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 ) = ⟨ c o s ( t ) , s i n ( t ) ⟩ C = \vec{r}(t) = \langle cos(t), sin(t) \rangle C = r ( t ) = ⟨ cos ( t ) , s in ( t )⟩ 0 ≤ t ≤ π 0 \leq t \leq \pi 0 ≤ t ≤ π
r ⃗ ′ ( t ) = ⟨ − s i n ( t ) , c o s ( t ) ⟩ \vec{r}^\prime (t) = \langle -sin(t), cos(t) \rangle r ′ ( t ) = ⟨ − s in ( t ) , cos ( t )⟩ ∫ 0 π 3 c o s 2 ( t ) s i n ( t ) ( − s i n ( t ) ) 2 + ( c o s ( t ) ) 2 d t \int_0^{\pi} 3 cos^2(t) sin(t) \sqrt{(-sin(t))^2 + (cos(t))^2}\ dt ∫ 0 π 3 co s 2 ( t ) s in ( t ) ( − s in ( t ) ) 2 + ( cos ( t ) ) 2 d t ∫ 0 π 3 c o s 2 ( t ) s i n ( t ) ⋅ 1 d t \int_0^{\pi} 3 cos^2(t) sin(t) \cdot 1\ dt ∫ 0 π 3 co s 2 ( t ) s in ( t ) ⋅ 1 d t Let u = c o s ( t ) u = cos(t) u = cos ( t ) d u = − s i n ( t ) d t du = -sin(t)\ dt d u = − s in ( t ) d t u ( 0 ) = c o s ( 0 ) ) 1 u(0) = cos(0) ) 1 u ( 0 ) = cos ( 0 )) 1 u ( π ) = c o s ( π ) = − 1 u(\pi) = cos(\pi) = -1 u ( π ) = cos ( π ) = − 1
− ∫ 1 − 1 3 u d u -\int_{1}^{-1} 3u\ du − ∫ 1 − 1 3 u d u ∫ − 1 1 3 u d u \int_{-1}^{1} 3u\ du ∫ − 1 1 3 u d u u 3 ∣ u = − 1 u = 1 = 2 u^3 \bigg\rvert_{u = -1}^{u = 1} = \boxed{2} 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 rectnagle will not be differentiable.
What we can do is divide C C C n n n
C = C 1 ∪ C 2 ∪ … ∪ C n C = C_1 \cup C_2 \cup \ldots \cup C_n C = C 1 ∪ C 2 ∪ … ∪ C n ∫ C f ( x , y ) d s = ∑ j = 1 n ∫ C j f ( x , y ) d s \int_C f(x, y) ds = \sum_{j = 1}^{n} \int_{C_j} f(x, y) ds ∫ C f ( x , y ) d s = 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 p a r a b o l a 2 x d s + ∫ C l i n e 2 x d s \int_C 2x\ ds = \int_{C_{parabola}} 2x\ ds + \int_{C_{line}} 2x\ ds ∫ C 2 x d s = ∫ C p a r ab o l a 2 x d s + ∫ C l in e 2 x d s Let’s start with the line:
r ⃗ ( t ) = ⟨ x 0 + t a , y 0 + t b ⟩ r ⃗ ( t ) = ⟨ 1 + t ( 1 − 1 ) , 1 + t ( 2 − 1 ) ⟩ r ⃗ ( t ) = ⟨ 1 , 1 + t ⟩ \vec{r}(t) = \langle x_0 + ta, y_0 + tb \rangle \newline
\vec{r}(t) = \langle 1 + t(1 - 1), 1 + t(2 - 1) \rangle \newline
\vec{r}(t) = \langle 1, 1 + t \rangle \newline r ( t ) = ⟨ x 0 + t a , y 0 + t b ⟩ r ( t ) = ⟨ 1 + t ( 1 − 1 ) , 1 + t ( 2 − 1 )⟩ r ( t ) = ⟨ 1 , 1 + t ⟩ r ⃗ ′ ( t ) = ⟨ 0 , 1 ⟩ ∣ 0 ≤ t ≤ 1 \vec{r}^\prime(t) = \langle 0, 1 \rangle \ | \ 0 \leq t \leq 1 r ′ ( t ) = ⟨ 0 , 1 ⟩ ∣ 0 ≤ t ≤ 1 ∫ C l i n e 2 x d s = ∫ 0 1 2 ⋅ 1 ⋅ 0 2 + 1 2 d t ∫ 0 1 2 d t = 2 \int_{C_{line}} 2x\ ds = \int_0^1 2 \cdot 1 \cdot \sqrt{0^2 + 1^2}\ dt \newline
\int_0^1 2\ dt = \boxed{2} ∫ C l in e 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 ⟩ \vec{r}(t) = \langle t, t^2 \rangle r ( t ) = ⟨ t , t 2 ⟩ r ⃗ ′ ( t ) = ⟨ 1 , 2 t ⟩ ∣ 0 ≤ t ≤ 1 \vec{r}^\prime(t) = \langle 1, 2t \rangle \ | \ 0 \leq t \leq 1 r ′ ( t ) = ⟨ 1 , 2 t ⟩ ∣ 0 ≤ t ≤ 1 ∫ C p a r a b o l a 2 x d s = 2 ∫ 0 1 t ⋅ 1 + ( 2 t ) 2 d t \int_{C_{parabola}} 2x\ ds = 2 \int_0^1 t \cdot \sqrt{1 + (2t)^2}\ dt \newline ∫ C p a r ab o l a 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 u(0) = 1 + (2 \cdot 0)^2 u ( 0 ) = 1 + ( 2 ⋅ 0 ) 2 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 \dfrac{2}{8} \int_1^5 \sqrt{u}\ du \newline 8 2 ∫ 1 5 u d u 1 4 u 3 2 3 2 ∣ u = 1 u = 5 \dfrac{1}{4} \dfrac{u^{\frac{3}{2}}}{\dfrac{3}{2}} \bigg\rvert_{u = 1}^{u = 5} \newline 4 1 2 3 u 2 3 u = 1 u = 5 1 4 2 3 u 3 2 ∣ u = 1 u = 5 \dfrac{1}{4} \dfrac{2}{3} u^{\frac{3}{2}} \bigg\rvert_{u = 1}^{u = 5} \newline 4 1 3 2 u 2 3 u = 1 u = 5 u 3 2 6 ∣ u = 1 u = 5 \dfrac{u^{\frac{3}{2}}}{6} \bigg\rvert_{u = 1}^{u = 5} \newline 6 u 2 3 u = 1 u = 5 5 3 2 − 1 6 \dfrac{5^{\frac{3}{2}} - 1}{6} 6 5 2 3 − 1 Therefore:
∫ C 2 x d s = ∫ C p a r a b o l a 2 x d s + ∫ C l i n e 2 x d s = 5 3 2 − 1 6 + 2 \int_C 2x\ ds = \int_{C_{parabola}} 2x\ ds + \int_{C_{line}} 2x\ ds = \boxed{\dfrac{5^{\frac{3}{2}} - 1}{6} + 2} ∫ C 2 x d s = ∫ C p a r ab o l a 2 x d s + ∫ C l in e 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 ) = ⟨ c o s ( t ) , s i n ( t ) ⟩ ∣ 0 ≤ π \vec{r}(t) = \langle cos(t), sin(t) \rangle \ | \ 0 \leq \pi r ( t ) = ⟨ cos ( t ) , s in ( t )⟩ ∣ 0 ≤ 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) \ | \ 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 - \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 − s i n ( t ) ) ⋅ s i n ( t ) ⋅ 1 d t k ∫ 0 π ( 1 − s i n ( t ) ⋅ s i n ( t ) d t k ∫ 0 π s i n ( t ) − s i n 2 ( t ) d t … k 4 − π 2 \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}} ∫ 0 π k ( 1 − s in ( t )) ⋅ s in ( t ) ⋅ 1 d t k ∫ 0 π ( 1 − s in ( t ) ⋅ s in ( t ) d t k ∫ 0 π s in ( t ) − s i n 2 ( t ) d t … k 2 4 − π m = ∫ 0 π k ( 1 − s i n ( t ) ) d t k ∫ 0 π ( 1 − s i n ( t ) ) d t k ∫ 0 π 1 − s i n ( t ) d t k [ t + c o s ( t ) ] ∣ t = 0 t = π k [ t + c o s ( 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
& k [t + cos(t)] \bigg\rvert_{t = 0}^{t = \pi} \newline
& \boxed{k(\pi - 2)}
\end{align*} m = ∫ 0 π k ( 1 − s in ( t )) d t k ∫ 0 π ( 1 − s in ( t )) d t k ∫ 0 π 1 − s in ( t ) d t k [ t + cos ( t )] t = 0 t = π k [ t + cos ( t )] t = 0 t = π k ( π − 2 ) y 0 = k 4 − π 2 k ( π − 2 ) y_0 = \dfrac{k \dfrac{4 - \pi}{2}}{k(\pi - 2)} \newline y 0 = k ( π − 2 ) k 2 4 − π y 0 = 4 − π 2 ( π − 2 ) y_0 = \dfrac{\dfrac{4 - \pi}{2}}{(\pi - 2)} \newline y 0 = ( π − 2 ) 2 4 − π y 0 = 4 − π 2 1 ( π − 2 ) y_0 = \dfrac{4 - \pi}{2} \dfrac{1}{(\pi - 2)} \newline y 0 = 2 4 − π ( π − 2 ) 1 y 0 = 4 − π 2 π − 4 y_0 = \dfrac{4 - \pi}{2\pi - 4} y 0 = 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 or y.
∫ C f ( x , y ) d x or ∫ C f ( x , y ) d y \int_C f(x, y)\ dx \text{ or } \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 \int_C f(x, y)\ dx = \int_a^b f(x(t), y(t)) x^\prime(t)\ dt ∫ 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 \int_C f(x, y)\ dy = \int_a^b f(x(t), y(t)) y^\prime(t)\ dt ∫ C f ( x , y ) d y = ∫ 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 ⟩ r ⃗ ( t ) = ⟨ − 5 + t ( 0 − ( − 5 ) ) , − 3 + t ( 2 − ( − 3 ) ) ⟩ r ⃗ ( t ) = ⟨ − 5 + 5 t , − 3 + 5 t ⟩ ∣ 0 ≤ t ≤ 1 \vec{r}(t) = \langle x_0 + ta, y_0 + tb \rangle \newline
\vec{r}(t) = \langle -5 + t(0 - (-5)), -3 + t(2 - (-3)) \rangle \newline
\vec{r}(t) = \langle -5 + 5t, -3 + 5t \rangle \ | \ 0 \leq t \leq 1 r ( t ) = ⟨ x 0 + t a , y 0 + t b ⟩ r ( t ) = ⟨ − 5 + t ( 0 − ( − 5 )) , − 3 + t ( 2 − ( − 3 ))⟩ r ( t ) = ⟨ − 5 + 5 t , − 3 + 5 t ⟩ ∣ 0 ≤ t ≤ 1 r ⃗ ′ ( t ) = ⟨ 5 , 5 ⟩ ∣ 0 ≤ t ≤ 1 \vec{r}^\prime(t) = \langle 5, 5 \rangle \ | \ 0 \leq t \leq 1 r ′ ( t ) = ⟨ 5 , 5 ⟩ ∣ 0 ≤ t ≤ 1 ∫ C y 2 d x = ∫ 0 1 ( − 3 + 5 t ) 2 ⋅ 5 d t = … = 35 3 \int_C y^2\ dx = \int_0^1 (-3 + 5t)^2 \cdot 5\ dt = \ldots = \dfrac{35}{3} ∫ C y 2 d x = ∫ 0 1 ( − 3 + 5 t ) 2 ⋅ 5 d t = … = 3 35 ∫ C x d y = ∫ 0 1 ( − 5 + 5 t ) ⋅ 5 d t = … = − 25 2 \int_C x\ dy = \int_0^1 (-5 + 5t) \cdot 5\ dt = \ldots = -\dfrac{25}{2} ∫ C x d y = ∫ 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 3 r ⃗ ( t ) = ⟨ x ( t ) , y ( t ) , z ( t ) ⟩ ∣ a ≤ t ≤ b C \in \mathbb{R}^3 \newline
f: C \rarr \mathbb{R}^3 \newline
\vec{r}(t) = \langle x(t), y(t), z(t) \rangle \ | \ a \leq t \leq b C ∈ R 3 f : C → R 3 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