There have been multiple accounts created with the sole purpose of posting advertisement posts or replies containing unsolicited advertising.
Accounts which solely post advertisements, or persistently post them may be terminated.
FiniteBanjo , 3 months ago (edited 3 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 = π⋅100⋅10 V = π1000 BONUS SOLUTION: V =∫0^10^ A⋅h dh A = ∫0^10^ 2πr dr V= ∫0^10^ ∫0^10^ h⋅2πr dr dh h is a constant for A’s integral so we can safely move it into V’s integral 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. EDIT: Hang on. I’m wrong with that height integral. Can somebody help remind me?
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 = π⋅100⋅10
V = π1000
BONUS SOLUTION:
V =∫0^10^ A⋅h dh
A = ∫0^10^ 2πr dr
V= ∫0^10^ ∫0^10^ h⋅2πr dr dh
h is a constant for A’s integral so we can safely move it into V’s integral
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
It goes a lot deeper but I’m not bored enough for that, yet.
EDIT: Hang on. I’m wrong with that height integral. Can somebody help remind me?