Electricity-And-Magnetism · Unit 1 · Video 2 · Interactive Practice

Integrals as Sums: Along a Line and a Curved Path

IKey Formulas

Formula	Name	What it adds up
$\displaystyle\int_A^B \lambda(x)\,dx \equiv \lim_{\Delta x_i \to 0} \sum_{i=1}^{m} \lambda(x_i)\,\Delta x_i$	Integral as a limit of a sum	Density at a point of each slice × slice width
$Q = \displaystyle\int_A^B \lambda\,dx = \lambda(B - A)$	Uniform line charge	Constant density × length
$Q = \displaystyle\int_A^B kx^2\,dx = \frac{k\,(B^3 - A^3)}{3}$	Varying density $\lambda(x) = kx^2$	Each $dx$ weighted by the density at its own position
$L = \displaystyle\int_A^B \|d\vec{s}\|, \quad \|d\vec{s}\| = \sqrt{dx^2 + dy^2}$	Length along a curved path	Magnitudes of tangent displacement elements

Key Insight: $\int$ is a stretched S for summa: cut the region into small elements, weight each by what it carries, and add them up in the limit of vanishing elements.

IIVisualization 1 — Slice Sums Become the Integral

For $\lambda(x) = 3x^2$ on $0 \le x \le 2$ m, the slice sums close in on $Q = 8$ C as the slices shrink.

Slices M 4 Sample point xᵢ

IIIVisualization 2 — Uniform Versus Varying Density

Each element carries $dQ = \lambda\,dx$ : a constant density gives density × length, a varying one weights by position.

Density

IVVisualization 3 — Displacement Elements Along a Curve

Along a curve each piece $\Delta\vec{s}$ has length $\sqrt{\Delta x^2 + \Delta y^2}$ ; their sum approaches $L$ , longer than the chord.

Pieces

💡 The line integral electromagnetism actually uses weights each $d\vec{s}$ by a vector field through a dot product, $\int \vec{F} \cdot d\vec{s}$ — that comes next.

VQuiz Questions

Problem 1 · Uniformly Charged Rod

Given: A rod lies on the $x$ -axis from $x = 0.5$ m to $x = 2.5$ m with uniform linear charge density $\lambda = 3\ \mu\text{C/m}$ — find the total charge $Q$ .

✅ Correct! A constant comes out of the integral:

Q = \lambda(B - A) = 3 \times 2 = 6\ \mu\text{C}

❌ Not quite.

3 \times 2.5

uses the coordinate of the end, not the length. The rod is

B - A = 2

m long.

❌ Not quite.

\lambda

is charge per metre, so charge = density × length, not density ÷ length.

❌ Not quite. Each element carries

\lambda\,dx

; summing over the rod gives

\lambda

times its length

B - A

Show solution

Each element $dx$ carries $dQ = \lambda\,dx$ . With $\lambda$ constant:

Q = \int_{0.5}^{2.5} \lambda\,dx = \lambda \int_{0.5}^{2.5} dx = \lambda\,(2.5 - 0.5) = 3\ \mu\text{C/m} \times 2\ \text{m} = 6\ \mu\text{C}

This is exactly density × length, as reading the integral as a sum predicts.

Problem 2 · Density That Grows With Position

Given: A line from $A = 1$ m to $B = 2$ m carries $\lambda(x) = kx^2$ with $k = 6\ \text{C/m}^3$ — find the total charge $Q$ .

✅ Correct!

Q = \frac{k}{3}(B^3 - A^3) = 2 \times (8 - 1) = 14

❌ Close, but…

(B - A)^3 \ne B^3 - A^3

. Evaluate the antiderivative

kx^3/3

at each endpoint separately, then subtract.

❌ Close, but… the antiderivative of

x^2

x^3/3

, not

x^3

— the factor

\tfrac{1}{3}

is missing.

❌ Not quite.

\lambda(B) \times (B - A) = 24 \times 1

treats the whole line as if it had the density at its far end. The density varies, so each

dx

must be weighted by

\lambda

at its own position.

❌ Not quite. Integrate

kx^2

from

1

2

Show solution

Q = \int_1^2 kx^2\,dx = k\left[\frac{x^3}{3}\right]_1^2 = \frac{6\,(2^3 - 1^3)}{3} = 2 \times 7 = 14\ \text{C}

Sanity check: the density runs from $\lambda(1) = 6$ C/m to $\lambda(2) = 24$ C/m over $1$ m, so $Q$ must lie between $6$ C and $24$ C — and $14$ C does.

Problem 3 · Checking the Formula by Hand

Given: $\lambda(x) = 3x^2$ C/m on $0 \le x \le 2$ m, exact charge $Q = 8$ C. With $4$ slices the midpoint sum is $7.875$ C (off by $0.125$ C) — find the midpoint sum with $2$ slices, and predict the error with $8$ slices.

Midpoint sum with 2 slices of width 1 m?

Error of the midpoint sum with 8 slices?

✅ Correct! Errors

0.5 \to 0.125 \to 0.03125

C: halving

\Delta x

cuts the midpoint error by

4

, and the sums close in on

8

❌ That is the left-edge sum. The midpoints of

[0,1]

and

[1,2]

are

x = 0.5

and

x = 1.5

❌ That is the right-edge sum. The midpoints of

[0,1]

and

[1,2]

are

x = 0.5

and

x = 1.5

❌ Check the sum. A finite sum of slices is not yet the integral; evaluate

3x^2

x = 0.5

and

x = 1.5

and multiply each by

\Delta x = 1

❌ Close, but… compare

M = 2

and

M = 4

: the error went from

0.5

0.125

, a factor of

4

, not

2

❌ Check the pattern. Errors

0.5

(

M = 2

) and

0.125

(

M = 4

) shrink by a factor of

4

each time the slices halve.

Show solution

Two slices, $\Delta x = 1$ m, midpoints $0.5$ and $1.5$ :

\sum \lambda(x_i)\,\Delta x_i = 3(0.5)^2(1) + 3(1.5)^2(1) = 0.75 + 6.75 = 7.5\ \text{C}

The error is $8 - 7.5 = 0.5$ C. With $4$ slices it is $0.125$ C — four times smaller when $\Delta x$ halves. For this density the midpoint error is exactly $\tfrac{1}{2}(\Delta x)^2$ , so with $8$ slices ( $\Delta x = 0.25$ m):

8 - 7.96875 = \tfrac{1}{2}(0.25)^2 = 0.03125\ \text{C}

The error never reaches $0$ for finite $M$ ; only the limit $\Delta x \to 0$ gives exactly $8$ C.

Problem 4 · Charge on a Curved Path

Given: A wire bent into a semicircle of radius $R = 0.5$ m runs from $A$ to $B$ , the two ends of a diameter. It carries uniform $\lambda = 2\ \mu\text{C/m}$ — find $Q = \int_A^B \lambda\,|d\vec{s}|$ .

✅ Correct! The elements

|d\vec{s}|

add up to the arc length

\pi R = 0.5\pi

m, so

Q = 2 \times 0.5\pi = \pi \approx 3.14\ \mu\text{C}

❌ Not quite.

\lambda \times 2R

uses the straight distance

|AB| = 1

m. Length is measured along the curve, which is longer than the chord.

❌ Close, but…

2\pi R

is a full circle; the wire is only half of one.

❌ Not quite.

\lambda \pi R^2

multiplies a line density by an area. A line charge is summed over length elements

|d\vec{s}|

❌ Not quite. Sum

\lambda\,|d\vec{s}|

along the arc.

Show solution

A constant density comes out of the integral, leaving the length of the path:

Q = \int_A^B \lambda\,|d\vec{s}| = \lambda \int_A^B |d\vec{s}| = \lambda L

The path is half a circle, so $L = \pi R = 0.5\pi$ m (versus the chord $2R = 1$ m). Then

Q = 2\ \mu\text{C/m} \times 0.5\pi\ \text{m} = \pi\ \mu\text{C} \approx 3.14\ \mu\text{C}

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