Vector calculus. Line integrals of the first kind

What it is

A line integral of the first kind is the integral of a function with respect to arc length:

C \int f (x, y, z) d s .

It does not depend on the direction in which the curve is traversed: $d s > 0$ always. Physically it is the mass of a wire with linear density $f$ . With $f \equiv 1$ the integral gives the length of the arc.

How to compute it

Everything comes down to one formula. If the curve is parametrised as $r (t) = (x (t), y (t), z (t))$ , $t \in [a, b]$ , then

d s = ∣ r^{'} (t) ∣ d t = x^{'2} + y^{'2} + z^{'2} d t,

and the integral becomes an ordinary definite one:

C \int f d s = a \int b f (x (t), y (t), z (t)) x^{'2} + y^{'2} + z^{'2} d t .

The steps:

Parametrise the curve. If it is given as $y = y (x)$ , take $x$ itself as the parameter; then $d s = 1 + y^{'2} d x$ .
Find the limits of the parameter from the endpoints.
Write out $∣ r^{'} (t) ∣$ and simplify the expression under the root — in textbook problems it almost always folds into a perfect square.
Substitute the parametrisation into $f$ and integrate.

Polar coordinates

If the curve is given in polar coordinates, $r = r (φ)$ , then

d s = r^{2} + r^{'2} d φ .

The root often folds into something like $∣ cos (φ /2) ∣$ — the absolute value must not be lost: where the function changes sign, the integral without it is understated or vanishes altogether.

Mass and centre of mass

The mass of a wire with linear density $ρ$ is $m = \int_{C} ρ d s$ , and the coordinates of its centre of mass are the first moments divided by the mass: $x_{c} = \frac{1}{m} \int_{C} x ρ d s$ , and likewise $y_{c}$ and $z_{c}$ . For a homogeneous curve the density cancels, and the centre of mass depends on the shape alone.

Symmetry

If the curve is mapped to itself by a permutation of the variables or by a reflection, the integrals of the corresponding functions are equal: $\oint x^{2} d s = \oint y^{2} d s$ , and the integral of an odd function over a symmetric curve is zero. Sometimes this replaces the parametrisation altogether.

Common mistakes

Setting the limits "clockwise". That is meaningless here: $a < b$ always, and the direction of traversal does not affect the answer. A negative answer for a non-negative $f$ means the limits are swapped.
Forgetting the factor $∣ r^{'} (t) ∣$ and integrating $f d t$ . This is the most common mistake: the result is an integral over nothing in particular.
Not simplifying the root. An expression like $1 + 2 c t^{2} + c^{2} t^{4} = 1 + c t^{2}$ is a standard set-up; if the root does not fold, the derivatives are most likely wrong.
A circle given implicitly ( $x^{2} + y^{2} = 2 x$ ) — complete the square first, find the centre and the radius, and only then parametrise.

Problems

Problem 1 (The arc length of a space curve). The curve $C$ is the intersection of the surfaces $x^{2} = 3 y$ , $2 x y = 9 z$ .

Find the length of the arc of this curve from the point $O (0; 0; 0)$ to the point $M (3; 3; 2)$ .

Solution.

Arc length is the integral of the first kind of one: $L = \int_{C} d s$ . So a parametrisation is needed.

Parametrisation. Both surfaces express $y$ and $z$ through $x$ , so take $x = t$ itself as the parameter. The first equation gives $y = t^{2} /3$ . Substitute into the second:

2 t \cdot \frac{t ^{2}}{3} = 9 z ⟹ z = \frac{2 t ^{3}}{27} .

Altogether

r (t) = (t; \frac{t ^{2}}{3}; \frac{2 t ^{3}}{27}) .

Limits. The point $O$ corresponds to $t = 0$ . For $M$ : at $t = 3$ we get $y = 3$ and $z = 2 \cdot 27/27 = 2$ — all three coordinates agree, so $t$ runs from $0$ to $3$ .

The length element.

r^{'} (t) = (1; \frac{2 t}{3}; \frac{2 t ^{2}}{9}),

∣ r^{'} (t) ∣ = 1 + \frac{4 t ^{2}}{9} + \frac{4 t ^{4}}{81} = \frac{81 + 36 t ^{2} + 4 t ^{4}}{9} .

The radicand is a perfect square: $81 + 36 t^{2} + 4 t^{4} = (9 + 2 t^{2})^{2}$ . This is always worth checking: if the root does not fold, the derivatives are most likely wrong. So

∣ r^{'} (t) ∣ = \frac{9 + 2 t ^{2}}{9} = 1 + \frac{2 t ^{2}}{9} .

Integration.

L = \int_{0}^{3} (1 + \frac{2 t ^{2}}{9}) d t = [t + \frac{2 t ^{3}}{27}]_{0}^{3} = 3 + \frac{54}{27} = 3 + 2 = 5.

Answer: $L = 5$ .

Problem 2 (An integral of the first kind over a shifted circle). Compute

\frac{1}{π} Γ \int (x - y) d s,

where $Γ$ is the circle $x^{2} + y^{2} = 2 x$ .

Solution.

Bring the circle to canonical form. The equation is given implicitly, so first complete the square: $x^{2} - 2 x + y^{2} = 0 \Rightarrow (x - 1)^{2} + y^{2} = 1$ .

This is the circle with centre $(1; 0)$ and radius $1$ .

Parametrisation. $x = 1 + cos θ$ , $y = sin θ$ , $θ \in [0; 2 π]$ .

The length element.

d s = (- sin θ)^{2} + (cos θ)^{2} d θ = d θ .

This is a general property of a circle of radius $R$ : $d s = R d θ$ , and here $R = 1$ .

Integration.

\frac{1}{π} \int_{0}^{2 π} (1 + cos θ - sin θ) d θ .

The integrals of $cos θ$ and $sin θ$ over a full period are zero, so only the constant remains:

\frac{1}{π} \cdot 2 π = 2.

What this result means: the mean value of $x - y$ over the circle equals the abscissa of the centre, that is $1$ , and the length of the circle is $2 π$ . The factor $1/ π$ in the statement is chosen so that the answer comes out whole.

Answer: $2$ .

Problem 3 (The mass and centre of mass of a helical wire). A wire has the shape of one turn of the helix $x = 3 cos t$ , $y = 3 sin t$ , $z = 4 t$ , $t \in [0; 2 π]$ , and its linear density at each point equals the $z$ -coordinate: $ρ = z$ .

Find the mass of the wire and the $z$ -coordinate of its centre of mass.

Solution.

The mass of a wire with linear density $ρ$ is a line integral of the first kind, $m = \int_{C} ρ d s$ , and a coordinate of the centre of mass is the first moment divided by the mass: $z_{c} = \frac{1}{m} \int_{C} z ρ d s$ .

The length element. $r^{'} (t) = (- 3 sin t; 3 cos t; 4)$ , whence $∣ r^{'} (t) ∣ = 9 sin^{2} t + 9 cos^{2} t + 16 = 5$ and $d s = 5 d t$ . The speed along a helix is constant — its characteristic property, and it keeps the integrals simple.

The mass. On the curve $ρ = z = 4 t$ :

m = \int_{0}^{2 π} 4 t \cdot 5 d t = 20 \cdot \frac{( 2 π ) ^{2}}{2} = 40 π^{2} .

The first moment.

\int_{C} z ρ d s = \int_{0}^{2 π} 16 t^{2} \cdot 5 d t = 80 \cdot \frac{( 2 π ) ^{3}}{3} = \frac{640 π ^{3}}{3} .

The centre of mass.

z_{c} = \frac{640 π ^{3} /3}{40 π ^{2}} = \frac{16 π}{3} .

A sense check: the turn is $8 π$ high, and with a uniform density the centre of mass would be halfway up, at $4 π$ . The density grows with height, so the centre of mass has moved up: $16 π /3 \approx 5.33 π > 4 π$ .

Answer: $m = 40 π^{2}$ , $z_{c} = \frac{16 π}{3}$ .

Problem 4 (The length of a cardioid). Find the length of the cardioid given in polar coordinates by $r = 2 (1 + cos φ)$ .

Solution.

The length element in polar coordinates. The curve $r = r (φ)$ is $x = r (φ) cos φ$ , $y = r (φ) sin φ$ . Differentiating gives $x^{'2} + y^{'2} = r^{2} + r^{'2}$ , so

d s = r^{2} + r^{'2} d φ .

The radicand. Here $r^{'} = - 2 sin φ$ , and

r^{2} + r^{'2} = 4 (1 + cos φ)^{2} + 4 sin^{2} φ = 8 (1 + cos φ) = 16 cos^{2} \frac{φ}{2} .

So $d s = 4 cos \frac{φ}{2} d φ$ — with the absolute value. That is the trap in this problem: on $[0; 2 π]$ the function $cos (φ /2)$ changes sign at $φ = π$ , and if the absolute value is lost, the integral over the full turn comes out as zero.

Symmetry. The cardioid is symmetric about the $O x$ axis, so it is enough to take the upper half, $φ \in [0; π]$ , where $cos (φ /2) ⩾ 0$ , and double it:

L = 2 \int_{0}^{π} 4 cos \frac{φ}{2} d φ = 8 \cdot [2 sin \frac{φ}{2}]_{0}^{π} = 16.

Answer: $L = 16$ .

Problem 5 (The centre of mass of an astroid arc). Find the coordinates of the centre of mass of the homogeneous arc of the astroid $x = cos^{3} t$ , $y = sin^{3} t$ that lies in the first quadrant.

Solution.

For a homogeneous arc the density cancels, and the centre of mass is determined by the geometry alone: $x_{c} = \frac{1}{L} \int_{C} x d s$ , $y_{c} = \frac{1}{L} \int_{C} y d s$ , where $L$ is the length of the arc.

The length element. $x^{'} = - 3 cos^{2} t sin t$ , $y^{'} = 3 sin^{2} t cos t$ , so

x^{'2} + y^{'2} = 9 sin^{2} t cos^{2} t (cos^{2} t + sin^{2} t) = 9 sin^{2} t cos^{2} t .

Its square root is $3∣ sin t cos t ∣$ . In the first quadrant $t \in [0; π /2]$ both factors are non-negative, so $d s = 3 sin t cos t d t$ without the absolute value — the choice of quadrant is exactly what settles the sign.

The length. $L = \int_{0}^{π /2} 3 sin t cos t d t = \frac{3}{2}$ .

The first moment.

\int_{C} x d s = \int_{0}^{π /2} cos^{3} t \cdot 3 sin t cos t d t = 3 \int_{0}^{π /2} cos^{4} t sin t d t = 3 \cdot \frac{1}{5} = \frac{3}{5} .

Hence $x_{c} = \frac{3/5}{3/2} = \frac{2}{5}$ . The arc is symmetric about the line $y = x$ (the substitution $t \to π /2 - t$ swaps $x$ and $y$ ), so $y_{c} = x_{c}$ .

Note that the centre of mass of an arc does not lie on the arc itself — the point $(2/5; 2/5)$ is between the curve and the origin.

Answer: $(\frac{2}{5}; \frac{2}{5})$ .

Problem 6* (An integral over a circle in space without a parametrisation). The curve $Γ$ is the intersection of the sphere $x^{2} + y^{2} + z^{2} = 9$ with the plane $x + y + z = 0$ .

Compute $\oint_{Γ} (x^{2} + y) d s$ .

Solution.

This circle can be parametrised, but awkwardly: it needs an orthonormal basis in an inclined plane. The problem is solved without a parametrisation — by symmetry.

What the curve is. The plane $x + y + z = 0$ passes through the centre of the sphere, so $Γ$ is a great circle of radius $R = 3$ centred at the origin. Its length is $L = 2 π R = 6 π$ .

The first term. Both the sphere and the plane are unchanged by any permutation of the variables $x, y, z$ . So the curve is mapped to itself as well, and with it the integrals are equal:

\oint_{Γ} x^{2} d s = \oint_{Γ} y^{2} d s = \oint_{Γ} z^{2} d s .

Their sum is known without computation: on the curve $x^{2} + y^{2} + z^{2} = 9$ , so

3 \oint_{Γ} x^{2} d s = \oint_{Γ} 9 d s = 9 \cdot 6 π = 54 π, \oint_{Γ} x^{2} d s = 18 π .

The second term. $\oint_{Γ} y d s = L \cdot y_{c}$ , where $y_{c}$ is the $y$ -coordinate of the centre of mass of the homogeneous circle. The centre of mass of a circle is its centre, the origin, so $\oint_{Γ} y d s = 0$ .

Result. $\oint_{Γ} (x^{2} + y) d s = 18 π$ .

The permutation trick works only when the curve really is symmetric: for the plane $x + 2 y + z = 0$ the integrals of $x^{2}$ and $y^{2}$ would already differ.

Answer: $18 π$ .

What it is

How to compute it

Polar coordinates

Mass and centre of mass

Symmetry

Common mistakes

Problems

line-integral-first-kind.ipynb