Calculus - Integration: Volume by Rotating an Area (3 of 10) Ex. 3: y=x^2,y=x About the x-axis
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- Опубліковано 23 сер 2024
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In this video I will find the volume bounded by y=x^2,y=x about the x-axis
holy shit!!!!
I thought I would never understand this topic.... but I DID!!!!
thanks a lot man!!
Hey! I just wanted to say thank you so much for taking the time to make these videos. I have a physics final and calc 2 this week and you've saved my life with these videos. I now understand everything better. You make it easier to understand, I love your videos THANK YOU!!
Thanks for sharing and good luck on your finals.
Thank you very much, it is very kind of you.
The post- rotation image was hard to imagine. I wouldn't have come up with a cone as the final image.
Great job Mr. Michel.....Thanks
I hope you do not mind me saying, but the second zero in the evaluation should have a negative sign in front of it. Not that it changes the final result.
Since + 0 = - 0 = 0 it doesn't matter. But for any other value of course, it needs to be a negative. All comments are always welcome.
Around 6:30: am I right that the whole subtraction should be enclosed by brackets? because the lower limit (although zero in this specific case) has to be multiplied by pi as well.
Find the surface area due to the rotation of the area between f(x)=x^3 and g(x)=x
Here is an easy and interesting way to solve such a problem: CALCULUS 3 CH 7.1 PAPPUS-GULDINUS THEOREM ua-cam.com/video/hVYE7XBhJ8Y/v-deo.html
thank you so much sir but what i observed from the volume of the element is your meant the thickness is just # since the area itself is just (y2-y1)dx
Find surface area by rotate y=erf(x) about x axis
x from -1 to 1
What, if they ask for the volume swept out? Is it the same thing?
SOOOOOOO helpful sir!
Glad it helped! 🙂
Amazing explanation ,thanks teacher!
Thank you! 😃
thank you so much
Why this volume not equal when i rotated about y-axis ?
i suppose that the rotation of the same curve about y-axis and x-axis are equal ?????
No, it depends on the shape of the object.
At the radius of the circle, you write (x^2-x^4),I know it right, Could you tell me why (x-x^2)^2 is wrong.?
If you multiply out (x - x^2)^2, you will get a different result. (Try it)
Thank you very much. I mean, x-x^2 also is the radius of the circle, right?
x^2 - x = 0 is the equation you get when you solve the 2 equations simultaneously. The solution to that equation represent where the 2 functions cross.
thank you! it is a great video
You are welcome!
Thank you sir
Its so hard to tell if its a washer
Thank you, really helped a lot. I'm not even native but i could understand you very well, congratulations!