Introduction to the surface integral | Multivariable Calculus | Khan Academy

Introduction to the surface integral

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Multivariable Calculus on Khan Academy: Think calculus. Then think algebra II and working with two variables in a single equation. Now generalize and combine these two mathematical concepts, and you begin to see some of what Multivariable calculus entails, only now include multi dimensional thinking. Typical concepts or operations may include: limits and continuity, partial differentiation, multiple integration, scalar functions, and fundamental theorem of calculus in multiple dimensions.

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[broudwauy]

Awesome explanation. Remembering that the cross product was the area of the parallelogram defined by the two vectors was an absolute revelation. I had absolutely no idea why the fundamental vector product showed up in surface integrals.

[whoisbobby]

Khan, you are literally the reason I went from cal 1- to linear and diff. You are the best!

[gianlucacastro]

That was the only thorough explanation I could find as to way the surface integral takes this cross product the way it does, thank you!!

[wontpower]

The cross product in the integral reminds me of the Jacobian. The area of dsdt doesn't quite match up with the surface area of the differential surface area in x y and z, so you need to multiply by abs(dr/ds x dr/dt) in order to account for that difference.

[malefetsanekoalane]

I have finally figured out what makes your explanations so great. You answer questions before they are asked. You replace computational ability with deep understanding, which is education for life. May you be spared for other generations.

[DiomedesStrosMkai]

That's why Khan is awesome.. A lot of the time my frustration comes when the book just skips a step and leaves me thinking '"what the hell?". Usually it's explained somewhere, but it's always off the cuff or several chapters back. This way it's *thoroughly* explained.. I'm not missing *anything*. Every piece of the puzzle is shown

[DavideSLiuni]

This man is the only reason of why I'm passing calculus in university

[vukstojiljkovic]

Thank you for this. They never did something like this, they just introduced it and started

[legendarysannin]

Please connect your videos with a link on the screen so we can jump from the video before. Very good explanation!

[FlippyBrown]

Sir, am I right to say that every single point on the region is associated to a parallelogram. And we are adding all the parallelograms(its area) associated to every single point on the region, in order to find the surface area?

[ManGo]

dude..i do not know how much you're getting paid for this..
but it's not enough..
how you explained this is outright brilliant. something my teachers always fall short of.

[oneinabillion]

Sal, I remember in linear algebra u said that. U will be remembered.

[rheejoan]

"now lets go do this for t!!!...i am running out of colors..."
Love this man

[sandracordoba]

You are indeed a brilliant person when it comes to explain this concepts. All my respect

[vaggs]

Okay short explaination. If we take the double integral of a function that is dependent on three variables, then we get the volume of a shape. This shape is a 3d shape. the bottom of the shape is the projection of the surface onto the zy plane, just like if we had a torch and lit it from exactly above. the top of the shape are the "hills that are formed" from the values of the fucntions.

Now what the surface integral calculates is the bottom of the same shape, but with a twist. the top stays the same but the bottom becomes the projection onto another surface that is not the zy plane but rather maybe a diagonal surface or something of that sort.

[gotnerdy]

Thanks for this and also making the video downloadable. Also what software/hardware are you using for this?

[anmolabhayjain]

Can anyone provide me with an explanation as to why during bijective transformation the boundary surface of one closed curve forms the boundary surface of the other closed curve or closed volume? That is why the boundary in u-v plane is also the boundary in x-y(or x, y, z for closed volume) plane of the transformed curve. Also assume that the transformation has a continuous first partial derivative over u and v.

[Barak]

You sir are amazing!
You teach so well and in so many topics and I can just thank you infinite amount of times!

[TennisGvy]

@Liaomiao That's if you are integrating a function in two directions. Integrating a function adds another dimension so to speak.

[lucidmath]

Sal can you please do some vids on intuitively explaining the volume integral?