For the disc method you say we multiply by dy to get the volume of the disc, but what do we actually multiply by? Dy is definitely not some scalar we multiply the integrand by, and it seems to me like we just write it down in the formula, but completely ignore it when actually calculating the volume, and if we don't multiply by the height of the disc we don't actually get a volume at all. Also, for the shell method why multiply r and h by 2pi, when what we get when we fold out the shell is a rectangle? Also what do you mean by width for the rectangle?
Responding to your second question just in case someone else has the same: 2*pi*radius gives the circumference of the shell, which is equivalent to the width of the rectangle created by "unfolding" the shell. So, 2*pi*radius of the shell is the width of the rectangle, multiplied by the height gives the area of the rectangle - which is also the area of the shell. Then we multiply by the thickness of the shell, which is delta x, as it approaches zero for the purposes of integration.
Really two completely different methods. I'm not sure why we are comparing them so closely. Both try to sum up volume whose limit as you go to infinity is the volume we are looking for but they are two different ways of approaching the problem. I'd probably separate the topics completely. Say here was a cool way we were solving the problems yesterday. Now we're going to solve them using the shell method today.
R and r should be functions of y, not functions of x. You should have used 2.pi.x.h.dx rather than 2.pi.r.h.dx. You should have used different limits in the x and y axes instead of using a and b for both. But all in all a good introduction.
Mr Woo, you've saved my HSC.
The one dislike in this video is from someone who went looking for more videos, realized this is a gem and eventually came back :]
Best explanation by far
Wow! Amazing video. Thanks for sharing such great insight.
Your videos are amazing, thank you!
the moment he brought out the second cylinder, he changed the history of maths
Great video !
For the disc method you say we multiply by dy to get the volume of the disc, but what do we actually multiply by? Dy is definitely not some scalar we multiply the integrand by, and it seems to me like we just write it down in the formula, but completely ignore it when actually calculating the volume, and if we don't multiply by the height of the disc we don't actually get a volume at all.
Also, for the shell method why multiply r and h by 2pi, when what we get when we fold out the shell is a rectangle? Also what do you mean by width for the rectangle?
Responding to your second question just in case someone else has the same:
2*pi*radius gives the circumference of the shell, which is equivalent to the width of the rectangle created by "unfolding" the shell. So, 2*pi*radius of the shell is the width of the rectangle, multiplied by the height gives the area of the rectangle - which is also the area of the shell. Then we multiply by the thickness of the shell, which is delta x, as it approaches zero for the purposes of integration.
Really two completely different methods. I'm not sure why we are comparing them so closely.
Both try to sum up volume whose limit as you go to infinity is the volume we are looking for but they are two different ways of approaching the problem.
I'd probably separate the topics completely. Say here was a cool way we were solving the problems yesterday. Now we're going to solve them using the shell method today.
I saw two dislikes?
R and r should be functions of y, not functions of x. You should have used 2.pi.x.h.dx rather than 2.pi.r.h.dx. You should have used different limits in the x and y axes instead of using a and b for both. But all in all a good introduction.
Japinha lindo
do better.