วีดีโอ

Why are you not preparing before recording a video? It is a pain to watch it!

nice.

❤

thank you, your video helped me showing where i was wrong

Not apart of whatever this is but good luck guys!

Your explaination is very good.

Fire🔥

Great video. How would randomize the questions instead?

how to draw an arbitrary triangle and label the vertices as ABC?

Thank you sir for explanation solution in my book was very messed up

thank you for these videos!

Should the problem state: compute (2y^3)*dx - (2x^3)*dy instead of (2y^3)*dy - (2x^3)*dx? Cause if we correctly apply greens theorem, it would be an integral of zero.

This was much needed, thanks bro.

Is there any way to fill a polygon?

Horse shit, a length is not equal to a surface area. Why do you omit the units of the integrals? Common sense over rules your chicken scratches and muddled thinking. Get a brain. Did you get your college degree from a cereal box or comic book or is the university that you are attending have professors walking around the campus with their private parts showing?

u dont use magnitude. u use r prime t and dot product it instead

yo thank you for the video my textbook skips over a lot of steps and this was helpful

Hi, I'm interesting to use this class to make my class exam. I'm having a problem to create a exame using this class. I'm trying to make a exame with following structure: ... \begin{document} \begin{questions} \question[1] What is the result of ? \begin{randomizechoices} \choice 1 \CorrectChoice 2 \choice 3 \choice 4 \end{randomizechoices} \question[1] What is the result of ? \begin{randomizechoices} \choice 1 \CorrectChoice 2 \choice 3 \choice 4 \end{randomizechoices} \end{questions} \end{document} when I compile I got 2@correctchoice at the top of the page, and the error: Extra \endcsname. Did you received this error? If yes, what could be me error? Kind regards!

Thank you very much! I really like how you teach this: failing and showing how to overcome those failures. Hope to see a lot more.

Good job!

Good luck with your course!

The algorithm hast brought me hither

wtf no idea. confused as f

thank you robert

Over some half dead skeleton looking George Washington face (with a check mark under his alive eye signifying his undead alertness)

I think this is good on youtube. Can fast forward, skip around, much better than zoom. I think this was worth your time. Good explanations.

How to add video or gif to beamer latex

Really well explained! But after you parameterize a surface and found your partial derivatives. You can take the cross product and get the normal vector right but what if lets say in a question they require a positive z-component but your normal has a negative z-component. Can you just multiply it by -1? What I meant to say is how can be sure that a scalar product with the partial derivatives will result in 0? In this case I am talking about a plane.

thank you sir

My brother in christ the answer is 16/9

🔥Awesome🔥 Sir where are you from

How are you doing that with writing on some invisible whiteboard!

All of the vector stuff and cross product stuff was not necessary to work the problem. It will only confuse people. Just take the partial derivatives, plug them into the SA integral and you're done.

Thanks✨✨

Great example!

Hi! I know this is old, but I just want to thank you for sharing this video. I was stuck on a similar problem, and it took watching it all the way through to realize my issue was I forgot to include the extra factor of r for polar coordinates. For the problem in question, the given cylinder was x^2+y^2=1 and the surface area turned out to be (2pi/3)(2sqrt(2)-1). It's always beneficial to take the time to actually watch a worked example. Thanks! c:

great video

thanks. that was very clear.

this helped a lot! thank you :)

If in doubt, it equals 0

There should not be a "3" in the final integral! Sorry!

Will the cross products on the test be this involved?

I love the small existential crisis after you call it an "art"

It should be sqrt(4-x^2-y^2) not sqrt(2-x^2 - y^2)

I can’t add (or subtract); as mentioned below, the points are (0,0,-3) and (3/4,0,0)

Is Q1 (0, 0, -3) ? And Q2 (3/4, 0, 0) ?

As pointed out below, the bottom limit on the z integral should be z=x. So, just change that in the two spots in your notes. The correct answer is 352/15.

www.symbolab.com/solver/step-by-step/%5Cint_%7B0%7D%5E%7B4%7D%20%5Cint_%7Bx%7D%5E%7B2%5Csqrt%7Bx%7D%7D%20%5Cint_%7B0%7D%5E%7B8-z%7D%20%5Cleft(x%5Cright)dydzdx

Go here: th-cam.com/video/pd1vNjxT2O0/w-d-xo.html to see a corrected version.