Introduction
In this part we’ll cover a new concept with integrals, the so-called line integral.
Definition 1 (Line integral) A curve, C C C r ⃗ ( t ) = ⟨ x ( t ) , y ( t ) ⟩ \vec{r}(t) = \langle x(t), y(t) \rangle r ( t ) = ⟨ x ( t ) , y ( t )⟩ t ∈ [ a , b ] t \in [a, b] t ∈ [ a , b ] f : C → R f: C \rarr \mathbb{R} f : C → R
Assuming x ( t ) x(t) x ( t ) y ( t ) y(t) y ( t )
∫ C f ( x , y ) d s = ∫ a b f ( x ( t ) , y ( t ) ) x ′ ( t ) 2 + y ′ ( t ) 2 d t . \int_C f(x, y) \, ds = \int_a^b f(x(t), y(t)) \sqrt{x^\prime(t)^2 + y^\prime(t)^2} \, dt. ∫ C f ( x , y ) d s = ∫ a b f ( x ( t ) , y ( t )) x ′ ( t ) 2 + y ′ ( t ) 2 d t . Intuition (Geometrical sense) Just as an integral over an interval [ a , b ] [a, b] [ a , b ]
Example 1 Compute ∫ C 3 x 2 y d s \int_C 3x^2 y \, ds ∫ C 3 x 2 y d s
Where C = r ⃗ ( t ) = ⟨ cos ( t ) , sin ( t ) ⟩ C = \vec{r}(t) = \langle \cos(t), \sin(t) \rangle C = r ( t ) = ⟨ cos ( t ) , sin ( t )⟩ 0 ≤ t ≤ π 0 \leq t \leq \pi 0 ≤ t ≤ π
r ⃗ ′ ( t ) = ⟨ − sin ( t ) , cos ( t ) ⟩ \vec{r}^\prime (t) = \langle -\sin(t), \cos(t) \rangle r ′ ( t ) = ⟨ − sin ( t ) , cos ( t )⟩ ∫ 0 π 3 cos 2 ( t ) sin ( t ) ( − sin ( t ) ) 2 + ( cos ( t ) ) 2 d t = ∫ 0 π 3 cos 2 ( t ) sin ( t ) ⋅ 1 d t \begin{align*}
& \int_0^{\pi} 3 \cos^2(t) \sin(t) \sqrt{(-\sin(t))^2 + (\cos(t))^2} \, dt \newline
= {} & \int_0^{\pi} 3 \cos^2(t) \sin(t) \cdot 1 \, dt
\end{align*} = ∫ 0 π 3 cos 2 ( t ) sin ( t ) ( − sin ( t ) ) 2 + ( cos ( t ) ) 2 d t ∫ 0 π 3 cos 2 ( t ) sin ( t ) ⋅ 1 d t Let u = cos ( t ) u = \cos(t) u = cos ( t ) d u = − sin ( t ) d t du = -\sin(t) \, dt d u = − sin ( t ) d t u ( 0 ) = cos ( 0 ) = 1 u(0) = \cos(0) = 1 u ( 0 ) = cos ( 0 ) = 1 u ( π ) = cos ( π ) = − 1 u(\pi) = \cos(\pi) = -1 u ( π ) = cos ( π ) = − 1
− ∫ 1 − 1 3 u 2 d u = ∫ − 1 1 3 u 2 d u = u 3 ∣ u = − 1 u = 1 = 2 \begin{align*}
-\int_{1}^{-1} 3u^2 \, du & = \int_{-1}^{1} 3u^2 \, du \newline
& = u^3 \bigg\rvert_{u = -1}^{u = 1} = \boxed{2}
\end{align*} − ∫ 1 − 1 3 u 2 d u = ∫ − 1 1 3 u 2 d u = u 3 u = − 1 u = 1 = 2 Non-differentiable lines
There may be some cases where the parameterization isn’t differentiable.
Imagine we have a simple rectangular form. The four corners of the rectangle will not be differentiable.
What we can do is divide C C C n n n
C = C 1 ∪ C 2 ∪ … ∪ C n , ∫ C f ( x , y ) d s = ∑ j = 1 n ∫ C j f ( x , y ) d s . \begin{align*}
C & = C_1 \cup C_2 \cup \ldots \cup C_n, \newline
\int_C f(x, y) \, ds & = \sum_{j = 1}^{n} \int_{C_j} f(x, y) \, ds.
\end{align*} C ∫ C f ( x , y ) d s = C 1 ∪ C 2 ∪ … ∪ C n , = j = 1 ∑ n ∫ C j f ( x , y ) d s . Example 2 Evaluate ∫ C 2 x d s \int_C 2x \, ds ∫ C 2 x d s C = x 2 C = x^2 C = x 2 ( 0 , 0 ) (0, 0) ( 0 , 0 ) ( 1 , 1 ) (1, 1) ( 1 , 1 ) ( 1 , 1 ) (1, 1) ( 1 , 1 ) ( 1 , 2 ) (1, 2) ( 1 , 2 )
∫ C 2 x d s = ∫ C parabola 2 x d s + ∫ C line 2 x d s \int_C 2x \, ds = \int_{C_{\text{parabola}}} 2x \, ds + \int_{C_{\text{line}}} 2x \, ds ∫ C 2 x d s = ∫ C parabola 2 x d s + ∫ C line 2 x d s Let’s start with the line:
r ⃗ ( t ) = ⟨ x 0 + t a , y 0 + t b ⟩ = ⟨ 1 + t ( 1 − 1 ) , 1 + t ( 2 − 1 ) ⟩ = ⟨ 1 , 1 + t ⟩ , r ⃗ ′ ( t ) = ⟨ 0 , 1 ⟩ , 0 ≤ t ≤ 1. \begin{align*}
\vec{r}(t) & = \langle x_0 + ta, y_0 + tb \rangle \newline
& = \langle 1 + t(1 - 1), 1 + t(2 - 1) \rangle \newline
& = \langle 1, 1 + t \rangle, \newline
\vec{r}^\prime(t) & = \langle 0, 1 \rangle, \quad 0 \leq t \leq 1.
\end{align*} r ( t ) r ′ ( t ) = ⟨ x 0 + t a , y 0 + t b ⟩ = ⟨ 1 + t ( 1 − 1 ) , 1 + t ( 2 − 1 )⟩ = ⟨ 1 , 1 + t ⟩ , = ⟨ 0 , 1 ⟩ , 0 ≤ t ≤ 1. ∫ C line 2 x d s = ∫ 0 1 2 ⋅ 1 ⋅ 0 2 + 1 2 d t = ∫ 0 1 2 d t = 2 \begin{align*}
\int_{C_{\text{line}}} 2x \, ds & = \int_0^1 2 \cdot 1 \cdot \sqrt{0^2 + 1^2} \, dt \newline
& = \int_0^1 2 \, dt = \boxed{2}
\end{align*} ∫ C line 2 x d s = ∫ 0 1 2 ⋅ 1 ⋅ 0 2 + 1 2 d t = ∫ 0 1 2 d t = 2 Now for the parabola:
r ⃗ ( t ) = ⟨ t , t 2 ⟩ , r ⃗ ′ ( t ) = ⟨ 1 , 2 t ⟩ , 0 ≤ t ≤ 1. \begin{align*}
\vec{r}(t) & = \langle t, t^2 \rangle, \newline
\vec{r}^\prime(t) & = \langle 1, 2t \rangle, \quad 0 \leq t \leq 1.
\end{align*} r ( t ) r ′ ( t ) = ⟨ t , t 2 ⟩ , = ⟨ 1 , 2 t ⟩ , 0 ≤ t ≤ 1. ∫ C parabola 2 x d s = 2 ∫ 0 1 t ⋅ 1 + ( 2 t ) 2 d t \int_{C_{\text{parabola}}} 2x \, ds = 2 \int_0^1 t \cdot \sqrt{1 + (2t)^2} \, dt ∫ C parabola 2 x d s = 2 ∫ 0 1 t ⋅ 1 + ( 2 t ) 2 d t Let u = 1 + ( 2 t ) 2 u = 1 + (2t)^2 u = 1 + ( 2 t ) 2 d u = 8 t d t du = 8t \, dt d u = 8 t d t u ( 0 ) = 1 + ( 2 ⋅ 0 ) 2 = 1 u(0) = 1 + (2 \cdot 0)^2 = 1 u ( 0 ) = 1 + ( 2 ⋅ 0 ) 2 = 1 u ( 1 ) = 1 + ( 2 ⋅ 1 ) 2 = 5 u(1) = 1 + (2\cdot 1)^2 = 5 u ( 1 ) = 1 + ( 2 ⋅ 1 ) 2 = 5
2 8 ∫ 1 5 u d u = 1 4 u 3 2 3 2 ∣ u = 1 u = 5 = 1 4 2 3 u 3 2 ∣ u = 1 u = 5 = u 3 2 6 ∣ u = 1 u = 5 = 5 3 2 − 1 6 . \begin{align*}
\dfrac{2}{8} \int_1^5 \sqrt{u} \, du & = \dfrac{1}{4} \dfrac{u^{\frac{3}{2}}}{\dfrac{3}{2}} \bigg\rvert_{u = 1}^{u = 5} \newline
& = \dfrac{1}{4} \dfrac{2}{3} u^{\frac{3}{2}} \bigg\rvert_{u = 1}^{u = 5} \newline
& = \dfrac{u^{\frac{3}{2}}}{6} \bigg\rvert_{u = 1}^{u = 5} \newline
& = \dfrac{5^{\frac{3}{2}} - 1}{6}.
\end{align*} 8 2 ∫ 1 5 u d u = 4 1 2 3 u 2 3 u = 1 u = 5 = 4 1 3 2 u 2 3 u = 1 u = 5 = 6 u 2 3 u = 1 u = 5 = 6 5 2 3 − 1 . Therefore:
∫ C 2 x d s = ∫ C parabola 2 x d s + ∫ C line 2 x d s = 5 3 2 − 1 6 + 2 \int_C 2x \, ds = \int_{C_{\text{parabola}}} 2x \, ds + \int_{C_{\text{line}}} 2x \, ds = \boxed{\dfrac{5^{\frac{3}{2}} - 1}{6} + 2} ∫ C 2 x d s = ∫ C parabola 2 x d s + ∫ C line 2 x d s = 6 5 2 3 − 1 + 2 Example 3 Imagine we have a wire that is described by the following parameterization.
r ⃗ ( t ) = ⟨ cos ( t ) , sin ( t ) ⟩ , 0 ≤ t ≤ π . \vec{r}(t) = \langle \cos(t), \sin(t) \rangle, \quad 0 \leq t \leq \pi. r ( t ) = ⟨ cos ( t ) , sin ( t )⟩ , 0 ≤ t ≤ π . The density proportional to the line y = 1 y = 1 y = 1
Which means:
ρ ( x , y ) = k ( 1 − y ) , k ∈ R . \rho(x, y) = k(1 - y), \quad k \in \mathbb{R}. ρ ( x , y ) = k ( 1 − y ) , k ∈ R . Find the center of mass.
x 0 = 0 due to symmetry x_0 = 0 \quad \text{due to symmetry} x 0 = 0 due to symmetry y 0 = ∫ C ρ ( x , y ) ⋅ y d s ∫ C ρ ( x , y ) d s y_0 = \dfrac{\int_C \rho(x, y) \cdot y \, ds}{\int_C \rho(x, y) \, ds} y 0 = ∫ C ρ ( x , y ) d s ∫ C ρ ( x , y ) ⋅ y d s ∫ 0 π k ( 1 − sin ( t ) ) ⋅ sin ( t ) ⋅ 1 d t = k ∫ 0 π ( 1 − sin ( t ) ) ⋅ sin ( t ) d t = k ∫ 0 π sin ( t ) − sin 2 ( t ) d t = … = k 4 − π 2 \begin{align*}
& \int_0^{\pi} k(1 - \sin(t)) \cdot \sin(t) \cdot 1 \, dt \newline
= {} & k \int_0^{\pi} (1 - \sin(t)) \cdot \sin(t) \, dt \newline
= {} & k \int_0^{\pi} \sin(t) - \sin^2(t) \, dt \newline
= {} & \ldots \newline
= {} & \boxed{k \dfrac{4 - \pi}{2}}
\end{align*} = = = = ∫ 0 π k ( 1 − sin ( t )) ⋅ sin ( t ) ⋅ 1 d t k ∫ 0 π ( 1 − sin ( t )) ⋅ sin ( t ) d t k ∫ 0 π sin ( t ) − sin 2 ( t ) d t … k 2 4 − π m = ∫ 0 π k ( 1 − sin ( t ) ) d t = k ∫ 0 π ( 1 − sin ( t ) ) d t = k ∫ 0 π 1 − sin ( t ) d t = k [ t + cos ( t ) ] ∣ t = 0 t = π = k ( π − 2 ) \begin{align*}
m & = \int_0^{\pi} k(1 - \sin(t)) \, dt \newline
& = k \int_0^{\pi} (1 - \sin(t)) \, dt \newline
& = k \int_0^{\pi} 1 - \sin(t) \, dt \newline
& = k [t + \cos(t)] \bigg\rvert_{t = 0}^{t = \pi} \newline
& = \boxed{k(\pi - 2)}
\end{align*} m = ∫ 0 π k ( 1 − sin ( t )) d t = k ∫ 0 π ( 1 − sin ( t )) d t = k ∫ 0 π 1 − sin ( t ) d t = k [ t + cos ( t )] t = 0 t = π = k ( π − 2 ) y 0 = k 4 − π 2 k ( π − 2 ) = 4 − π 2 ( π − 2 ) = 4 − π 2 1 ( π − 2 ) = 4 − π 2 π − 4 . \begin{align*}
y_0 & = \dfrac{k \dfrac{4 - \pi}{2}}{k(\pi - 2)} \newline
& = \dfrac{\dfrac{4 - \pi}{2}}{(\pi - 2)} \newline
& = \dfrac{4 - \pi}{2} \dfrac{1}{(\pi - 2)} \newline
& = \dfrac{4 - \pi}{2\pi - 4}.
\end{align*} y 0 = k ( π − 2 ) k 2 4 − π = ( π − 2 ) 2 4 − π = 2 4 − π ( π − 2 ) 1 = 2 π − 4 4 − π . Line integrals with respect to x or y
So far we have integrated with respect to the arc length. But we can integrate with respect to x x x y y y
∫ C f ( x , y ) d x or ∫ C f ( x , y ) d y \int_C f(x, y) \, dx \quad \text{or} \quad \int_C f(x, y) \, dy ∫ C f ( x , y ) d x or ∫ C f ( x , y ) d y ∫ C f ( x , y ) d x = ∫ a b f ( x ( t ) , y ( t ) ) x ′ ( t ) d t , ∫ C f ( x , y ) d y = ∫ a b f ( x ( t ) , y ( t ) ) y ′ ( t ) d t . \begin{align*}
\int_C f(x, y) \, dx & = \int_a^b f(x(t), y(t)) x^\prime(t) \, dt, \newline
\int_C f(x, y) \, dy & = \int_a^b f(x(t), y(t)) y^\prime(t) \, dt.
\end{align*} ∫ C f ( x , y ) d x ∫ C f ( x , y ) d y = ∫ a b f ( x ( t ) , y ( t )) x ′ ( t ) d t , = ∫ a b f ( x ( t ) , y ( t )) y ′ ( t ) d t . Example 4 Evaluate ∫ C y 2 d x + x d y \int_C y^2 \, dx + x \, dy ∫ C y 2 d x + x d y
Where C C C ( − 5 , − 3 ) (-5, -3) ( − 5 , − 3 ) ( 0 , 2 ) (0, 2) ( 0 , 2 )
Let’s rewrite our integral:
∫ C y 2 d x + x d y = ∫ C y 2 d x + ∫ C x d y \int_C y^2 \, dx + x \, dy = \int_C y^2 \, dx + \int_C x \, dy ∫ C y 2 d x + x d y = ∫ C y 2 d x + ∫ C x d y Let’s find the parameterization for the line:
r ⃗ ( t ) = ⟨ x 0 + t a , y 0 + t b ⟩ = ⟨ − 5 + t ( 0 − ( − 5 ) ) , − 3 + t ( 2 − ( − 3 ) ) ⟩ = ⟨ − 5 + 5 t , − 3 + 5 t ⟩ , 0 ≤ t ≤ 1 , r ⃗ ′ ( t ) = ⟨ 5 , 5 ⟩ , 0 ≤ t ≤ 1. \begin{align*}
\vec{r}(t) & = \langle x_0 + ta, y_0 + tb \rangle \newline
& = \langle -5 + t(0 - (-5)), -3 + t(2 - (-3)) \rangle \newline
& = \langle -5 + 5t, -3 + 5t \rangle, \quad 0 \leq t \leq 1, \newline
\vec{r}^\prime(t) & = \langle 5, 5 \rangle, \quad 0 \leq t \leq 1.
\end{align*} r ( t ) r ′ ( t ) = ⟨ x 0 + t a , y 0 + t b ⟩ = ⟨ − 5 + t ( 0 − ( − 5 )) , − 3 + t ( 2 − ( − 3 ))⟩ = ⟨ − 5 + 5 t , − 3 + 5 t ⟩ , 0 ≤ t ≤ 1 , = ⟨ 5 , 5 ⟩ , 0 ≤ t ≤ 1. ∫ C y 2 d x = ∫ 0 1 ( − 3 + 5 t ) 2 ⋅ 5 d t = … = 35 3 , ∫ C x d y = ∫ 0 1 ( − 5 + 5 t ) ⋅ 5 d t = … = − 25 2 . \begin{align*}
\int_C y^2 \, dx & = \int_0^1 (-3 + 5t)^2 \cdot 5 \, dt = \ldots = \dfrac{35}{3}, \newline
\int_C x \, dy & = \int_0^1 (-5 + 5t) \cdot 5 \, dt = \ldots = -\dfrac{25}{2}.
\end{align*} ∫ C y 2 d x ∫ C x d y = ∫ 0 1 ( − 3 + 5 t ) 2 ⋅ 5 d t = … = 3 35 , = ∫ 0 1 ( − 5 + 5 t ) ⋅ 5 d t = … = − 2 25 . ∫ C y 2 d x + x d y = ∫ C y 2 d x + ∫ C x d y = 35 3 − 25 2 = − 5 6 \int_C y^2 \, dx + x \, dy = \int_C y^2 \, dx + \int_C x \, dy = \dfrac{35}{3} - \dfrac{25}{2} = \boxed{-\dfrac{5}{6}} ∫ C y 2 d x + x d y = ∫ C y 2 d x + ∫ C x d y = 3 35 − 2 25 = − 6 5 Line integrals with three variables
There is no difference in two or three variables:
C ⊂ R 3 , f : C → R , r ⃗ ( t ) = ⟨ x ( t ) , y ( t ) , z ( t ) ⟩ , a ≤ t ≤ b . \begin{align*}
& C \subset \mathbb{R}^3, \newline
& f: C \rarr \mathbb{R}, \newline
& \vec{r}(t) = \langle x(t), y(t), z(t) \rangle, \quad a \leq t \leq b.
\end{align*} C ⊂ R 3 , f : C → R , r ( t ) = ⟨ x ( t ) , y ( t ) , z ( t )⟩ , a ≤ t ≤ b . ∫ C f ( x , y , z ) d s = ∫ a b f ( x ( t ) , y ( t ) , z ( t ) ) x ′ ( t ) 2 + y ′ ( t ) 2 + z ′ ( t ) 2 d t . \int_C f(x, y, z) \, ds = \int_a^b f(x(t), y(t), z(t)) \sqrt{x^\prime(t)^2 + y^\prime(t)^2 + z^\prime(t)^2} \, dt. ∫ C f ( x , y , z ) d s = ∫ a b f ( x ( t ) , y ( t ) , z ( t )) x ′ ( t ) 2 + y ′ ( t ) 2 + z ′ ( t ) 2 d t .