integral calculus 17-feb-2017 - university of pittsburghjordan/chem1000-s17/int-calc.pdf ·...
TRANSCRIPT
Integral calculusline integrals
From calculus, in the case of a single variable
1
01 0 , where f
x
x
dFF F x F x f x dx xdx
Suppose , ,dF M x y dx N x y dy
1 1,x y
0 0,x y
There is an infinite number ofdifferent paths from 0 0 1 1, ,x y x y to
1 1
00
, ,x
x y
ydF M x y x dx y x yN dy
Now, consider the case that two variables are at play.
Note how in the dx integralwe expressed y in terms of xand in the dy integral, weexpressed x in terms of y.
Rev. 2/27/2019
Example 2 3 , 3 4M x y N x y
integrate along from (x,y) = (0, 3) to (2,7) 2 3y x
2 7
0 3
2 7
0 3
2 22 70 3
32 3 2 3 4 32
11 98 92 2
8 11 992 2 2 2
11 9 11 916 18 49 7 9 34 2 4 2
126
c
ydF x x dx y dy
x dx y dy
x yx y
if dF is an exact differential, then
1 1 0 0, ,c c
y x
F Fdf dx dy F x y F x yx y
if dF not an exact differential, the integral depends on the path
2 7
0 3
2 2 2 70 3
2 3
2 6 4
9 6 2
4 19 42 2(49) 18
3 4
18
9
126
x y dx
x dx y dy
x x y
x y dy
y
(3,7)
(2,3)(0,3)
The integral is independent of the path, i.e., only depends on the endpoints.
The example given above corresponds to an exact differential.
We already integrated this along the most directpath. Now lets integrate from x = 0 to 2, keepingy = 3, and then from y = 3 to 7, keeping x = 2.
same as before
Example of an inexact differential
ab
b
(2,2)
2,0(0,0)
du dx xdy
2 2
0 02
2 20 0
2 2 2 20 00 0
2 2 42
2 2 6
c a
c b
dx xdy dx ydy
yx
dx xdy dx dy x y
There are also line integrals with three independent variables.
path a is along the liney = x
, , , , , ,c
M x y z dx N x y z dy P x y z dz
Path a
Path b
(y = x)
Example: du yzdx xzdy xydz
This is an exact differential (with u = xyz)Exact differentials are important in thermodynamics
reversible processes proceed infinitely slowlycyclic processes: begin and end
in same state
ex. 0 0 0 1 1 1 0 0 0, , , , , ,P T V P T V P T V
A line integral for a cyclic process is written as du
0du if du is an exact differential
If du is not an exact differential 0du obvious such a process is irreversible, since goingaround a cycle does not fully restore the systemto its initial state.
For a gas (or liquid) revd PdV
This is not an exact differential
rev rev revcd not
Example: 1 mol of ideal gas at T = 500 K, V = 20.0 l
(a) expand at constant T to V = 40.0 l(b) cool at constant volume to 300 K(c) compress to V = 20 l at T = 300 K(d) heat to T= 500 K at V = 20 l
40
20
2
1
401.0 8.3145 500 288220
rev
n
nRTPdV dVV
V JKnRT n K JV mol
(a)
mol
PV nRTusing
Step (b) , so work = 00V
Step (c) 1729revnRTPdV dV JV
Step (d) since there is no volume change 0rev
To go around the cycle, 1153rev J
cdq q dq = heat transferred in a reversible process
It is also not an exact differential
dU dq dU q
so for a cyclic process q = ‐for the process worked out above, heat had to be provided
,rev pdq C dT constant P and no reaction or phase change, Cp = heat capacity
adiabatic process: 0addq
ad addU d
Double integrals
2 2
1 1
,a b
a bI f x y dxdy
When doing the first integral, treatthe other variable as a constant
2
0 0
22
00
2 2
0
3 2 2
0
4
42
2
23 2
a b
ba
a
a
x xy dydx
xyx y dx
bx b x dx
bx b x
Example
Double integrals
3 2 2
0 0
332 2
00
33 3 3
0 0
4 40
2
3
273 9 213
21 214 4
a x
xa
a a
a
x xy y dydx
yx y xy dx
xx x dx x dx
x a
A somewhat more complicated case is when the upper limit of the inner integral depends on the variable used for the outer integration.Example
Look at bottom of objecty ranges from 0 to x2 ‐ 4 (limits of )
If upper surface were flat (parallel to xy plane), with f = 2
2
32 0 2 2 222 4 2
2 2 4 2 4 21.33x
xdydx x dx x
If the upper surface were flat with f = 6, the volume would be 63.9
Actually, the upper surface varies from 2 to 6 according to f = 2‐y
22
22 0 2 042 4 2
22 4 22 22 2
2 2
52 4 2 3 222
2 22
4 8 162 4 2 82 2
1 66 16 162 10 3
xx
yV y dydx y dx
x x xx dx x dx
xx x dx x x
A more challenging example
In QM, for the H atom and for a particle in a 3‐D box, we encounter triple integrals
2 2 2
1 1 1
2 , ,Prob a b c
a b cx y z dxdydz
A special case is when the integral factors
2 2 2
1 1 1
2 2 2
1 1 1
a b c
a b c
a b c
a b c
f x g y h z dzdydx
f x dx g y dy h z dz
It may be easier to do an integralin polar or spherical coordinates
This is the case for the H atom.
Probability of finding a particle in the specified volume (rectangular box)
Integrals involving polar coordinates 2 2 2a x y aBe Be
ha
2
2
22 2
0 0
2 2 20 0
2
2
a
a
B e d d
B e d
Ba
Note, we pick up a factor of whenintegrating in polarcoordinates
Normalize2 21
2B aB
a
1f ha
if = a height = 0if = 0 height = h
2 3 2 2 20 2 0
01 2 2
2 3 2 3 3aa
a a hah d d h ha a
2 2
0 0 0, , sinf r r drd d
Integrals in spherical coordinatesHere the integral is over all space. 2 is max angle for and is the max anglefor
2 integrated over all space should give 1.