~it was bumped over to the next page so I recopied it and added my answer for the question and was wondering if my integration was correct since you said
DrWls: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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Q--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
~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 ]=
1,386.252g
~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??
You have the integration right, I did not check the calc work. The boundry values apply to the density (2.7,9.3) and you applied them correctly.
Thank you Bob =D
Can someone please check my Calculus work and see if it is correct?
THANK YOU =D
It is right.
The only thing that can be improved is to substitute the numbers including the units for the quantities. Then you never have to think about the units, the maths itself will take care of that.
Okay then thank you I'll check that =D
To check if your integration is correct, let's go through it step by step.
a) To identify the constants B and C required in the expression p = B + Cx to describe the variable density, you need to use the given information about the density changing from 2.70 g/cm^3 to 19.3 g/cm^3.
At x = 0, the density is 2.70 g/cm^3. This gives us the equation:
2.70 = B + C(0)
2.70 = B
At x = 14 cm, the density is 19.3 g/cm^3. This gives us another equation:
19.3 = B + C(14)
Solving these equations simultaneously will give us the values of B and C.
b) Now let's move on to the integration. The formula for the mass of the rod is given by:
m = ∫(pdV) = ∫(pAdx) = ∫[B + Cx]Adx
You correctly removed the constant A from the integral, as it can be pulled out as a constant. So, the integral becomes:
m = 9.00 ∫[B + Cx]dx
Now we can integrate this expression. The antiderivative of B is Bx, and the antiderivative of Cx is Cx^2/2. So, integrating the expression will give us:
m = 9.00[Bx + Cx^2/2]
To evaluate the integral within the limits of integration (0 and 14 cm), we substitute the values of x into the expression and subtract the value at the lower limit from the value at the upper limit.
m = 9.00[B(14) + C(14)^2/2 - (B(0) + C(0)^2/2)]
Simplifying this, we get:
m = 9.00[14B + 98C - 0]
Since we found that B = 2.70 earlier, we can substitute this value into the equation:
m = 9.00[14(2.70) + 98C]
Now we need to use the information given in the question, which states that the density is 9.30 g/cm^3 when p = 14 cm. Note that this information does not affect the integration process. It is just a consistency check to verify that our expression for density is correct.
Finally, we have the expression for mass in terms of C:
m = 9.00[14(2.70) + 98C] = 9.00[37.8 + 98C]
Multiplying this out:
m = 339.6 + 882C
From here, we cannot determine the exact value of the mass without the value of C. You would need to know the specific value of C to calculate the numeric result.
In summary, your integration process was correct, but you missed an important step in solving for the constants B and C using the given density values. Remember to substitute the values of x into the integrated expression and apply the limits of integration to calculate the final result.