’ve rotated it a little bit so you can

see [UNINTELLIGIBLE] case. This right here– let me get

some dark colors out –that right there is the x axis, that

in the back is the y axis, and the vertical axis

is the z axis. And this is actually 2, this

is 1 right here, y equal 1 is right there, so this

is graphed that way. So if I were to graphs this

contour in the xy plane, it would be under this graph and

it would go like something like this— let me see if I can

draw it –it would look something like this. This would be on the xy plane. This is the same exact graph,

f of x is equal to xy. This is f of x; f of

xy is equal to xy. That’s both of these,

I just rotated it. In this situation that right

there is now the x axis. I rotated to the left,

you can kind of imagine. That right there is the x axis,

that right there is the y axis– it was rotated closer

to me –that’s the z axis. And then this curve, if I were

to draw in this rotation, is going to look like this: when t

is equal to 0, we’re at x is equal to 1, y is equal to 0,

and it’s going to form a unit circle, or half or quarter

of a unit circle like that. And when t is equal to pi over

2, we’re going to get there. And what we want to do

is find the area of the curtain that’s defined. So let’s see, let’s raise

a curtain from this curve up to f of xy. So if we keep raising walls

from this up to xy, we’re going to have a wall looks

something like that. Let me shade it in, color

it in so it looks a little bit more substantive. So a wall that looks

something like that. If I were to try to do it here

this would be under the ceiling, but the wall

look something like that right there. We want to find

the area of that. We want to find the area of

this right here where the base is defined by this curves, and

then the ceiling is defined by this surface here, xy, which I

graphed and I rotated in two situations. Now in the last video we came

up with a, well, you could argue whether it’s simple, but

the is, well, let’s just take small arc lengths– change in

arc lengths, and multiply them by the height at that point. And those small change in arc

lengths, we called them ds, and then the height is just

f of xy at that point. And we’ll take an infinite sum

of these, from t is equal to 0 to t will equal pi over 2, and

then that should give us the area of this wall. So we said is, well, to figure

out the area of that we’re just going to take the integral from

t is equal to o to t is equal to pi over 2– it doesn’t make

a lot of sense when I write it like this –of f of xy times–

or let me even better, instead of writing f of xy, let me just

write the actual function. Let’s get a little

bit more concrete. So f of xy is xy times– so the

particular xy –times the little change in our arc

length at that point. I’m going to be very

hand-wavy here. This is all a little bit

review of the last video. And we figured out in the last

video this change in arc length right here, ds, we figured out

that we could rewrite that as the square root of the dx

versus– or the derivative of x with respect to t squared

–plus the derivative of y with respect to t squared, and

then all of that times dt. So I’m just rebuilding

the formula that we got in the last video. So this expression can be

rewritten as the integral from t is equal to 0 to t is

equal to pi over 2 times xy. But you know what? Right from the get go we

want everything eventually be in terms of t. So instead of writing x

times y, let’s substitute the parametric form. So instead of x let’s

write cosine of t. That is x. x is equal to cosine

of t on this curve. That’s how we define x, in

terms of the parameter t. And then times y, which

we’re saying is sine of t. That’s our y; all I

did is rewrote xy in terms of t times ds. ds is this; it’s the square

root of the derivative of x with respect to t squared plus

the derivative of y with respect to t squared. All of that times dt. And now we just have to find

these two derivatives. And it might seem really hard,

but it’s very easy for us to find the derivative of x with

respect to t and the derivative of y with respect to t. I can do it right or down here. Let me lose our graphs

for a little bit. We know that the derivative of

x with respect to t is just going to be: what’s the

derivative of cosine of t? Well, it’s minus sine of t. And the derivative of

y was respect to t? Derivative of a sine of

anything is the cosine of that anything. So it’s cosine of t. And we can substitute these

back into this equation. So remember, we’re just trying

to find the area of this curtain that has our curve here

as kind of its base, and has this function, this

surface as it’s ceiling. So we go back down here,

and let me rewrite this whole thing. So this becomes the integral

from t is equal to o to t is equal to pi over 2– I don’t

like this color –of cosine of t, sine of t, cosine times

sine– that’s just the xy –times ds, which is this

expression right here. And now we can write this as–

I’ll go switch back to that color I don’t like –the

derivative of x with respect to t is minus sine of t, and we’re

going to square it, plus the derivative of y with respect to

t, that’s cosine of t, and we’re going to square it– let

me make my radical a little bit bigger –and then all

of that times dt. Now this still might seem like

a really hard integral until you realize that this right

here, and when you take a negative number and you squared

it, this is the same thing. Let me rewrite, do this

in the side right here. Minus sine of t squared plus

the cosine of t squared, this is equivalent to sine of t

squared plus cosine of t squared. You lose the sign information

when you square something; it just becomes a positive. So these two things

are equivalent. And this is the most

basic trig identity. This comes straight out of the

unit circle definition: sine squared plus cosine squared,

this is just equal to 1. So all this stuff under

the radical sign is just equal to 1. And we’re taking the square

root of 1 which is just 1. So all of this stuff right

here will just become 1. And so this whole crazy

integral simplifies a good bit and just equals the square root

of t equals 0 to t is equal to pi over 2 of– and I’m going to

switch these around just because it will make it a

little easier in the next step –of sine of t times

cosine of t, dt. All I did, this whole thing

equals 1, got rid of it, and I just switched

the order of that. It’ll make the next up a

little bit easier to explain. Now this integral– You say

sine times cosine, what’s the antiderivative of that? And the first thing you should

recognize is, hey, I have a function or an expression here,

and I have its derivative. The derivative of

sine is cosine of t. So you might be able to a u

substitution in your head; it’s a good skill to be

able to do in your head. But I’ll do it very

explicitly here. So if you have something

that’s derivative, you define that something as u. So you say u is equal to sine

of t and then du, dt, the derivative of u with respect

to t is equal to cosine of t. Or if you multiply both sides

by the differential dt, if we’re not going to be too

rigorous, you get du is equal to cosine of t, dt. And notice right

here I have a u. And then cosine of t, dt,

this thing right here, that thing is equal to d of u. And then we just have to

redefine the boundaries. When t is equal to 0– I mean

so this thing is going to turn into the integral –instead of

t is equal to 0, when t is equal to 0 what is u equal to? Sine of 0 is 0, so this

goes from u is equal to 0. When t is pie over 2

sine of pi over 2 is 1. So when t is pi over

2, u is equal to 1. So from is equal to 0

to u is equal to 1. Just redid the boundaries

in terms of u. And then we have instead

of sign of t, I’m going to write u. And instead of cosine of t, dt,

I’m just going to write du. And then this is a super-easy

integral in terms of u. This is just equal to: the

antiderivative of u is u 1/2 times u squared– we just

raised the exponent and then divided by that raised exponent

–so 1/2 u squared, and we’re going to evaluate

it from 0 to 1. And so this is going to be

equal to 1/2 times 1 squared minus 1/2 times 0 squared,

which is equal to 1/2 times 1 minus 0, which is equal to 1/2. So we did all that work and

we got a nice simple answer. The area of this a curtain– we

just performed a line integral –the area of this curtain

along this curve right here is– let me do it in a

darker color –on 1/2. You know, if this was in

centimeters, it would be 1/2 centimeters squared. So I think that was you know,

a pretty neat application of the line integral.

thanx. day wat **

Woo Sal!!! Yay!!

Keep the calculus/differential equations coming.

It's pretty much because of you that I am at this level now and you and, although Its just has to be done – reading proper materials and doing practice questions – getting your take on the concepts is invaluable.

If anyone wants to visualize vector and potential fields I suggest you to go to falstadcom. There is a java applet on this page which can generate these fields.

when i graduate college and make my fortune, i will donate a gazillion dollars to khan academy. sal is my favorite edumacator.

wow when I retire, I think I am going to join khanacademy

Thank you so much, these videos are about to save my education. I was struggling with line and surface integrals, and some physics electromagnetic stuff, The past few videos were great, and I really get the stuff now. It is really a great thing you are doing for people.

When you plot x(t) and y(t) on the 3d graphic, the quarter circle should begin and end on the x dimension extremity and the y dimension extremity of the cube, as in the function z=xy those extremities represent x=1 and y=1.

@pmcmena1 no because he reordered the bounds to fit the u function; at the end of the video it wasn't "from t=0 to t=π/2", but rather "from u=0 to u=1".

@pmcmena1 basic stuff

hanks a lot. Great explanation and example. Your teaching skill is great.

Thanks a lot. Great explanation and example. Your teaching skill is great.

You Sir, are awesome.

You have a real gift for teaching.

Very rare

I had to pause the video saying that this is amazing!!!

Thanku very much sir for this and so many other videos..u are a great teacher..

this video explains how radon transform work in ct scan…

thanks a ton..u made such a complicated thing as easy as it can be…

your waaaaayyyy too slow, people aint watching to learn how to integrate, they're watching to learn about line integral, but the explanations are superb, very intuitive 😛

Yes

tip: try to hold shift while drawing a line to make it straight!

it should work on all drawing applications…

you are a master of teaching!

Thanks Man (Morocco)

Thankyou ever so much sir. This whole YouTube teaching programme of yours is extremely benevolent. Thankyou.

I learn a lot from it.:)

It seems strange to me but even when I feel sleepy and I see his lectures I can understand even then!

I followed everything until the end – why didn't you replace u with sin(t) at the end of your u substitution and keep your bounds of integration in terms of t?

I hope when my teacher eventually sees these videos he realizes that he's a incompetent union protected lower and commits hari kari

tell u the truth..i studied this in 1975 and am now 59…Now I understand the damn line integral.

I love it when the subtitles say "UNINTELLIGIBLE" in all caps.

@ the end, let me do it in a darker color… chooses light purple

lol

Wow…that's great

What do the colors in the 3-D plane stand for?

pretty sure they represent slopes…

Why not mention that the factor you replace ds with is actually the length of c'(t) ?

I used that method as well.

sin(t) cos(t) can be integrated to -1/2(cos(t))^2 or 1/2(sin(t))^2. The answer comes out the same both ways.

if i consider a non parameterized function….like say i simply have y=x^2 ……should i parameterize it??

what software or device is used to make this videos?

I wish my professors would teach like this!

Thank thank

You make wonderful curly brackets!

I'm counting on your videos to get a 7 on calculus

Thank you Sir.

Thank you very much. That was very well explained.

befor i finish the video thnx realy it helped

I LOVE YOU MAN!!!!!!!!!!!!(writing in a couple of hours!!!

If only I watched these videos before I started cramming for finals… Learning is so much easier when you aren't stressed haha

x=cost haha

"Ill go switch back to that color I dont like" LOL

Do contour integration

😀

hueuha

you can also integrate sint*cost dt by converting it to 1/2*sin(2t) dt right? I did and got the same answer.

Thanks!

Excellent video ! For those who are familiar with Matlab or GNU Octave , here is a simple script that will help you to actually see the "curtain" that he refers to .

clear all;

clc;

t=[0:0.001:pi/2];

x=sin(t);

y=cos(t);

z=(x.*y);

stem3(x,y,z);

xlabel("sin(t)");

ylabel("cos(t)");

zlabel("f(x,y)");

thank you so much Sal…

Thank you <3

Sal….u just saved my exam which is 3hrs from now…thanks a bunch 🙂

ALL WE DID TO GET A FUCKING 1/2……..THAT IS SOMETHING THAT PISSES ME OFF

ANYWAY…..SUPERB WAY OF EXPLAINING STUFF IN A SIMPLE AND LUCIS MANNER

hi to everyone

sir just I want to know how you got this limits and how can I find it in other examples ?thanks

i love you

I like the software you used in the video very much, can you give me the name of it?

at 10:53 , for small method we can multiply and divide by 2. so integral part will be 2sint*cost that is equal to sin2t.then integration of sin2t= -cos2t/2 .

pretty neat

.. So what is Green's theorem?

We just performed a line integral! F*ck yeah! 😀

Once again, Sal was able to teach me in 10 minutes what my college professor failed to in 2 hours.

awesome…visualization

thank very much, you are an amazing teacher

It's fantastic how you summarize a one hour lecture in 15 minutes

Amazing job. I haven't even started calc 3 yet but I like to jump ahead. Very nicely done it managed to click and I have a very good understanding in line integrals now!

Why you dont resubstutute sint in place of u

Sal. It seems you plugged the equation for the curve into where I thought you said the equation for the ceiling should be. I thought you said the ceiling was x times y but you've made it the exact equation for the floor.What if you have a problem where the equation for the ceiling is completely different than the floor? In other words, the f(x,y) (ceiling) should only be x times y not cos t times sin of t. What am I missing here, please? You cannot define the ceiling as a function of t, only the floor curve.

What happened to Grant?

12:18 I don't understand why the du gets dropped

9:45 giant OH moment

Your videos are amazing !

kindly sir write on white board….its not understandable for everyone

can anyone tell me what software he used to trace the f(x;y) curve please?

What I don't get is how you get a plane from a line. All those green and yellow rectangles on the plane where never even on the line to begin with. Please help someone.

How can we virtualized the line integral if f(x,y) is a 2d vector, example Electric field vector

The Khan academy drinking game: every time he says something about "not too rigorous" or "hand waving" you throw back a shot.

2:08 Sal said sqrt(2),sqrt(2) but meant sqrt(3)/2,sqrt(3)/2

Been doing line surface and volume integrals for 2 months now only to realize now that the line integral is the area of the surface about the line. Face palm?

I understand the concept, proof of the formula and the results. But my question is, how is this useful in real life applications?

could you recommend any book that I can practice solving more problems

amazing…….

Why did he use substitution instead of the product rule for the integrations of sin(t) *cos(t)

or you could just call it sin2t/2.

I find it quite difficult to visualise

why don't we just use double integral? looks like it can give the same area too

dS (differential arc)is fundamentally non linear in the picture but it being infinitesimally small, it can safely be assumed to be linear thus allowing its approximation by Pythagoras theorem. Is this point of view correct?

you are the best I wish 60 years ago my teacher can do this

Can't we use the dot product between the function and ds? It will make it more simpler, isn't it?

If you switch the variables back to t’s after the substitution you get -1/2, why’s that?

Can you help me with this question?

The value of the integral of the function g(x,y)=4

x^3+10y^4 along the straight line segment from the point (0,0) to the point (1,2) in the xy-plane is?

My answer with your method is coming 33*(5^0.5). ds part is coming as 5^0.5.

The correct answer is being shown as 33 only.

Need help?????

imagine if you didn't let him take a darker color..

Sir, you just earned my respect! I wish I saw this before my first exam T.T Nonetheless, I'm now looking forward to the next exam!