Which Of The Following Examples Satisfy The Hypotheses Of The Extreme Value Theorem On The Given Interval?A.

1) The line integral value is : ∫F dr = 6

2) The line integral value is : ∫F dr = 6

3) The line integral value is : ∫F dr = 6

Here we are given:

\(F.dr = (3x + 2y)dx + (2x -2z)dy + (3x -2y)dz,\)

where\(\vec{F}\) is a conservative field

So,

f(x, y , z) = \(\int\limits (3x + 2y)dx + (2x -2z)dy + (3x -2y)dz,\)

f(x , y , z) = (3zx +2yx) + (2xy – 2zy) + (3xz – 2yz) + c(x , y , z)

\(f(x, y, z) = (6xz + 4xy – 4yz) + c(x, y, z)\)

Now substitute the values of x , y ,z ,

Line segment from (0, 0 , 0) to (1,1 ,1)

∫F dr = f(1,1,1) – f(0,0,0)

= |6 + 4 – 4 |- |0 + 0 – 0|

∫F dr = 6

Line segment from (0,0,0) to (0,0,1) to (1,1,1)

First we take (0,0,0) to (0,0,1)

\(\int _c \vec{F}.\vec{dr}=f(0,0,1)-f(0,0,0) =[6(0)(1)+4(0)(0)-4(0)(1)]-[6(0)(0)+4(0)(0)-4(0)(0)] =[0+0-0]-[0+0-0]\)

∫F dr = 0

\(\Rightarrow \int _{c}\vec{F}.\vec{dr}=0+6 [F.dr = 6\)

Line segment from (0,0,0) to (1,0,0) to (1,1,0) to (1,1,1)

First we take (0,0,0) to (1,0,0)

\(\int _c \vec{F}.\vec{dr}=f(1,0,0)-f(0,0,0) =[6(1)(0)+4(1)(0)-4(0)(0)]-[6(0)(0)+4(0)(0)-4(0)(0)] =[0+0-0]-[0+0-0]\int _{c}\vec{F}.\vec{dr}=0\)

Next we take (1,0,0) to (1,1,0)

\(\int _c \vec{F}.\vec{dr}=f(1,1,0)-f(1,0,0) = [6(1)(0) + 4(1)(1) − 4(1)(0)] – [6(1)(0) + 4(1)(0) — 4(0)(0)] = [0+4-0] – [0+0-0]\int _{c}\vec{F}.\vec{dr}=4\)

Lastly we take (1,1,0) to (1,1,1)

\(F.dr = f(1,1,1) ƒ(1,1,0) = [6(1)(1) +4(1)(1) − 4(1)(1)] – [6(1)(0) + 4(1)(1) — 4(1)(0)] =[6+4-4]-[0+4-0]\int _{c}\vec{F}.\vec{dr}=2\)

Adding the three results we get

\(\Rightarrow \int _{c}\vec{F}.\vec{dr}=0+4+2\\\\\int\limits F.dr = 6\)

Therefore we see that the Line integral for the three cases comes out to be same between the initial and final points since it is independent of the path taken.

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