Evaluate $\int_{(0,1)}^{(1,2)}\left(x^2+y\right) d x+\left(y^2+x\right) d y$
i. A straight line from $(0,1)$ and $(1,2)$
ii. A straight line from $(0,1)$ and $(1,1)$ and from $(1,1)$ to $(1,2)$
iii. The parabolic $x=t, y=t^2+1$
Answer (1)
Parameterize the straight line from ( 0 , 1 ) to ( 1 , 2 ) as x = t and y = 1 + t , where 0 ≤ t ≤ 1 , and evaluate the line integral, obtaining 3 14 .
Parameterize the straight line from ( 0 , 1 ) to ( 1 , 1 ) as x = t and y = 1 , where 0 ≤ t ≤ 1 , and the straight line from ( 1 , 1 ) to ( 1 , 2 ) as x = 1 and y = t , where 1 ≤ t ≤ 2 , and evaluate the line integral, obtaining 3 14 .
Parameterize the parabolic path x = t , y = t 2 + 1 as x = t and y = t 2 + 1 , where 0 ≤ t ≤ 1 , and evaluate the line integral, obtaining 3 14 .
Conclude that the line integral is path-independent and equals 3 14 in all three cases.
Explanation
Problem Setup We are asked to evaluate the line integral ∫ ( 0 , 1 ) ( 1 , 2 ) ( x 2 + y ) d x + ( y 2 + x ) d y along three different paths:
i. A straight line from ( 0 , 1 ) to ( 1 , 2 ) .
ii. A straight line from ( 0 , 1 ) to ( 1 , 1 ) and then a straight line from ( 1 , 1 ) to ( 1 , 2 ) .
iii. The parabolic path x = t , y = t 2 + 1 .
Path i: Straight Line i. Straight line from (0,1) to (1,2)
Parameterize the line as x = t and y = 1 + t , where 0 ≤ t ≤ 1 . Then d x = d t and d y = d t . Substituting these into the integral, we get:
∫ 0 1 ( t 2 + 1 + t ) d t + (( 1 + t ) 2 + t ) d t = ∫ 0 1 ( t 2 + 1 + t + 1 + 2 t + t 2 + t ) d t
= ∫ 0 1 ( 2 t 2 + 4 t + 2 ) d t = [ 3 2 t 3 + 2 t 2 + 2 t ] 0 1 = 3 2 + 2 + 2 = 3 2 + 4 = 3 14
Path ii: Two Straight Lines ii. Straight line from (0,1) to (1,1) and from (1,1) to (1,2)
Path 1: (0,1) to (1,1)
Parameterize the line as x = t and y = 1 , where 0 ≤ t ≤ 1 . Then d x = d t and d y = 0 . Substituting these into the integral, we get:
∫ 0 1 ( t 2 + 1 ) d t + ( 1 2 + t ) ( 0 ) = ∫ 0 1 ( t 2 + 1 ) d t = [ 3 1 t 3 + t ] 0 1 = 3 1 + 1 = 3 4
Path 2: (1,1) to (1,2)
Parameterize the line as x = 1 and y = t , where 1 ≤ t ≤ 2 . Then d x = 0 and d y = d t . Substituting these into the integral, we get:
∫ 1 2 ( 1 2 + t ) ( 0 ) + ( t 2 + 1 ) d t = ∫ 1 2 ( t 2 + 1 ) d t = [ 3 1 t 3 + t ] 1 2 = ( 3 8 + 2 ) − ( 3 1 + 1 ) = 3 7 + 1 = 3 10
Adding the results of the two integrals, we get 3 4 + 3 10 = 3 14 .
Path iii: Parabola iii. Parabolic path x = t , y = t 2 + 1
Parameterize the path as x = t and y = t 2 + 1 , where 0 ≤ t ≤ 1 . Then d x = d t and d y = 2 t d t . Substituting these into the integral, we get:
∫ 0 1 ( t 2 + t 2 + 1 ) d t + (( t 2 + 1 ) 2 + t ) ( 2 t ) d t = ∫ 0 1 ( 2 t 2 + 1 ) d t + ( t 4 + 2 t 2 + 1 + t ) ( 2 t ) d t
= ∫ 0 1 ( 2 t 2 + 1 + 2 t 5 + 4 t 3 + 2 t + 2 t 2 ) d t = ∫ 0 1 ( 2 t 5 + 4 t 3 + 4 t 2 + 2 t + 1 ) d t
= [ 3 1 t 6 + t 4 + 3 4 t 3 + t 2 + t ] 0 1 = 3 1 + 1 + 3 4 + 1 + 1 = 3 5 + 3 = 3 14
Final Answer In all three cases, the line integral evaluates to 3 14 . This suggests that the vector field is conservative, and the line integral is path-independent.
Examples
Line integrals are used in physics to calculate the work done by a force field along a path. For example, if you want to know how much energy it takes to move an object through an electric or gravitational field, you would use a line integral. This concept is also used in engineering to design pipelines and electrical circuits, ensuring that energy and fluids are efficiently transported.
