You are the greatest RUclips professor ever. You are really good at teaching thanks a lot. I’m taking a summer course in calc 2 and your videos are amazing. When I get my engineering degree I’ll be sure to donate to you for all the help you given me I truly appreciate it
Thank you, I really appreciate the kind words and support! Glad these videos have been able to help you. I wish you the best as you work towards your degree! :)
Hi. Love your videos. One suggestion I have is to invert your videos, as the white background is blinding on my tv. Thanks and keep up the great videos.
Thanks for the feedback! Glad you like the videos! I currently do not have any plans to change their style, however I can look into releasing alternate versions in the future with inverted colors, like you suggested. I’ll keep it in mind. Hope you continue to find the videos to be helpful regardless! :)
Hi! Great video! But in the book Calculus: Early Transcendentals the formula that they use for SA is is the integral from a to b 2pi f(x) sqrt(1+f'(x)^2)dx. Is the f(x) the same thing as r(x)? Because for the example at 19:40 they would not be the same thing I think?
Correct, f(x) and r(x) are not always the same. They can be, but it depends on the problem and if the right conditions are met. More specifically, if the variable you are working in terms of (x or y) matches the axis you are revolving around, then f(x) and r(x) are likely the same. So (for example) in example 1, they are the same because we are revolving around the x-axis and working in terms of x (both x, they match!). But in example 2 (at 19:40), we are working in terms of x but revolving around the y-axis (x and y, they do not match). So, if you wanted r(x) and f(x) (or r(y) and g(y) if working in terms of y) to always be the same in the formula as your textbook defines it, you would always need to work in terms of x when revolving around the x-axis, and always work in terms of y when revolving around the y-axis. However, I would not recommend this, as often times switching to work in terms of y in these types of problems is not very convenient when it comes to actually solving the integral. I talk about that a little but when introducing the formulas towards the beginning of the video. Hope this helps clear up any confusion!
This was a helpful video,thank you.... but to avoid spending too much time one of these questions during tests or exams just convert the bounds to be in terms of u and substitute into u and there you have your answer.
Glad it was helpful! And yes, if you prefer to convert the bounds to be in terms of u you can totally do it that way. It was always my personal preference to integrate in terms of u and then get back into terms of x, which is why I do it a lot in my videos, but you don't have to do it that way!
In this video we are calculating surface area of surfaces of revolution, which is different than volume. We want to find the total amount of area around the surface of the shape. So, we have to include L, the arc length of the curve as part of this calculation, since it helps describe what the curvature of the shape will be. The arc length L includes dx in its definition/formula, so dx it is still part of the surface area formula. L is not replacing dx, it includes dx. They are both part of the formula. The L formula only calculates arc length of a curve, but by multiplying it by the radius r(x) and 2π, we can find the area of a surface formed by revolving that curve around an axis (since that revolution will form circles if you were to look at cross sections of the surface). Does this help?
The function we are working with in that example is f(x)=x^3, and since we are revolving that curve around the x-axis, the curve itself will be the radius of the surface of revolution. So no matter where you are along the curve for a particular value of x, the radius will be given by x^3. So that's why the radius function r(x) is x^3. Hope this helps!
Most underrated channel over, your videos are amazing man
Thank you, I really appreciate that. Glad the videos are helpful for you! :)
You are the greatest RUclips professor ever. You are really good at teaching thanks a lot. I’m taking a summer course in calc 2 and your videos are amazing. When I get my engineering degree I’ll be sure to donate to you for all the help you given me I truly appreciate it
Thank you, I really appreciate the kind words and support! Glad these videos have been able to help you. I wish you the best as you work towards your degree! :)
Your videos are helping me so much!
underrated channel
Hi. Love your videos. One suggestion I have is to invert your videos, as the white background is blinding on my tv. Thanks and keep up the great videos.
Thanks for the feedback! Glad you like the videos! I currently do not have any plans to change their style, however I can look into releasing alternate versions in the future with inverted colors, like you suggested. I’ll keep it in mind. Hope you continue to find the videos to be helpful regardless! :)
@@JKMath You could even start a new channel called jk math inverted. Apply an inverted filter to your old videos and re-upload them.
@@JKMath Never change
thank you soo muchh broo, the playlist really helps me a lot in studying for finals
You're very welcome! Best wishes on your finals!
Hi! Great video! But in the book Calculus: Early Transcendentals the formula that they use for SA is is the integral from a to b 2pi f(x) sqrt(1+f'(x)^2)dx. Is the f(x) the same thing as r(x)? Because for the example at 19:40 they would not be the same thing I think?
Correct, f(x) and r(x) are not always the same. They can be, but it depends on the problem and if the right conditions are met. More specifically, if the variable you are working in terms of (x or y) matches the axis you are revolving around, then f(x) and r(x) are likely the same. So (for example) in example 1, they are the same because we are revolving around the x-axis and working in terms of x (both x, they match!). But in example 2 (at 19:40), we are working in terms of x but revolving around the y-axis (x and y, they do not match). So, if you wanted r(x) and f(x) (or r(y) and g(y) if working in terms of y) to always be the same in the formula as your textbook defines it, you would always need to work in terms of x when revolving around the x-axis, and always work in terms of y when revolving around the y-axis. However, I would not recommend this, as often times switching to work in terms of y in these types of problems is not very convenient when it comes to actually solving the integral. I talk about that a little but when introducing the formulas towards the beginning of the video. Hope this helps clear up any confusion!
This was a helpful video,thank you.... but to avoid spending too much time one of these questions during tests or exams just convert the bounds to be in terms of u and substitute into u and there you have your answer.
Glad it was helpful! And yes, if you prefer to convert the bounds to be in terms of u you can totally do it that way. It was always my personal preference to integrate in terms of u and then get back into terms of x, which is why I do it a lot in my videos, but you don't have to do it that way!
why not multiply it by dx instead of l just like in volume
In this video we are calculating surface area of surfaces of revolution, which is different than volume. We want to find the total amount of area around the surface of the shape. So, we have to include L, the arc length of the curve as part of this calculation, since it helps describe what the curvature of the shape will be. The arc length L includes dx in its definition/formula, so dx it is still part of the surface area formula. L is not replacing dx, it includes dx. They are both part of the formula. The L formula only calculates arc length of a curve, but by multiplying it by the radius r(x) and 2π, we can find the area of a surface formed by revolving that curve around an axis (since that revolution will form circles if you were to look at cross sections of the surface). Does this help?
sorry sir but i don’t understand why the radius is x cube in the example 1
The function we are working with in that example is f(x)=x^3, and since we are revolving that curve around the x-axis, the curve itself will be the radius of the surface of revolution. So no matter where you are along the curve for a particular value of x, the radius will be given by x^3. So that's why the radius function r(x) is x^3. Hope this helps!