Loved this video. My mind was blown when you made a quick visual proof for (a+b)^2=a^2+2ab+b^2, I knew how to use the theorem but I never thought about drawing the square. I also like how instead of saying "dx" you say " a lil bit of X" it really helps grasp the concept.
Thank you! I’m so glad there were nuggets that helped you. For all integrals, you can say that stuff equals the integral of a little bit of stuff, and build up from there.
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!
Im over here just trying to figure out how high I should make the marking on my styrofoam cup with base radus 1" top radius 1.625" and depth 4.3" so i can get exactly 11 in^3 of liquid. Depressing im an engineer and can't figure this out on my own, but i just ended up finding the equation relating the midpoint radius to the volume and interatively plugged in numbers until I got a solution. I couldn't find a mesuring cup.
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.
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.
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.
Loved this video. My mind was blown when you made a quick visual proof for (a+b)^2=a^2+2ab+b^2, I knew how to use the theorem but I never thought about drawing the square. I also like how instead of saying "dx" you say " a lil bit of X" it really helps grasp the concept.
Thank you! I’m so glad there were nuggets that helped you. For all integrals, you can say that stuff equals the integral of a little bit of stuff, and build up from there.
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!!
Thank you for this thorough and clear explanation.
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!
Great video thanks Raelene, really clear and succinct explanation of the setup.
Thank you Promar. It is a totally transferable approach.
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!
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.
i got this for homework. didnt get this in two hours (I'm a 11th grader don't attack me). watching this video helped a lot. thanks
Great method , never thought of using straight line equation , thanks alot
Thank you so much Raelene. Best wishes
Thank you
Very good derivation. I will return to your TH-cam channel for more of your tutorials.
Thank you - there are more to come!
Thank you so much, you save the day❤️
Awesome video. Thank you so much
just amazing
Very beautiful explanation.
Thank you so much.
wow thank you so much. this was great
great explanation, thank you
Im over here just trying to figure out how high I should make the marking on my styrofoam cup with base radus 1" top radius 1.625" and depth 4.3" so i can get exactly 11 in^3 of liquid.
Depressing im an engineer and can't figure this out on my own, but i just ended up finding the equation relating the midpoint radius to the volume and interatively plugged in numbers until I got a solution.
I couldn't find a mesuring cup.
I imagine measuring cups in cubic inches aren’t very common, either. You would need to convert to mL or ounces.
Thanks it really helps
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.
Gr8 video
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.
You deserve a great ......👍👍👍
An excellent explanation. Thanks a lot.
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
Ok now I know that its better do It ALL for x axis
I thought she would only use geometric formuls...