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

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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?