Calculus III:

Line integrals with respect to the
coordinates x, y, and z

1. Definition and method

In this sectio, we will determine the line integrals with respect to x, or y, or both x and y.

Let's start with a two-dimensional curve C with parameterization:

x = x(t), y = y(t)     a ≤ t ≤ b.

The line integral of f with respect to x is:

∫_C f(x,y) dx = ∫_a^b f(x(t), y(t)) x'(t) dt

The line integral of f with respect to y is:

∫_C f(x,y) dy = ∫_a^b f(x(t), y(t)) y'(t) dt

With x'(t) = dx(t)/dt, and y'(t) = dy(t)/dt

Note that the only notational difference between these two and the line integral with respect to arc length is the differential. These have a dx or dy while the line integral with respect to arc length has a ds.

So when evaluating line integrals be careful to first note which differential you’ve got so you don’t work the wrong kind of line integral.

These two integral often appear together in the form :

∫_C P(x,y) dx + ∫_C Q(x,y) dy

Example 1

Let's evaluate ∫_C x²y dx + cos (Ï€y/2) dy where C is the line segment from (0,1) to (2,4).

The parameterization of the curve is:

x(t) = 0(1 - t) + t (2) = 2t
y(t) = 1(1 - t) + t (4) = 3t + 1

With :

0 ≤ t ≤ 1

We have :

dx = 2 dt , y = 3 dt

Therefore:

∫_C x²y dx + cos (Ï€y/2) dy =
∫₀¹ 4t²(3t + 1) (2 dt) + cos ((3t + 1)Ï€/2) (3 dt) =
8 ∫₀¹ (3t³ + t²) dt + 3 ∫₀¹ cos ((3t + 1)Ï€/2) dt =
8((3/4)t⁴ + (1/3)t³)|₀¹ + 3 (2/3Ï€) sin ((3t + 1)Ï€/2)|₀¹ =
8((3/4) + (1/3)) + (2/Ï€) [ sin ((3 + 1)Ï€/2) - sin ((1)Ï€/2)] =
26/3 + (2/Ï€) [0 - 1] =
26/3 - 2/Ï€.

∫_C x²y dx + cos (Ï€y/2) dy = 26/3 - 2/Ï€

Example 2

We know that changing the direction of the curve for a line integral with respect to arc length doesn’t change the value of the integral. Let’s see what happens with line integrals with respect to x and/or y. Let's evaluate ∫_C x²y dx + cos (Ï€y/2) dy where C is the line segment from (2,4) to (0,1).

The parameterization of the curve is:

x(t) = 2(1 - t) + t (0) = 2 - 2t
y(t) = 4(1 - t) + t (1) = 4 - 3t

With :

0 ≤ t ≤ 1

We have :

dx = - 2 dt , y = - 3 dt

Therefore:

∫_C x²y dx + cos (Ï€y/2) dy =
∫₀¹(2 - 2t)²(4 - 3t) (- 2 dt) + cos ((4 - 3t)Ï€/2) (- 3 dt) =
8 ∫₀¹(t - 1)²(3t - 4) dt - 3 cos ((3t - 4)Ï€/2) dt =
8 ((3/4)t⁴ - (10/3)t³ - (11/2)t² - 4t]|₀¹) dt - 3 (2/3Ï€) sin((3t - 4)Ï€/2)|₀¹ =
8((3/4)t⁴ - (10/3)t³) - (11/2)t² - 4t]|₀¹ dt - 3 (2/3Ï€) sin((3t - 4)Ï€/2)|₀¹ =
8 (- 13/12) - (2/Ï€) [- 1 - 0] =
- 26/3 + 2/Ï€

∫_C x²y dx + cos (Ï€y/2) dy = - 26/3 + 2/Ï€

So, switching the direction of the curve leads to the opposite sign of the value from the first example 1. In fact this will always happen with these kinds of line integrals:

If C is any curve then:

∫_-C f(x,y) dx = - ∫_C f(x,y) dx , and
∫_-C f(x,y) dy = - ∫_C f(x,y) dy

With the combined form of these two integrals, we have:

∫_-C P(x,y) dx + Q(x,y) dy = - ∫_C P(x,y) dx + Q(x,y) dy

2. Line integrals over three-dimensional curves

We can extend the line integrals formulas to three-dimensional curves;

The line integral of f with respect to x is:

∫_C f(x,y,z) dx = ∫_a^b f(x(t), y(t), z(t)) x'(t) dt

The line integral of f with respect to y is:

∫_C f(x,y,z) dy = ∫_a^b f(x(t), y(t), z(t)) y'(t) dt

The line integral of f with respect to z is:

∫_C f(x,y,z) dz = ∫_a^b f(x(t), y(t), z(t)) z'(t) dt

With

x'(t) = dx(t)/dt, y'(t) = dy(t)/dt , and z'(t) = dz(t)/dt
a ≤ t ≤ b

As with the two-dimensional space these three will often occur together and have the form:

∫_C P(x,y,z) dx + ∫_C Q(x,y,z) dy + ∫_C R(x,y,z) dz .