Thank you so much for your video! I was gone for a day and my teacher didn't give me the notes for this yet, but still gave me the work for it, and this helps a lot. Truly a life saver!
k.ly Because the graph is bounded by the relation x = y^3 and the vertical line x = 8. To find the shell height, we need to subtract the 'lower' function from the 'higher' one, or in this case, the one further to the left from the one further to the right.
That confused me also, but it's because x=8 is where the right side of the rectangle is bounded at. Since we're integrating in terms of y, we subtract the right (outer function) minus the left to find the shell height.
Not sure why nobody has replied to this yet, but I'll reply just in case others have the same question. Its 8-y^3 because you're looking for the height of the shell which is the distance from the point on the curve to the outer limit, which is 8. say y^3 = 1, then the shell's height would be 7. If it was just y^3, the height of the shell would end up being 1, which wouldn't work.
For further clarification we are taking the area between the two functions and rotating it about an axis. To get the area in between you need to take the area of the upper function (x= 8) minus the area of the lower function (x= y^3). Hence the 8 - y^3.
Because with the cylindrical shell method you should set up the cylinder parallel to the axis of rotation. We were rotating about the x-axis so it would make the most sense to write it in terms of y. If you wanted to keep it in terms of x with this particular problem the disk method would be the easiest approach.
Things are happening that I do not comprehend.
Whoever wrote my textbook needs lessons from you. Thank you!
THANK YOU. I got out of class and did not understand much of this. I like how you label everything.
Times the DY
Times the DY
Times the DY
Billy Diesel (Times the DY)^3
its crazy that 2013 videos save me the day before exam
Thank you so much for your video! I was gone for a day and my teacher didn't give me the notes for this yet, but still gave me the work for it, and this helps a lot. Truly a life saver!
Even MIT videos aren't clear on this chapter. Thanks a lot.
Watching this gave me that aha moment I was looking for absolutely love Kahn Academy
3 hours for my BC exam. Wish me luck:)
how did it go?
@@hussainalhammadi2790 got a 4/5 (advanced placement)
Thank you very much .👍👍🔝🔝
Thank you so much :)
It was a very great help!
Wait, quick question, why is it 8 minus whatever x is got this one and not for others?
k.ly Because the graph is bounded by the relation x = y^3 and the vertical line x = 8. To find the shell height, we need to subtract the 'lower' function from the 'higher' one, or in this case, the one further to the left from the one further to the right.
Thanks Khan, this was very helpful.
Thanks you so much! I understand now!
Very helpful. thanks
why did you not include the top and bottom area of the cylinder to get the volume?
Why is it 8-y^3 and not just y^3?
That confused me also, but it's because x=8 is where the right side of the rectangle is bounded at. Since we're integrating in terms of y, we subtract the right (outer function) minus the left to find the shell height.
Not sure why nobody has replied to this yet, but I'll reply just in case others have the same question. Its 8-y^3 because you're looking for the height of the shell which is the distance from the point on the curve to the outer limit, which is 8. say y^3 = 1, then the shell's height would be 7. If it was just y^3, the height of the shell would end up being 1, which wouldn't work.
For further clarification we are taking the area between the two functions and rotating it about an axis. To get the area in between you need to take the area of the upper function (x= 8) minus the area of the lower function (x= y^3). Hence the 8 - y^3.
you are saving my life!!!!
brother went full Corpse Husband at 6:50
Why is the radius of the cylindrical shell y? Why isn't it cube root of y? Bc the shell actually touches the graph..
1st again. this is easy
Can someone tell me what program he was using to draw in this video?
Thank you on forehand.
The function looks like a pork chop I would see on Looney Tunes. Funny function
You are so awesome :D
i love you.
Why doesn't he ever just keep it in terms of X?
Because with the cylindrical shell method you should set up the cylinder parallel to the axis of rotation. We were rotating about the x-axis so it would make the most sense to write it in terms of y. If you wanted to keep it in terms of x with this particular problem the disk method would be the easiest approach.
Elle c
First !!!
second? :o