• @[email protected]
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    6
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    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 = πr2h

    V = π⋅102⋅10

    V = π⋅100⋅10

    V = π1000


    BONUS SOLUTION:

    V =∫010 A⋅h dh

    A = ∫010 2πr dr

    V= ∫010010 h⋅2πr dr dh

    h is a constant for A’s integral so we can safely move it into V’s integral

    V= ∫010 h⋅∫010 2πr dr dh

    π is a constant so we can safely remove it from A’s integral

    A = π⋅∫010 2r dr

    A = π⋅[r2]010

    A = π⋅( [102] - [02] )

    A = π102

    A = π100

    V = ∫010 h⋅π100 dh

    π100 is a constant so we can safely remove it from V’s integral

    V = π100⋅∫010 h dh

    V = π100⋅[h]010

    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?

    • @benignintervention
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      38 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