Comment on Let π = 5
FiniteBanjo@lemmy.today 8 months ago
I would be a smartass and leave Pi as a factor throughout and in the answer. I’m used to doing that in Calculus anyways.
V = πr^2^h
V = π⋅10^2^⋅10
V = π1000
BONUS SOLUTION:
V =∫~0~^10^ A⋅h dh
A = ∫~0~^10^ 2πr dr
V= ∫~0~^10^ h⋅∫~0~^10^ 2πr dr dh
π is a constant so we can safely remove it from A’s integral
A = π⋅∫~0~^10^ 2r dr
A = π⋅[r^2^]~0~^10^
A = π⋅( [10^2^] - [0^2^] )
A = π10^2^
A = π100
V = ∫~0~^10^ h⋅π100 dh
π100 is a constant so we can safely remove it from V’s integral
V = π100⋅∫~0~^10^ h dh
V = π100⋅[h]~0~^10^
V = π100⋅([10] - [0])
V = π100⋅10
V = π1000
It goes a lot deeper but I’m not bored enough for that, yet.
benignintervention@lemmy.world 8 months ago
If you really wanted to be through you’d start at a point, integrate out along dr for a line, then integrate in a circle through dtheta to derive the area before doing the rest