Physics, Calculus(alot of stuff together)= HELP!!

A rod extending between x=0 and x= 14.0cm has a uniform cross-
sectional area A= 9.00cm^2. It is made from a continuously changing
alloy of metals so that along it's length it's density changes
steadily from 2.70g/cm^3 to 19.3g/cm^3.

a) Identify the constants B and C required in the expression p= B +
Cx to describe the variable density.

b) The mass of the rod is given by
m= integral(pdV)= integral(pAdx)= integral 14cm/0 (14cm above integral sign and 0 below integral)(B+Cx)*(9.00cm^2)dx

(below the firs integral it says all material and below the
second integral it says all x

~this part a has gotten me confused as to what to do at all and the second part has gotten me even more confused!!


Thank YOU!

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  1. I forgot to include what it says under the integration:

    Carry out the integration to find the mass of the rod

    Now about that..what in the equation for the integral is the mass???
    it's not like they have a m anywhere in there..unless it's p but I started thinking it was to represent density..

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  2. Yes, p represents density in this case. The mass of the rod is
    A*(Integral of) p (x) dx
    where A is the cross sectional area of the rod.
    For p(x), use the function you derive in part (a).
    p(x) = A + bx
    p(0) = A = 2.70 g/cm^3 (That already tells you what A is)
    p(14) = A + B*14 = 19.3 g/cm^3
    Solve for B.

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  3. first of all I don't know how to use the #s given to put into the equation or do something else with them.

    I know...


    density changes from: 2.70g/cm^3 to 19.3g/cm^3

    How do I put it into p= B+Cx??

    would x= 14-0??
    C and B I'm not sure what they represent except that you said the mass is equal to B right?

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  4. and since p=m/V does that play a role in this as well?

    Again I have no idea what C represents.

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  5. I forgot t use the expression B + Cx for p, and used A + Bx instead. The method of solution is the same. In your case, B = 2.70, and C = 1.186
    p(x) = 2.70 + 1.186 x
    Note that p = 2.70 g/cm^3 when x = 0 and 9.30 g/cm^3 when p = 14 cm, as required.

    Now integrate A p(x) dx for the total mass

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  6. for this I got

    m= A integral 14\0 (B+Cx)dx
    (took out the A contstant A=9.00cm^2)

    m= 9.00 integral 14\0 (2.70+1.186x)dx =

    9.00(2.70x + 0.593x^2)|14 0 =
    (the 14 is above and 0 below the | sign)

    9.00[ 2.70(14) + 0.593(14)^2 ]=


    ~I'm not sure if I did the integration right and also I'm confused as to why you said
    "note that p= 2.70g/cm^3 when x=0 and 9.30g/cm^3 when p= 14cm, as required" does this apply to what I plug into the integration ?? I would think it wouldn't since I have to put in 14 and 0 in anyways and the 0 cancels out and all is left is the 14 right?

    Is this alright what I got??

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