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Professor Organic Chemistry Tutor, thank you for an exceptional video/lecture on The Surface Area of Revolution of Parametric Equations about the Xand Y axis. In some Calculus textbooks, Surface Area of Revolution problems are computed in different ways. This is an error free video/lecture on TH-cam TV with the Organic Chemistry Tutor.
Certain books represent it that way. The importance of this method is to realize what y is representing. He knew it represented the integrating variable which is why he brought y into the integrand. Well, technically, he didn't bring y into the integrand, he brought t^2. As long as you know wwhat your integrating with respect to, than it's okay
it's more to show you that when you are taking the SA of something being revolved around an axis, it is really because you are taking the length(the function) and multiplying it by the circumference of a circle to get the SA. I think I could have worded that better but I think u know what I mean.
Right. "y" should never have been outside of the integral to begin with. He makes a "correction" by bringing it into the integral eventually, but that is an "illegal" move. The only time a variable can be multiplied into an integral-or factored out of an integral-is when the variable is unrelated to the differential. In this case, y is very related to the differential, dt, and should have been in the integral from the beginning.
be careful when you working with a interval with a negative number in it when you remove the sqrt of t^2 make sure you put absolute value and then you have to work with two interval when is negative you will put negative sign before and when is positive example [-1,1] after removing sqrt of t^2 that`s will equal to interval [-1,0] -t + [0,1] t . this is very important when you dealing with a change sign in interval
2:24 this is very very wrong and misleading, you can't just put functions in and out of an integral like they are constants. That should have not been outside the integral in the first place. The real formulas have the x(t) and the y(t) INSIDE the integrals. Please take down this video.
The t should never have been outside the integral sign to start with. The incremental surface area dA= 2pi.y.dl,where dl is the sqrt of [ (dx/dt)^2 +(dy/dt)^2]. The entire expression then needs to be integrated with the only constant being 2pi since y,dx/dt and dy/dt are all functions of t. This reduces to A=2pi.integral [y(t).sqrt{(dx/dt)^2 +(dy/dt)^2}dt]
plz redo this examples you not allow to take the r(x) out of the integration you are changing the definition of the Area of surface of revolution . it work for you this time because you have to cancel the t but all the time , please change otherwise people will make their exams wrong thank you!
I don't ever want to see you pull a variable out like this again. You have done a lot of good work, but some stuff is unforgivable. You were treading dangerous water here.
be careful when you working with a interval with a negative number in it when you remove the sqrt of t^2 make sure you put absolute value and then you have to work with two interval when is negative you will put negative sign before and when is positive example [-1,1] after removing sqrt of t^2 that`s will equal to interval [-1,0] -t + [0,1] t . this is very important when you dealing with a change sign in interval
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Final Exams and Video Playlists: www.video-tutor.net/
Full-Length Math & Science Videos: www.patreon.com/mathsciencetutor/collections
Professor Organic Chemistry Tutor, thank you for an exceptional video/lecture on The Surface Area of Revolution of Parametric Equations about the Xand Y axis. In some Calculus textbooks, Surface Area of Revolution problems are computed in different ways. This is an error free video/lecture on TH-cam TV with the Organic Chemistry Tutor.
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Why did you place the Y outside of the integral? It is a variable, not a constant. You cannot take out "y" if it has variables
Certain books represent it that way. The importance of this method is to realize what y is representing. He knew it represented the integrating variable which is why he brought y into the integrand. Well, technically, he didn't bring y into the integrand, he brought t^2. As long as you know wwhat your integrating with respect to, than it's okay
it's more to show you that when you are taking the SA of something being revolved around an axis, it is really because you are taking the length(the function) and multiplying it by the circumference of a circle to get the SA. I think I could have worded that better but I think u know what I mean.
Right. "y" should never have been outside of the integral to begin with. He makes a "correction" by bringing it into the integral eventually, but that is an "illegal" move. The only time a variable can be multiplied into an integral-or factored out of an integral-is when the variable is unrelated to the differential. In this case, y is very related to the differential, dt, and should have been in the integral from the beginning.
@@brennanbeell4008 no, it's not okay to use ambigous nonstandard notation. There is no reason why the y should be outside.
Find surface area by rotate y=erf(x) about x axis
x from -1 to 1
be careful when you working with a interval with a negative number in it when you remove the sqrt of t^2 make sure you put absolute value and then you have to work with two interval when is negative you will put negative sign before and when is positive example [-1,1] after removing sqrt of t^2 that`s will equal to interval [-1,0] -t + [0,1] t . this is very important when you dealing with a change sign in interval
Find the surface area due to the rotation of the area between f(x)=x^3 and g(x)=x
2:28??? I don't think you can do that with variables, only constants
2:24 this is very very wrong and misleading, you can't just put functions in and out of an integral like they are constants. That should have not been outside the integral in the first place. The real formulas have the x(t) and the y(t) INSIDE the integrals. Please take down this video.
why after the t move into the integral, t does not change?
The t should never have been outside the integral sign to start with. The incremental surface area dA= 2pi.y.dl,where dl is the sqrt of [ (dx/dt)^2 +(dy/dt)^2]. The entire expression then needs to be integrated with the only constant being 2pi since y,dx/dt and dy/dt are all functions of t. This reduces to A=2pi.integral [y(t).sqrt{(dx/dt)^2 +(dy/dt)^2}dt]
What happened to 18 at 9:35 ?
he includes it back in @ 10:43
@@jam34786 Thank you.
why did you not unsub your U's before you evaluated?
x=acos(theta)
y=bsin(theta)
a=6378
b=6357
Surface area of revolution of this curve is more practical
Hi, a quostion. Why when you divede t^3 by 18t you only remain with t^2 and not t^2/18?
he forgot it. And it did it at 10:55
You lost me there at 2: 38 I mean isn't the t outside the integrals to be excluded from intergrating😏
plz redo this examples you not allow to take the r(x) out of the integration you are changing the definition of the Area of surface of revolution . it work for you this time because you have to cancel the t but all the time , please change otherwise people will make their exams wrong thank you!
That’s how the formula is depending on the axis of rotation check your textbooks
Jeeeesus huge ass process
Sy is confusing ahhh.
I don't ever want to see you pull a variable out like this again. You have done a lot of good work, but some stuff is unforgivable. You were treading dangerous water here.
Disliked, because you started with writing y outside the integral, which is mathematically unsound.
Your a potato
Ok yuri renner
yo mama is mathematically unsound
@@mikealgraham5549 yo wth🤣🤣
don’t hate on my brotha organic chemistry tutor
getting the wrong answer
be careful when you working with a interval with a negative number in it when you remove the sqrt of t^2 make sure you put absolute value and then you have to work with two interval when is negative you will put negative sign before and when is positive example [-1,1] after removing sqrt of t^2 that`s will equal to interval [-1,0] -t + [0,1] t . this is very important when you dealing with a change sign in interval