I was spending so long solving this exact problem, and I said "this is getting really messy, I must be doing something wrong" Turns out I was on the right track all along, and calculus is just tedious XD Great explanation, you made this really easy to understand, thank you
Thank you! All calculus problems of accumulations start off this way: stuff equals the integral of a little bit of stuff! You may find it more enjoyable, as I do, to think of calculus (and mathematics generally) as meditative rather than tedious.
@@RaeleneMaths Thank you very much hahaha. I'm spending lots of hours on my Mechanics of Materials problem specifically about the deformation caused by an applied load and self-weight on a hanging bar with different diameter. Due to my ego, I forcefully tried to solve it by intuition and stock knowledge of disk method, trigo, and algeb but I became frustrated and went on yt hahaha. Thank you again you're so good on teaching!!
Expired Clan thank you very much! I aim to build understanding on a few versatile and powerful cornerstones. Like a capsule wardrobe but for mathematics!
I accidentally wrote y=R for my upper bound, because I misread the h in green as an R. In the problem I had, h and R both had a measurement of 4, so the volume worked out to be the same. However when I looked at the problem later, my bounds didn't make any sense. Glad I caught my mistake.
Hi, can I ask on how you can find the height wherein the volume of the frustum is equal? Eg, the total volume is 100 m^3, how can I know the height where I can get 50 m^3? Thank you in advance
Do you know the two radii of the frustum, r and R? If not, then you cannot solve for a specific height of the frustum using the formula V = πh(r^2 + rR + R^2)/3. You can figure out the height of the frustum as a ratio of the original (full, large) cone's height. If the full cone's volume is 100 m^3 and the frustum's volume is 50 m^3, then the volume of the small removed cone is also 50 m^3, which is 1/2 of the full cone's volume. 1/2 * V_full = V_small 1/2* π/3*HR^2 = π/3*h*r^2 1/2* HR^2 = h*r^2, and so by similarity H/(cuberoot2) * R/(cuberoot2) * R/(cuberoot2) = h*r*r so h, the height of the small removed cone is H/(cuberoot2), which means that the height of the frustum is the full cone's height minus the small cone's height: h_frustum = H - h = H - H/(cuberoot2) = H(1 - 1/(cubert(2)) So h_frustum ~ 0.206*H, which is about 20% of the original full cone's height.
Thank you for this thorough and clear explanation.
OMG................ that was a Perfect Video on finding the volume of frustum using Calculus !!! Thank you... YOU are VERY GOOD!! :D ..
Thank you! I’m glad you enjoyed it. It’s a repeatable process for so many calculations involving integrals.
Thank you so much Raelene. Best wishes
Awesome video. Thank you so much
I was spending so long solving this exact problem, and I said "this is getting really messy, I must be doing something wrong"
Turns out I was on the right track all along, and calculus is just tedious XD
Great explanation, you made this really easy to understand, thank you
Thank you! All calculus problems of accumulations start off this way: stuff equals the integral of a little bit of stuff! You may find it more enjoyable, as I do, to think of calculus (and mathematics generally) as meditative rather than tedious.
@@RaeleneMaths Thank you very much hahaha. I'm spending lots of hours on my Mechanics of Materials problem specifically about the deformation caused by an applied load and self-weight on a hanging bar with different diameter. Due to my ego, I forcefully tried to solve it by intuition and stock knowledge of disk method, trigo, and algeb but I became frustrated and went on yt hahaha. Thank you again you're so good on teaching!!
Great method , never thought of using straight line equation , thanks alot
Great video thanks Raelene, really clear and succinct explanation of the setup.
Thank you Promar. It is a totally transferable approach.
great explanation, thank you
Thank you so much, you save the day❤️
thank you. You help me a lot. Clear that i can understand tho im just 12. But Im rlly into maths!! brilliant video !
It’s wonderful that you love math and that you’re learning integral calculus at 12 years old! Keep learning this beautiful subject!
Your understanding of math is nearly unprecedented. Wow, great video.
Expired Clan thank you very much! I aim to build understanding on a few versatile and powerful cornerstones. Like a capsule wardrobe but for mathematics!
wow thank you so much. this was great
just amazing
Very good derivation. I will return to your UA-cam channel for more of your tutorials.
Thank you - there are more to come!
Very beautiful explanation.
Thank you so much.
you're so good at explaining it's so easy to understand. looking forward for more tutorials from you!
Many thanks; I'm happy it is clear for you.
Thanks it really helps
Gr8 video
An excellent explanation. Thanks a lot.
Thank you
I accidentally wrote y=R for my upper bound, because I misread the h in green as an R. In the problem I had, h and R both had a measurement of 4, so the volume worked out to be the same. However when I looked at the problem later, my bounds didn't make any sense. Glad I caught my mistake.
Hi, can I ask on how you can find the height wherein the volume of the frustum is equal? Eg, the total volume is 100 m^3, how can I know the height where I can get 50 m^3? Thank you in advance
Do you know the two radii of the frustum, r and R? If not, then you cannot solve for a specific height of the frustum using the formula V = πh(r^2 + rR + R^2)/3.
You can figure out the height of the frustum as a ratio of the original (full, large) cone's height. If the full cone's volume is 100 m^3 and the frustum's volume is 50 m^3, then the volume of the small removed cone is also 50 m^3, which is 1/2 of the full cone's volume.
1/2 * V_full = V_small
1/2* π/3*HR^2 = π/3*h*r^2
1/2* HR^2 = h*r^2, and so by similarity
H/(cuberoot2) * R/(cuberoot2) * R/(cuberoot2) = h*r*r
so h, the height of the small removed cone is H/(cuberoot2), which means that the height of the frustum is the full cone's height minus the small cone's height:
h_frustum = H - h = H - H/(cuberoot2) = H(1 - 1/(cubert(2))
So h_frustum ~ 0.206*H, which is about 20% of the original full cone's height.
You deserve a great ......👍👍👍
Ok now I know that its better do It ALL for x axis
thanks for the awesome explanation guess I'm just dumb
Can I clarify anything?
Did you forget to keep the "h" in the numerator of hR/h?
nvm you just corrected yourself haha
@@laurengilmore1307 yes, just after introducing it, I forgot it! But picked it up again a minute later
I thought she would only use geometric formuls...