This is a seen modelling and problem solving question for a test and is really important. I have tried for hours but these questions are really getting to me. Here is my working out, so could someone please check it for me. I have:

Area = CDEF + BCFG - DAFE - ABGF

CDEF = 1.8 x 0.3 = 0.54

ABGF = 0.9 x 0.3 = 0.27

BCFG = f (0.72/V) limits of b = 2.5 and a = 0.5

= 0.72ln(V)

= [(0.72ln(2.5)) - (0.72ln(0.5))]

= 1.258795297

DAFE = f (1.8/V) limits of b = 0.4 and a = 0.2

= 1.8ln(V)

= [(1.8ln(0.4)) - (1.8ln(0.2))]

= 1.247884925

Area = 0.54 + 1.158795297 - 1.1247664925 - 0.27

=0.181130372 * 100

=18.1130372

But the real answer is 118.7

2. Originally Posted by zapparage This is a seen modelling and problem solving question for a test and is really important. I have tried for hours but these questions are really getting to me. Here is my working out, so could someone please check it for me. If it is the shaded area that you want to solve by integration, then here is one way.

An element of the area, dA = [(V of curve BC) -(V of curve DA)]*dP
dA = [1.8/P -0.72/P]dP
dA = [1.08/P]dP

According to the figure, P goes from 0.6 to 1.8,
so, integrating from P = 0.6 to P = 1.8,
A = INT(0.6 to 1.8)[1.08/ P]dp
A = (1.08)[ln(P)]|(0.6 to 1.8)
A = (1.08)[ln(1.8) -ln(0.6)]
A = 1.1865 -----------------**

That is if P and V are of the same power of 10.
But according to the figure,
P is in 10^5
while V is in 10^(-3)
So if you multiply 10^5 by 10^(-3), you will get 10^2
Therfore, A must be multiplied by 10^2
So, A = 1.1865 *10^2 = 118.65 joules

3. Wow thanks a lot. How would that work for the one below, of which I'm also having a great deal of trouble with? 4. Please can someone help me with the second. I am going mental.

5. Originally Posted by zapparage Wow thanks a lot. How would that work for the one below, of which I'm also having a great deal of trouble with? Since the figure is similar to the previous one, then the procedure is similar.
Get a horizontal dA whose length is (V of curve CB minus V of curve DA), and whose depth is dP.
The limits of integration, according to the graph, is from P=1.0 to P=3.0.

Your main problem is how to express V in terms of P.

I will do the V of curve CB:
P = 0.575 +5.79e^(-0.29V)
e^(-0.29V) = (P -0.575)/5.79
Take the natural log of both sides,
-0.29V = ln(P -0.575) -ln(5.79)
V = -(1/0.29)ln(P -0.575) +(1/0.29)ln(5.79)
V = -3.44828ln(P -0.575) +6.05563 --------------(i)

For the V of curve DA, following the above,
V = -(1/1.44)ln(P -0.88) +(1/1.44)ln(8.965)
V = -0.69444ln(P -0.88) +1.52314 ---------------(ii)

So,
dA = [(-3.44828ln(P -0.575) +6.05563) -(-0.69444ln(P -0.88) +1.52314)]*dP
dA = [-3.44828ln(P -0.575) +0.69444ln(P -0.88) +4.53249]dP

And,
A = INT.(1 to 3)[-3.44828ln(P -0.575) +0.69444ln(P -0.88) +4.53249]dP

Umm, that's a long one.
You can continue from here.
Use INT[ln(u)]du = u*ln(u) -u +C ------***

You get A = 7.19421 from that.
Then, actual A = 7.19421 * 10^2 = 719.42 joules --------answer.

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