Shell method for rotating around vertical line | AP Calculus AB | Khan Academy
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- čas přidán 8. 01. 2013
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Introducing the shell method for rotation around a vertical line. Created by Sal Khan.
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4:50 am on Tuesday, May 25, 2021 is when I finally understood Cylindrical shells. Khan Academy, you were the right teacher for me on this.
3:59 am October 14, right there with you
12:30 am on Monday, Feb 6, 2023 for me!!!
Thanks Khan Academy.
yup same, day before my final lol
Feb 1, 2024 for me!!!!
I cannot express how thankful I'm to you, your videos are more than amazing, I'm taking calculus II class with a professor that cannot speak English. So I have to teach myself. These videos have been the tool I'm using along with my book. Thanks a lot!
sir, you are clearly an artist. We also would like to see your artworks.
"something... Something... not too dissimilar to this." I love it!
Him: this is challenging my art skills
also him: *draws graphs that I have so much difficulty drawing*
I wish my professor explained it like this. Your explanation makes it so much easier to understand.
Nice video thank you! Other videos on youtube did not really explain what is shell method. From your explanation I immediately understood. Thank you
Thank you! this video made me easier to imagine the logic behind the formula
Truly helped me to visualise!!! Thank you!!
Awesome SAL!!!! This is amazing!
Makes sense to me now. Thanks
TY prof Sal, my professor in integral calculus does not thought this topic...THIS IS A BIG HELP :)
Thank you so much for these videos Sal...
great explanation of the concept for the formula!
thank you for explaining it so much better than my teacher
Great explanation!
thank you!!
I LOVE YOU ( art skills ) ♥♥♥
your art skills are amazing
had to watch this twice before finally getting the concept... on the day of my exam
You're amazing!
To the point, clear and concise. Thanks!
do you still remember what the shell method is???? it’s been 10 years starr1169, have you been exercising your math?
thank you
28/5/2024 I will have my calc final exam ,wish me the best guys and Mr Khan I don't really know how to thanks you sir ,❤❤thanks so much u just save me 😊
Great graph
Thanks for explaining why we are doing it this way instead of just introducing it.
My class is a Hybrid class, so it goes much faster, and I have to learn stuff on my own. Kind of a pain sometimes, but loading this alleviated my headache.
dear khan academy.... i love you
All i need is someone who is skilled at art.
ya this is pretty easy i found that calc 1 was harder people say otherwise understanding whats happening took me a few videos but you can also memorize the formula and what to do in some situations the problem is if theres a new problem it might not work the same way and you might need to understand where you get the formula to solve the new problem
What do you mean? Remember, the definite integral is really just the Riemann sum, after you take the limit as n approaches infinity. So, it doesn't really matter where the "midpoint" is. It could even be argued that an infinitely thin rectangle (with thickness dx) has no midpoint. Unless you're referring to something else...
How come I found this to be the easier method? Everyone else in my class hated this, but I used it whenever I could.
Currently learning washer disker/ shell method right now. From what I can say, the shell method definitely has an easier formula, but is harder to visualize than washer and disk. I find the washer/disk to be much more intuitive.
I see it now
I knew there was something odd about this explanation. The rectangle is supposed to be positioned so that the midpoint of the top of the rectangle runs along f(x). Maybe it doesn't matter with the way that you've explained it, but that part was really confusing me.
0:26
The only part that confuses me is the radius part!
Why do you call "depth" of the rectangle instead of "width"?
I understand that the way he does it is much easier to work out, but I just wanted to see if my method worked, and it turned out that it only did when the rectangle was positioned so that the top's midpoint runs along the function. Then, I looked up the shell method on Google, and it turns out just about every college/high school teaches the shell method as the rectangle being in the center as I was saying. Now that I think about it, I remember my high school calc teacher saying that.
cool
shell la la..lol.. good video...thank god for khan...who got to calc 2 and didnt understand a damn thing until now!!!???...and I thought I was a math wiz
Vernard Wilson learn grammar
I mean, you're not supposed to immediately understand Calculus. I took a 9 week summer course for Calc 1. It was hard, sure. The entire time I actually didn't understand anything, but still did great on tests. Eventually, you understand Calculus. Everything else below is actually pretty easy.
What program is that? And Thanks for the great lesson.
windows xp paint.
I agree, but for some reason it doesn't work if this is solved a different way. For me, it is more intuitive to use the volumes of hollowed out cylinders to find the volume of revolution (the volume of a larger radius cylinder minus the smaller radius cylinder with the heights of both cylinders being equal to the value of the function at that current x value). This didn't give me the right answer at first, but then I tried it again with the rectangle in the middle. Unclear? Sorry 500 characters.
is the radius always going to be x when using this method? no matter the function or the interval?
Glyph no if it’s not revolving around and axis and instead revolving around a different line. Eg x=2 the radius would be 2 - x
sal khan W
Would that mean that if I have to rotate the shell around the value x = 4, the it would just be int(2pi(4-x)f(x)?
And if it is, why wouldn't it be x-4 inside the integrand?
What if I wanted to rotate it about a line such as x = 1 or x - -2?
Shift the function.
why is r equal to x? thanks!
It's the radius of the shape. It's half of the circular figure since it's rotated about y=0. So it's whatever the +x distance is on the graph
you just multiply the function by x in effect (and then 2pi)
can you come up with the formula for a doughnut using this method?
Just use the formula of a circle. Let's say that the origin is (0,0). You'd shift it so that it's not touching/intersecting your axis of reflection. Then continue.
what if sal khan was my teacher?
what do you mean by depth?
think of the "shell" as a hollow pipe. The thickness of the pipe is so small that we denote it as dx. we need this dx in order to find the volume of the shell, because we need depth*2pi*radius*height to find volume.
The depth is the x axis length. It's the width of each small rectangle between the curve of the function and the x axis. When you rotate the rectangle around a line you get a 3d object. The
thickness or the depth of the object is the width of the rectangle.
By the way, isn't this learned in Calc 2, because it's integral calculus? Not differential Calculus....
taking calc 2 now, learning this now
@@alexandersindorf5426 Maybe it's because I'm at a community college? Typically universities don't accept high school students.
❤️u
all dis business
I cant stand it when he repeats himself, it's like when you're playing a game and it lags or rubberbands.
I love how Sal wholly sounds like a white guy making a video but he's actually not
you just made this confusing. next.