(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A cylindrical coffee cup (8 cm in diameter and 10 cm tall) is filled to the brim

with coffee. Neglecting the weight of the cup, determine the torque at the handle

(2 cm from edge of cup 5 cm up from bottom of cup).

The easy way would be to just use the center of mass of the cup but that isnt the way its told to be done. So triple integrals it is.

density of water(ρ) = 1g/cm^{3}

2. Relevant equations

T =rxF

if R is from

3. The attempt at a solution

So to set up my coordinate's I put the origin in the center of the cylinder with x going out of the page y to the right and z up.

T =rxF

=rx mFg(Fg = force of gravity)

dT =rx dmFg(dm = ρdxdydz)

[itex]\int[/itex] [itex]\int[/itex][itex]\int[/itex] |r|ρgdzdydx

I think (tell me if I'm wrong) the bounds of the first integral (from left to right):

-4 to 4, -[itex]\sqrt{}[/itex]4-x^{2}to [itex]\sqrt{}[/itex]4-x^{2}, -5 to 5.

but where I'm having problem is finding an equation that relates r ( the vector from the handle to the cup) for each dm. If someone could help me out that would be great.

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# Homework Help: Using Triple integrals to solve torque around a point.

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