The line integral is the path of the function along a line having a continuous value.
The integral solved gives the value 111.
The given question is
[tex]\int\limits^c {z}^2 \, dx+ {x}^2 \, dy+ {y} ^2\, dz[/tex]
where C is the line from (1, 0, 0) to (4, 1, 3)
Here
x→ 1⇒ 4
y→ 0⇒1
z→ 0⇒ 3
Let t be defined be the range 0≤ t ≤ 1
Then x= 4t+1 : dx= 4dt
y= t ": dy= dt
z= 3t : dz= 3dt
Putting the values
[tex]\int\limits^1_0{9t^2} \, 4dt + \int\limits^1_0 {16t^2+1+8t} \, 3dt +\int\limits^1_0 {t^2} \,3dt[/tex]
= [36(1)²- 36(0)²]+ 3 [16(1)² +1+8(1)] - 3 [16(0)² +1+8(0)] + [3(1)²- 3(0)² ]
= 36 +75-3+3
= 111
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