Line Integrals
We can represent a path or curve, 𝑐 , in the plane in the following 3 ways:
1. 𝑐⃗(𝑡) =< 𝑥 (𝑡 ), 𝑦(𝑡 ) > (vector form of a curve)
2. 𝑐⃗(𝑡) = 𝑥(𝑡)𝑖⃗ + 𝑦(𝑡)𝑗⃗ (another vector form of a curve)
3. 𝑥 = 𝑥 (𝑡 ), 𝑦 = 𝑦(𝑡 ) (parametric form of a curve)
In each case t will be in some interval (eg, 𝑎 ≤ 𝑡 ≤ 𝑏)
From first year calculus we know that the length of a curve, 𝑥 = 𝑥(𝑡 ), 𝑦 = 𝑦(𝑡 ),
𝑎 ≤ 𝑡 ≤ 𝑏, which we can also write as 𝑐⃗(𝑡) =< 𝑥 (𝑡 ), 𝑦(𝑡 ) >, is given by:
𝑏 𝑑𝑥 2 𝑑𝑦 2 𝑏
Length of curve= √
∫𝑐 𝑑𝑠=∫𝑎 ( 𝑑𝑡 ) + ( 𝑑𝑡 ) 𝑑𝑡 = ∫𝑎 |𝑐⃗′(𝑡)|𝑑𝑡
where 𝑐(𝑎) and 𝑐(𝑏) are the endpoint of the curve 𝑐, |𝑐⃗′(𝑡)| is the length of the
velocity vector of 𝑐⃗(𝑡), and the curve is 𝐶 1 , i.e., 𝑥 ′ (𝑡 ), and 𝑦′(𝑡) are continuous.
We define a line (or path) integral of a function 𝑓(𝑥, 𝑦) over a 𝐶 1 curve 𝑐, in the
plane by:
𝒃
∫𝒄 𝒇(𝒙, 𝒚)𝒅𝒔 = ∫𝒂 𝒇(𝒙(𝒕), 𝒚(𝒕))|𝒄
⃗⃗′(𝒕)|𝒅𝒕.
If the curve 𝑐 is in 3-space rather than the plane then we have:
𝒃
⃗⃗′ (𝒕)|𝒅𝒕
∫𝒄 𝒇(𝒙, 𝒚, 𝒛)𝒅𝒔 = ∫𝒂 𝒇(𝒙(𝒕), 𝒚(𝒕), 𝒛(𝒕))|𝒄
𝑏
Notice that if 𝑓(𝑥, 𝑦, 𝑧) = 1, then ∫𝑐 𝑓 (𝑥, 𝑦, 𝑧)𝑑𝑠 = ∫𝑐 𝑑𝑠= ∫𝑎 |𝑐⃗′(𝑡)|𝑑𝑡
which is just the length of the curve 𝑐.
, 2
Ex. Evaluate ∫𝑐 𝑦(𝑠𝑖𝑛𝑧)𝑑𝑠; where 𝑐 is the helix
𝑥 (𝑡) = 𝑐𝑜𝑠𝑡, 𝑦(𝑡 ) = 𝑠𝑖𝑛𝑡, 𝑧(𝑡 ) = 𝑡; where 0 ≤ 𝑡 ≤ 2𝜋.
𝑐
Notice we could write this curve as 𝑐⃗(𝑡) =< 𝑐𝑜𝑠𝑡, 𝑠𝑖𝑛𝑡, 𝑡 >; 0 ≤ 𝑡 ≤ 2𝜋.
𝑐⃗′(𝑡) =< −𝑠𝑖𝑛𝑡, 𝑐𝑜𝑠𝑡, 1 > ⟹ |𝑐⃗⃗′(𝑡)| = √𝑐𝑜𝑠2𝑡 + 𝑠𝑖𝑛2 𝑡 + 1 = √2.
𝑓(𝑥 (𝑡 ), 𝑦(𝑡 ), 𝑧(𝑡)) = 𝑦(𝑠𝑖𝑛𝑧) = (𝑠𝑖𝑛𝑡 )(𝑠𝑖𝑛𝑡 ).
𝑏
∫𝑎 𝑓 (𝑥 (𝑡 ), 𝑦(𝑡 ), 𝑧(𝑡))|𝑐⃗′(𝑡)|𝑑𝑡
2𝜋 2𝜋
= ∫0 (𝑠𝑖𝑛𝑡)(𝑠𝑖𝑛𝑡)(√2)𝑑𝑡 = √2 ∫0 (𝑠𝑖𝑛2 𝑡)𝑑𝑡.
1 cos(2𝑡) 1 cos(2𝑡)
Remember: 𝑠𝑖𝑛2 𝑡 = − and 𝑐𝑜𝑠 2 𝑡 = + ; we will need
2 2 2 2
these formulas often!
2𝜋 2𝜋
2
1 cos(2𝑡) 1 sin(2𝑡 ) 2𝜋
√2 ∫ (𝑠𝑖𝑛 𝑡)𝑑𝑡 = √2 ∫ ( − )𝑑𝑡 = √2( 𝑡 − )⃒
0 0 2 2 2 4 0
1
= √2 [(2 (2𝜋) − 0) − (0 − 0)] = √2𝜋.
We can represent a path or curve, 𝑐 , in the plane in the following 3 ways:
1. 𝑐⃗(𝑡) =< 𝑥 (𝑡 ), 𝑦(𝑡 ) > (vector form of a curve)
2. 𝑐⃗(𝑡) = 𝑥(𝑡)𝑖⃗ + 𝑦(𝑡)𝑗⃗ (another vector form of a curve)
3. 𝑥 = 𝑥 (𝑡 ), 𝑦 = 𝑦(𝑡 ) (parametric form of a curve)
In each case t will be in some interval (eg, 𝑎 ≤ 𝑡 ≤ 𝑏)
From first year calculus we know that the length of a curve, 𝑥 = 𝑥(𝑡 ), 𝑦 = 𝑦(𝑡 ),
𝑎 ≤ 𝑡 ≤ 𝑏, which we can also write as 𝑐⃗(𝑡) =< 𝑥 (𝑡 ), 𝑦(𝑡 ) >, is given by:
𝑏 𝑑𝑥 2 𝑑𝑦 2 𝑏
Length of curve= √
∫𝑐 𝑑𝑠=∫𝑎 ( 𝑑𝑡 ) + ( 𝑑𝑡 ) 𝑑𝑡 = ∫𝑎 |𝑐⃗′(𝑡)|𝑑𝑡
where 𝑐(𝑎) and 𝑐(𝑏) are the endpoint of the curve 𝑐, |𝑐⃗′(𝑡)| is the length of the
velocity vector of 𝑐⃗(𝑡), and the curve is 𝐶 1 , i.e., 𝑥 ′ (𝑡 ), and 𝑦′(𝑡) are continuous.
We define a line (or path) integral of a function 𝑓(𝑥, 𝑦) over a 𝐶 1 curve 𝑐, in the
plane by:
𝒃
∫𝒄 𝒇(𝒙, 𝒚)𝒅𝒔 = ∫𝒂 𝒇(𝒙(𝒕), 𝒚(𝒕))|𝒄
⃗⃗′(𝒕)|𝒅𝒕.
If the curve 𝑐 is in 3-space rather than the plane then we have:
𝒃
⃗⃗′ (𝒕)|𝒅𝒕
∫𝒄 𝒇(𝒙, 𝒚, 𝒛)𝒅𝒔 = ∫𝒂 𝒇(𝒙(𝒕), 𝒚(𝒕), 𝒛(𝒕))|𝒄
𝑏
Notice that if 𝑓(𝑥, 𝑦, 𝑧) = 1, then ∫𝑐 𝑓 (𝑥, 𝑦, 𝑧)𝑑𝑠 = ∫𝑐 𝑑𝑠= ∫𝑎 |𝑐⃗′(𝑡)|𝑑𝑡
which is just the length of the curve 𝑐.
, 2
Ex. Evaluate ∫𝑐 𝑦(𝑠𝑖𝑛𝑧)𝑑𝑠; where 𝑐 is the helix
𝑥 (𝑡) = 𝑐𝑜𝑠𝑡, 𝑦(𝑡 ) = 𝑠𝑖𝑛𝑡, 𝑧(𝑡 ) = 𝑡; where 0 ≤ 𝑡 ≤ 2𝜋.
𝑐
Notice we could write this curve as 𝑐⃗(𝑡) =< 𝑐𝑜𝑠𝑡, 𝑠𝑖𝑛𝑡, 𝑡 >; 0 ≤ 𝑡 ≤ 2𝜋.
𝑐⃗′(𝑡) =< −𝑠𝑖𝑛𝑡, 𝑐𝑜𝑠𝑡, 1 > ⟹ |𝑐⃗⃗′(𝑡)| = √𝑐𝑜𝑠2𝑡 + 𝑠𝑖𝑛2 𝑡 + 1 = √2.
𝑓(𝑥 (𝑡 ), 𝑦(𝑡 ), 𝑧(𝑡)) = 𝑦(𝑠𝑖𝑛𝑧) = (𝑠𝑖𝑛𝑡 )(𝑠𝑖𝑛𝑡 ).
𝑏
∫𝑎 𝑓 (𝑥 (𝑡 ), 𝑦(𝑡 ), 𝑧(𝑡))|𝑐⃗′(𝑡)|𝑑𝑡
2𝜋 2𝜋
= ∫0 (𝑠𝑖𝑛𝑡)(𝑠𝑖𝑛𝑡)(√2)𝑑𝑡 = √2 ∫0 (𝑠𝑖𝑛2 𝑡)𝑑𝑡.
1 cos(2𝑡) 1 cos(2𝑡)
Remember: 𝑠𝑖𝑛2 𝑡 = − and 𝑐𝑜𝑠 2 𝑡 = + ; we will need
2 2 2 2
these formulas often!
2𝜋 2𝜋
2
1 cos(2𝑡) 1 sin(2𝑡 ) 2𝜋
√2 ∫ (𝑠𝑖𝑛 𝑡)𝑑𝑡 = √2 ∫ ( − )𝑑𝑡 = √2( 𝑡 − )⃒
0 0 2 2 2 4 0
1
= √2 [(2 (2𝜋) − 0) − (0 − 0)] = √2𝜋.
