Really enjoyed the video. The only suggestions I would make is y would have been more natural to use than x. The second thing which would be helpful would have been to show why the relationship of d/a*sqrt(2) = x/h based on the assumption that the triangle was 45degrees (1,1,SQRT(2)) relation. So naturally if we have d/2 for the smaller triangles base the height is also d/2 which is also equal to in your case x and a*sqrt(2)/2 for the base of the larger triangle has the same height and this is equal to h. Thus (d/2)/(a*sqrt(2)/2) = x/h reduces to d/a*sqrt(2)=x/h as noted.
@2:36: I don't understand why you wouldn't look at the thing side on. It would really simplify things, using the more common formula for the area of a square (side length squared) and leaving out the whole sqrt(2) squaring and cancelling business. If b is the side length of the small square (that with diagonal d), then you can just say: b/a = x/h b = ax/h (similar triangles) => V = lim sum b^2 𝛿x = lim sum a^2 x^2/h^2 𝛿x = a^2/h^2 int x^2 dx = ...
Could somebody help me?: I tried using the same logic in this video to calculate the volume of a hemisphere (which I know is 2/3(Pi*r^3), I just wanted to derive it myself) - because like the square-based pyramid, a hemisphere has a similar cross-section (a circle). But I must've went wrong somewhere: When you look at a hemisphere side-on, it looks like a semicircle. In the video, Eddie changes the variable "x" - the height. In my workings, I changed the variable "theta" - the angle between the radius of the semicircle and the x-axis as you travel anticlockwise from the x-axis to the y-axis. I found this to be the easier variable to change; the height of a cross-section in the semicircle/hemisphere can be expressed as rSin(theta), and the radius of a cross-section circle can be expressed as rCos(theta) - I don't need to prove that, this is simply how the functions Sin and Cos are defined The area of a cross-section circle = Pi * (rCos(theta))^2 = Pi * r^2 * Cos(theta)^2 The "thickness" of this cross-section is the change in height, which I've just defined to be rSin(theta) And so the integral I wrote was: [Pi * r^3 * Cos(theta)^2 * Sin(theta)] upper and lower bound of 90 and 0 respectively (degrees) (Maybe my error is here - this is just what I thought is the correct integral in this method) I took Pi * r^3 out of the integral because they are constants and integrated Cos(theta)^2 * Sin(theta). This gives you -1/3 * cos(theta)^3 Plug in the upper bound (90) to get 0; plug in the lower bound (0) to get -1/3. 0 - -1/3 = 1/3 And finally, do Pi * r^3 * 1/3 So my final answer was, the volume of a hemisphere = 1/3(Pi*r^3) ...which is exactly half of the correct answer :( So I must be close right?! Where was my mistake? Please advise
I'm not positive, but I think the issue is that the thicknesses of your discs isn't uniform. Essentially, a small change in theta will result in a different thickness when theta is small vs. the thickness the same small change in theta will give you when theta is larger. This variability is problematic. Also, remember that when doing calculus it's vital to use radians, you aren't allowed to use degrees. The better way to do this problem is as a simple solid of revolution. Rotate a quarter circle around whichever axis you want and then slice it up into discs (i.e. miniature cylinders).
@@JohnSmith-rf1tx Hiya; I managed to figure it out myself in the end! I said the "thickness" of my cross-section circles could be expressed as rSin(theta) - not true, that is (as mentioned previously) the height that the circle is above the x-axis. The thickness is the CHANGE in this height, so I had to differentiate rSin(theta) first (which gets you rCos(theta)) before plugging it into the integral Which means I had to integrate Pi * r^3 * Cos(theta)^3 instead with respect to theta. Following the exact same steps from there gets: Volume of a hemisphere = 2/3(Pi*r^3) Which is correct!! :D - I was very pleased! So the method Eddie showed in this video CAN also be used to derive the formula for the volume of a [hemi]sphere (And yes I did actually use radians, I just said degrees in my comment in the hope that more people would understand it!)
Really enjoyed the video. The only suggestions I would make is y would have been more natural to use than x. The second thing which would be helpful would have been to show why the relationship of d/a*sqrt(2) = x/h based on the assumption that the triangle was 45degrees (1,1,SQRT(2)) relation. So naturally if we have d/2 for the smaller triangles base the height is also d/2 which is also equal to in your case x and a*sqrt(2)/2 for the base of the larger triangle has the same height and this is equal to h. Thus (d/2)/(a*sqrt(2)/2) = x/h reduces to d/a*sqrt(2)=x/h as noted.
Absolutely amazing
@2:36: I don't understand why you wouldn't look at the thing side on. It would really simplify things, using the more common formula for the area of a square (side length squared) and leaving out the whole sqrt(2) squaring and cancelling business.
If b is the side length of the small square (that with diagonal d), then you can just say:
b/a = x/h b = ax/h (similar triangles)
=> V = lim sum b^2 𝛿x = lim sum a^2 x^2/h^2 𝛿x = a^2/h^2 int x^2 dx = ...
Have you taken into consideration of the slant when doing so?
Helped me a looooot thank you :)
you are genius bro
Goat
how do you proof the formula of the volume of a pyramid with a rectangular base ?
Thanks
Could somebody help me?:
I tried using the same logic in this video to calculate the volume of a hemisphere (which I know is 2/3(Pi*r^3), I just wanted to derive it myself) - because like the square-based pyramid, a hemisphere has a similar cross-section (a circle).
But I must've went wrong somewhere:
When you look at a hemisphere side-on, it looks like a semicircle. In the video, Eddie changes the variable "x" - the height. In my workings, I changed the variable "theta" - the angle between the radius of the semicircle and the x-axis as you travel anticlockwise from the x-axis to the y-axis. I found this to be the easier variable to change; the height of a cross-section in the semicircle/hemisphere can be expressed as rSin(theta), and the radius of a cross-section circle can be expressed as rCos(theta) - I don't need to prove that, this is simply how the functions Sin and Cos are defined
The area of a cross-section circle = Pi * (rCos(theta))^2 = Pi * r^2 * Cos(theta)^2
The "thickness" of this cross-section is the change in height, which I've just defined to be rSin(theta)
And so the integral I wrote was: [Pi * r^3 * Cos(theta)^2 * Sin(theta)] upper and lower bound of 90 and 0 respectively (degrees)
(Maybe my error is here - this is just what I thought is the correct integral in this method)
I took Pi * r^3 out of the integral because they are constants and integrated Cos(theta)^2 * Sin(theta). This gives you -1/3 * cos(theta)^3
Plug in the upper bound (90) to get 0; plug in the lower bound (0) to get -1/3. 0 - -1/3 = 1/3
And finally, do Pi * r^3 * 1/3
So my final answer was, the volume of a hemisphere = 1/3(Pi*r^3)
...which is exactly half of the correct answer :(
So I must be close right?! Where was my mistake? Please advise
I'm not positive, but I think the issue is that the thicknesses of your discs isn't uniform. Essentially, a small change in theta will result in a different thickness when theta is small vs. the thickness the same small change in theta will give you when theta is larger. This variability is problematic. Also, remember that when doing calculus it's vital to use radians, you aren't allowed to use degrees.
The better way to do this problem is as a simple solid of revolution. Rotate a quarter circle around whichever axis you want and then slice it up into discs (i.e. miniature cylinders).
@@JohnSmith-rf1tx Hiya; I managed to figure it out myself in the end!
I said the "thickness" of my cross-section circles could be expressed as rSin(theta) - not true, that is (as mentioned previously) the height that the circle is above the x-axis. The thickness is the CHANGE in this height, so I had to differentiate rSin(theta) first (which gets you rCos(theta)) before plugging it into the integral
Which means I had to integrate Pi * r^3 * Cos(theta)^3 instead with respect to theta. Following the exact same steps from there gets: Volume of a hemisphere = 2/3(Pi*r^3)
Which is correct!! :D - I was very pleased!
So the method Eddie showed in this video CAN also be used to derive the formula for the volume of a [hemi]sphere
(And yes I did actually use radians, I just said degrees in my comment in the hope that more people would understand it!)