You are by far one of the best instructors of calculus online! Such clear explanations of the concepts and clear explanations of the calculations. Thank you so much!
I never understood this concept so well before...you have an amazing voice and the way you explain things looks so simple!! Thanks a ton...keep uploading more... :)
OMG.. I just started this Video and I'm at the 53 second mark ... and I had to Stop the Video so I could Post a comment!.. This Video is EXACTLY what I've been looking for!!.. THANK YOU KRISTA!!.. YOU're more of a QUEEN than a KING.. (doesn't the Queen have more power in Chess?) :D.. THANK YOU... I must have wasted 2 hours going through OTHER videos trying to figure out the Intuition of the LINE Integral.. and well.. I guess Fate saves the Best for LAST.. thank you Krista!!.. and depending on when you read this.. Merry Christmas.. :) ..
+Philip Y I'm really glad it helped, Philip! The line integral can be a tough thing to visualize, so I'm glad it's finally making sense. Merry Christmas to you too!
You are awesome when it comes to this calculus thing! I went through this semester without the aid of your videos much, and wish I had spent more time on those topics covered with the aid of your videos. Again Thank you for your time and efforts to place this information out where struggling students like myself can access it!
I like the patience and tenacity with which you explain how to evaluate this line integral example. You derived the first order parametric equations for each of the 3D coordinates well. I would like to see examples of line integrals of vector and scalar fields - such as work done by a vector field along a curve from point a to point b.
when t=0, f is at the starting point of the curve (0, 0, 0) when t=1, f is at the end point of the curve (1, 2, 3) so when you put from t=0 to t=1 into the parametric equations , you will have precisely the curve (line segment) described in the problem. Pretty much, we choose to parameterize from t=0 to t=1 to keep the integral simple.
Hi Krista King! I am glad to write to you again. I have another doubt. Why have we always, in parametrics equations, limit of integration of zero until one? Thank you!
A comment regarding interpretation. The function w=f(x,y,z) is in 4space, so it can’t be drawn as a curve in 3space. Instead, I would suggest interpreting the function f with 3 independent variables and a fourth, dependent variable as a linear density function, imagining that the density of the material that makes up the line segment (think “wire”) varies, depending on the spot. The units on the linear density function f (the integrand before the “ds”) would be something like “grams per centimeter”. The symbol “ds” represents a tiny length of the wire (units: cm). When these get multiplied, we see that the integral is adding up “grams”, to find total mass of the “wire” (line segment) of variable density.
integralCALC Woah! First TH-cam to ever reply to me! Well since you actually see these you videos have been saving me all year! My prof only teaches theory and doesn't do any examples. You have no idea how thankful I am! :D
If the z parameter is dropped we have the area over a line in the x-y plane under a surface f(x,y) which should be greater than the area between the surface defined and the integral of the differential of arc length of the line from (0,0,0) to (1,2,3). If f(x,y) is the plane z=2 and x=t, y=t, (t goes from 0 to 1) we get the rectangle ( made of two triangles) with area 2√2. But if the z parameter =2t and x=t and y=t, then the area between the surface f(x,y) =2 and the parametrized line would seem to be the triangle of area√2 or one half of the total rectangular area. Formally correct but we seem to be in an extra dimension.
There are plenty of line integral problems in which the bounds aren't from 0 to 1, however these problems don't pertain to line segments. Line integrals can be executed for any surface.
It can be correct, but how to be sure? What if the results is area of the curve between c and x or y -axis instead? If z=f(x,y) but here is z=f(x,y,z) which would be some type of recursive equation or something as that... Here f(x,y,z) does not present 3D-surface, but some type of potential in 3 dimensions. It should then select equipotential surface by setting x*e^(y*z)=constant to calculate the distance to this surface. If constant is zero that would represent the yz-plane. The results could be the area between rhe c and yz-plane rather than between c and z-axis. But i couldn't be sure.
I followed your calculation with great interest. However in this example the area between the function and the line (curtain) is about 71,62.. which is not the same like the line-integral which is 125,48.
integralCALC Hey I have a question, what if the line segment goes from (1,1,0) to (1,1,1) would x=0 , y=0, and z =1 ? Which in turn would mean that r(t)=tk
Why does the curtain connects with z axis and not x or y axis? Which part of the problem states that or would it not matter which axis it goes back to in your drawing?
wait what? so an integral of a function of two parameters isnt volume, but a shadow? and an integral of a function of 3 parameters gives us volume? WHAT?!
Are you sure that we take the "to-point" and subtract it from the "from-point" and multiply that by t? My book says something else! :( Straight-line segment is from (1,2,3) to (0, -1,1) and the book does it this way: r(t) = (i + 2j + 3k) + t(-i -3j - 2k) = (1-t)i + (2-3t)j + (3-2t)k but if i do if the way you decribe: x = (0-1)t = -t y= (-1-2)t = -3t z = (1 - 3)t = -2t
Depends on whether the vector field is conservative. If so, then yes. Think about gravity in physics, gravity is conservative so the work done on an object is the same no matter what path it took. As long as the object started at a and ended up at b, the work is the same. If the vector field is nonconservative something like you pushing a chair across the room, the work changes because the longer the arc length, the more friction you have to push against.
Can someone help me with that: C is given by x=t^2 y=t^3 z=t^2 Evaluate the integral under the region c . The integral is: Zdx+xdy+ydz. I have no idea what to do.. How to solve it :/
Unfortunately youtube is full of videos that prove how stupid people are in the 21st century. Thanks for making an exemption (and proving me how geek I am for once more by watching this!:D )!! Nice video! What's the program that you are using as for the blackboard?? Really liked that (although that it would be better if it was larger). I would like to make something similar as for structural design.
A 3D line integral does not calculate the area of ANYTHING --that drawing on the left at 3:06 is total nonsense! She knows how to do the computations but she has little insight or intuition into what they mean.
![เปิดใจ “ต้นหอม ศกุนตลา” หลังโดนพาดพิงธุรกิจที่เคยทำ และผลกระทบทำถูกปลดฟ้าผ่า! | แฉ 24 ต.ค.67 [2/3]](http://i.ytimg.com/vi/vCi5OIqlxaE/mqdefault.jpg)
You are by far one of the best instructors of calculus online! Such clear explanations of the concepts and clear explanations of the calculations. Thank you so much!
Awww, thank you so much, Dani! :D
To be more specific these are parametric equations: x = 0+(1-0)t; y= 0+(2-0)t; z = 0+(3-0)t
I wish I could see at least one of my instructors explaining like you. You are the best that I have never seen. Break a leg.
Your video just saved one of Ph.D candidates in far east. Thank you.
Your voice makes this easy to understand. Thank you
I never understood this concept so well before...you have an amazing voice and the way you explain things looks so simple!! Thanks a ton...keep uploading more... :)
I definitely will! Thank you for the support. I'm glad you're liking the videos. :)
best thing that has ever happened to youtube. thank u
OMG.. I just started this Video and I'm at the 53 second mark ... and I had to Stop the Video so I could Post a comment!.. This Video is EXACTLY what I've been looking for!!.. THANK YOU KRISTA!!.. YOU're more of a QUEEN than a KING.. (doesn't the Queen have more power in Chess?) :D.. THANK YOU... I must have wasted 2 hours going through OTHER videos trying to figure out the Intuition of the LINE Integral.. and well.. I guess Fate saves the Best for LAST.. thank you Krista!!.. and depending on when you read this.. Merry Christmas.. :) ..
+Philip Y I'm really glad it helped, Philip! The line integral can be a tough thing to visualize, so I'm glad it's finally making sense. Merry Christmas to you too!
+Krista King | CalculusExpert.com you are awesome
You are awesome when it comes to this calculus thing! I went through this semester without the aid of your videos much, and wish I had spent more time on those topics covered with the aid of your videos. Again Thank you for your time and efforts to place this information out where struggling students like myself can access it!
Arvin Cunningham Aw thanks! I'm happy I can help!
I like the patience and tenacity with which you explain how to evaluate this line integral example. You derived the first order parametric equations for each of the 3D coordinates well. I would like to see examples of line integrals of vector and scalar fields - such as work done by a vector field along a curve from point a to point b.
One picture is worth than million words!! , its proved by your lectures
thank you for your contribution to education
great job and keep it up!
+Siddhant Unavane Thank you very much!
Great job explaining what a Line Integral actually represents with your picture. Very helpful!
Eric Hendricks Thank you so much!
More helpful than my teacher! Thank you
+Ann L Glad I could help!
brilliant, what a great job of explaining all the steps
Thank you. May you be blessed with many views and subscribers
Saving my butt from one math class to the next i swear to god 😭 thank you!!!!
You're welcome, I'm so glad I'm able to help!!
when t=0, f is at the starting point of the curve (0, 0, 0)
when t=1, f is at the end point of the curve (1, 2, 3)
so when you put from t=0 to t=1 into the parametric equations , you will have precisely the curve (line segment) described in the problem.
Pretty much, we choose to parameterize from t=0 to t=1 to keep the integral simple.
love how straight to the point you are
Hi Krista King! I am glad to write to you again.
I have another doubt.
Why have we always, in parametrics equations, limit of integration of zero until one?
Thank you!
Because it's her wish.
This, just as the rest of your videos are, was extremely helpful and easy to understand. I might be able to pass Calc 3 yet
Aw thanks! I'm glad they're helping!
When you give the video a thumbs up before you even watch it cus you know it's going to be awesome. And it is. Thank you. :)
Awww thanks Dragonfly! :D
A comment regarding interpretation. The function w=f(x,y,z) is in 4space, so it can’t be drawn as a curve in 3space.
Instead, I would suggest interpreting the function f with 3 independent variables and a fourth, dependent variable as a linear density function, imagining that the density of the material that makes up the line segment (think “wire”) varies, depending on the spot. The units on the linear density function f (the integrand before the “ds”) would be something like “grams per centimeter”. The symbol “ds” represents a tiny length of the wire (units: cm). When these get multiplied, we see that the integral is adding up “grams”, to find total mass of the “wire” (line segment) of variable density.
Thank you so much. I love how in depth the example and explanation were.
You're welcome, I hope it helped!! :D
Thanks a lot krista , you are a saviour !!
I'm glad I could help!
This is so much more helpful than my professor! I learned more in 10 min from you than in 50 from him
:D
Now you understand why math is called the Queen of the Sciences. She's more artistic than the other guys.
Great job, Krista. Keep going.
Thanks Rich!
Amazing! Much better than my teacher! You saved me!
I'm so glad it helped!!
integralCALC Woah! First TH-cam to ever reply to me! Well since you actually see these you videos have been saving me all year! My prof only teaches theory and doesn't do any examples. You have no idea how thankful I am! :D
Kiley Niemeyer :D
This video was very helpful. Tnx a lot Krista!
Awesome! Thanks for letting me know. :)
Haha "loop back and start doing funky things" 10:32 love it :)
One small point. My understanding is, it is the curve C that is divided uo in small portions, not the f(x,y,z).
If the z parameter is dropped we have the area over a line in the x-y plane under a surface f(x,y) which should be greater than the area between the surface defined and the integral of the differential of arc length of the line from (0,0,0) to (1,2,3). If f(x,y) is the plane z=2 and x=t, y=t, (t goes from 0 to 1)
we get the rectangle ( made of two triangles) with area 2√2. But if the z parameter =2t and x=t and y=t, then the area between the surface f(x,y) =2 and the parametrized line would seem to be the triangle of area√2 or one half of the total rectangular area. Formally correct but we seem to be in an extra dimension.
how do you determine the limits of integration to be 0
Thanks alot. You are really amazing
Thank you for your help:)
I dont understand why bounds are always 0 to 1. Ive done plenty problems where the bounds are different
There are plenty of line integral problems in which the bounds aren't from 0 to 1, however these problems don't pertain to line segments. Line integrals can be executed for any surface.
Because everyone loves pi'e'
10.39 sec
Thank you Professor!!!!!
You're welcome, Paulo! :)
amazing xplaination ...........tnx to clear my concepts
Thanks! Glad you liked it!
What an amazing explanation!
Aw thanks!
Thank u ma'am ure brilliant
You're welcome, Sahil, I'm happy to help! :)
It can be correct, but how to be sure?
What if the results is area of the curve between c and x or y -axis instead?
If z=f(x,y) but here is z=f(x,y,z) which would be some type of recursive equation or something as that...
Here f(x,y,z) does not present 3D-surface, but some type of potential in 3 dimensions. It should then select equipotential surface by setting x*e^(y*z)=constant to calculate the distance to this surface. If constant is zero that would represent the yz-plane. The results could be the area between rhe c and yz-plane rather than between c and z-axis. But i couldn't be sure.
I followed your calculation with great interest. However in this example the area between the function and the line (curtain) is about 71,62.. which is not the same like the line-integral which is 125,48.
thank you Krista
You're welcome Esam! Glad it could help.
That is true. Thank you.
This is amazing👍
Thanks, sabarish, I'm glad you liked it! :D
Really helpful, thank you.
Thanks ma'am!!!
thank you so much ma'am.
She's good. Thank you loads.
I need you to further explain why the bounds of integration are from 0 to 1
If the equation was xyz^2 ds then do we have to have information about z=z(t) or not ? line segment from (0,0,0) to (1,2,3)
integralCALC Hey I have a question, what if the line segment goes from (1,1,0) to (1,1,1) would x=0 , y=0, and z =1 ? Which in turn would mean that r(t)=tk
the stuff under the square root is the same thing as r'(t) and x(t),y (t),z (t) is the same as F (r (t))
HI Kristen. I an not able to understand the very first bit how you replaced ds with dt??
this is amazing! thank you!
Why does the curtain connects with z axis and not x or y axis? Which part of the problem states that or would it not matter which axis it goes back to in your drawing?
Nice video!
YOU DA QUEEN!!
Shouldn't 3 variable scalar functions be in 4d space?
yeah, and that shadow thing for 2 variables... I'm not sure about that either.
😧Watching it in awe like "WHY DIDN'T MY PROFESSOR JUST EXPLAIN IT LIKE THAT?!?"
I'm so glad it helped! :)
@@kristakingmath ACED my class, AND got nearly 100% on my last two exams after tunin into your examples and explanations, Krista🤓👍🙏
Thank You!!
Great video
Thanks, Hampton! :D
integralCALC how do you know when to multiply f(r(t)) by the magnitude of r'(t), instead of just taking the dot product of f(r(t)) and r'(t) ??
wait what? so an integral of a function of two parameters isnt volume, but a shadow? and an integral of a function of 3 parameters gives us volume? WHAT?!
Are you sure that we take the "to-point" and subtract it from the "from-point" and multiply that by t? My book says something else! :(
Straight-line segment is from (1,2,3) to (0, -1,1)
and the book does it this way:
r(t) = (i + 2j + 3k) + t(-i -3j - 2k) = (1-t)i + (2-3t)j + (3-2t)k
but if i do if the way you decribe:
x = (0-1)t = -t
y= (-1-2)t = -3t
z = (1 - 3)t = -2t
Salar Chof yes that’s exactly what I was thinking. I’m pretty sure you are doing the right thing
Salar Chof n(
The video has the question labelled "from (0,0,0) to (1,2,3) instead of what you wrote
can u tell me why u took the value of X to the difference between the two x coordinates.
krista king you are amazing. you are the love of my life. i am from México
Oh My God this is awesome
Awesome!!!
THANK YOU !!!!!!!! I might pass now
:D
Wow never new we could just place what you equaled and just use the interval we started with instead of changeing the interval and
so are line integrals path dependent then?
Depends on whether the vector field is conservative. If so, then yes. Think about gravity in physics, gravity is conservative so the work done on an object is the same no matter what path it took. As long as the object started at a and ended up at b, the work is the same. If the vector field is nonconservative something like you pushing a chair across the room, the work changes because the longer the arc length, the more friction you have to push against.
thx! this helps a lot! :))
thanks
You're welcome!
why limits of integral got transformed to 0 to 1........... is it compulsion that these would always be 0 to 1.
I like your lecture
+Dhiman Roy Glad you liked it! Thanks for letting me know.
thanks, once again
;)
Can someone help me with that:
C is given by
x=t^2
y=t^3
z=t^2
Evaluate the integral under the region c .
The integral is:
Zdx+xdy+ydz.
I have no idea what to do..
How to solve it :/
Michael Van biezen has a video on that
you just can`t draw the graph of f(x y z)=x*e^(y*z) in 3D, it means that aria is`nt in your 3D graph, it is in 4th dimension.
I need.. vector calculus book by pc matthews.please do help me.soft copy.please.
Krista Im writing the Pope to nominate you patron saint of mathematics
That's okay with me! :)
the first 7 minutes didnt make any sense to me, after you started doing the question then i understood
Unfortunately youtube is full of videos that prove how stupid people are in the 21st century. Thanks for making an exemption (and proving me how geek I am for once more by watching this!:D )!! Nice video! What's the program that you are using as for the blackboard?? Really liked that (although that it would be better if it was larger). I would like to make something similar as for structural design.
+thessalonician I'm so glad you liked the video! I use a program called Sketchbook.
I use sketchbook as well. So the blackboard is probably a background. Many thanks!
+thessalonician Yep, the blackboard is just an image.
oh c'mon bro!
that's sweet
Can you help to solve exercises?
No, I'm not currently available for that.
very niccc
THICC
A 3D line integral does not calculate the area of ANYTHING --that drawing on the left at 3:06 is total nonsense! She knows how to do the computations but she has little insight or intuition into what they mean.
it represents the area under the curve.
if you're going to explain two different methods... why not use two vids?
Marry me and teach me math every day please
Cute voice