Good catch! My apologies on that mistake. You are correct, since we are revolving around the y-axis, the integral should be entirely in terms of y for the disk/washer method. Can't believe I didn't catch that when I edited this video, especially since I correctly set up the first integral in terms of y, but not the second. To fix the mistake, solve for x in the function y=x^2+1 to get x=√(y-1) as the inner radius and substitute that in for where I put x^2+1. And then of course, there should be a dy instead of dx at the end of the integral. Thanks for pointing this out, again, my apologies on this mistake and any confusion it may have caused.
This is a question on average radius r at 4:30. Normally r is just equal to x. Instead of x, can we just use (a+b)/2 which is the actual average radius? Conceptually, we can take the area bounded by f(x) from a to b multiplied by the thickness 2pi(r) where r = (a+b)/2.
No, the radius always needs to be defined in terms of x, it will not be constant. I see why you would think it could be (a+b)/2, but remember that in that diagram I drew, that was just ONE cylinder that is representative of many cylinders we are using to find the volume using the shell method. So each cylinder will have a different average radius. That's why we need to use x, as the average radius of each cylinder will be dependent on where those cylinders are located along the x-axis. You can test this out with the example problems in this video, if you change the radius from x to (a+b)/2 you will get a completely different answer, and it will be an incorrect answer. Hope this helps!
Since we are revolving around the y-axis, the height will be measured in terms of x, so you want to look up and down the area, not left to right. From this we can see that the height is between the x-axis and the curve, so the height is just the function that represents the curve, just like I labeled it in the video. Hope this helps!
@@ritvikindupuri2388 You could think of it that way, sure. But in general when working in terms of x, you look from up to down, and when working in terms of y, you look from right to left.
i think there is a mistake at the time 46:15 the therms should in y because the revolution about y axis by disc method
Good catch! My apologies on that mistake. You are correct, since we are revolving around the y-axis, the integral should be entirely in terms of y for the disk/washer method. Can't believe I didn't catch that when I edited this video, especially since I correctly set up the first integral in terms of y, but not the second. To fix the mistake, solve for x in the function y=x^2+1 to get x=√(y-1) as the inner radius and substitute that in for where I put x^2+1. And then of course, there should be a dy instead of dx at the end of the integral. Thanks for pointing this out, again, my apologies on this mistake and any confusion it may have caused.
@JKMath you are great go on sir
You explain so well! very organized and easy to understand 🙏🏻Thanks JK Math!
You're welcome!
THANK YOU SO MUCH! This help me a lot with my final test!
You're very welcome, hope your test went well! :)
Thanks! Watched khan academy first, and it finally clicked with this video
Why so less views 💀 on a masterpiece
Thank you so much
You're welcome!
This was amazing, thank you so much!!
You're welcome! Glad the video could help :)
This is a question on average radius r at 4:30. Normally r is just equal to x. Instead of x, can we just use (a+b)/2 which is the actual average radius? Conceptually, we can take the area bounded by f(x) from a to b multiplied by the thickness 2pi(r) where r = (a+b)/2.
No, the radius always needs to be defined in terms of x, it will not be constant. I see why you would think it could be (a+b)/2, but remember that in that diagram I drew, that was just ONE cylinder that is representative of many cylinders we are using to find the volume using the shell method. So each cylinder will have a different average radius. That's why we need to use x, as the average radius of each cylinder will be dependent on where those cylinders are located along the x-axis. You can test this out with the example problems in this video, if you change the radius from x to (a+b)/2 you will get a completely different answer, and it will be an incorrect answer. Hope this helps!
@@JKMath
Thanks for the reply.
It's cool
Just thought I'd point out that you don't even need to use an integral to find the area of the bottom part - it's a rectangle!
keep doing what you’re doing bro, this helps me so much
Thank you, I appreciate that. Glad the videos help! Will definitely continue to make more :)
For Example 1 why is the height not 1- x^3 as the height would be the distance from x=1 to y=x^3
Since we are revolving around the y-axis, the height will be measured in terms of x, so you want to look up and down the area, not left to right. From this we can see that the height is between the x-axis and the curve, so the height is just the function that represents the curve, just like I labeled it in the video. Hope this helps!
@@JKMath I see it’s mainly cause since X is a vertical line we have to look up to down rather than side to side (y=)
@@ritvikindupuri2388 You could think of it that way, sure. But in general when working in terms of x, you look from up to down, and when working in terms of y, you look from right to left.
@@JKMath oh i see
Like this comment if you are here from my class