# Line Integrals

Physics I · Differential Equations and Vector Calculus · https://tryals.app/learn/physics-i/line-integrals

## Scalar vs Vector Line Integrals

A **line integral** accumulates a field along a curve $C$. 

A **scalar line integral** integrates a scalar field $f$ along $C$:

$$\int_C f\,ds = \int_a^b f(\vec{r}(t))\,\|\vec{r}'(t)\|\,dt$$

A **vector line integral** integrates a vector field $\vec{F}$ along $C$, computing the **work** done:

$$\int_C \vec{F}\cdot d\vec{r} = \int_a^b \vec{F}(\vec{r}(t))\cdot\vec{r}'(t)\,dt$$

| Type | Integrand | Measure | Use Case |
| --- | --- | --- | --- |
| **Scalar** | $f(\vec{r})$ | $ds = \|\vec{r}'(t)\| dt$ | Mass, arc length |
| **Vector** | $\vec{F}(\vec{r})$ | $d\vec{r} = \vec{r}'(t) dt$ | Work, flow |

## Conservative Fields & Path Independence

A field $\vec{F}$ is **conservative** if $\vec{F} = \nabla\phi$ for some potential $\phi$. Then the integral depends only on **endpoints**:

$$\int_C \vec{F}\cdot d\vec{r} = \phi(\vec{r}_B) - \phi(\vec{r}_A)$$

**Test for conservativeness** in $\mathbb{R}^{3}$: verify $\nabla\times\vec{F} = \vec{0}$.

**Example**: Gravity work moving from $h_1$ to $h_2$:

$$W = -mg(h_2-h_1)$$

This is path-independent because gravity is conservative with $\phi = mgh$.

> **Common pitfall:** Path-independence requires a conservative field. Always verify $\nabla\times\vec{F}=\vec{0}$ on a simply connected domain before using endpoints.

## Practice questions

6 of this lesson's 14 practice questions, with answers. The full set is in the app.

### 1. Every vector field has the property that its line integral depends only on the endpoints.

**Answer:** False

**Why:** False. Only **conservative** fields are path-independent, equivalently, those with zero curl, or expressible as a gradient. A swirling field gives different answers along different paths and non-zero circulation round a closed loop. That is what separates a force storing potential energy from one that dissipates it.

Page: https://tryals.app/practice/physics-i/line-integrals/every-vector-field-has-the-property-that-its-line-integral-depends

### 2. Complete the statement with the correct term.

**Answer:** A vector field F is conservative if and only if its curl equals **zero**.

**Why:** $\nabla \times F$ = 0 is the necessary and sufficient condition (on simply connected domains).

Page: https://tryals.app/practice/physics-i/line-integrals/complete-the-statement-with-the-correct-term

### 3. A line integral in a conservative field depends only on the endpoints. What does that imply about a closed loop?

A. That the integral equals the area enclosed by the loop
B. That the loop must be traversed clockwise for the result to vanish
C. That the integral around any closed loop is zero
D. That conservative fields have no closed field lines

**Answer:** C. That the integral around any closed loop is zero

**Why:** Same endpoints, zero difference. A nonzero loop integral would let you gain energy by going round in a circle, which is why "conservative" and "has a potential" are the same statement.

Page: https://tryals.app/practice/physics-i/line-integrals/a-line-integral-in-a-conservative-field-depends-only-on-the

### 4. Which conditions guarantee that a vector field F in ℝ² is conservative?

A. ∮ F · $dr = 0$ for every closed curve
B. F has constant magnitude
C. $\partial P/\partial y$ = $\partial Q/\partial x$ (where $F = (P, Q)$)
D. $F = \nabla\varphi$ for some scalar function $\varphi$

**Answer:** A. ∮ F · $dr = 0$ for every closed curve; C. $\partial P/\partial y$ = $\partial Q/\partial x$ (where $F = (P, Q)$); D. $F = \nabla\varphi$ for some scalar function $\varphi$

**Why:** Equal mixed partials, existence of a potential, and zero circulation are all equivalent.

Page: https://tryals.app/practice/physics-i/line-integrals/which-conditions-guarantee-that-a-vector-field-f-in-is

### 5. You must compute the work $\int_C \vec{F}\cdot d\vec{r}$ along a complicated spiral path, and you notice $\vec{F} = \nabla f$ for some potential $f$. What is the fastest correct move?

A. The work is automatically zero for gradient fields
B. Parametrize the spiral and integrate term by term
C. Evaluate $f$ at the endpoints: work $= f(\text{end}) - f(\text{start})$
D. Replace the spiral by a straight line, the work is only approximately equal

**Answer:** C. Evaluate $f$ at the endpoints: work $= f(\text{end}) - f(\text{start})$

**Why:** Conservative fields make the path irrelevant: $\int_C \nabla f \cdot d\vec{r} = f(B) - f(A)$ exactly, spiral or not. This is why potential energy exists, and why "work done against gravity" depends on height gained, never on the route taken.

Page: https://tryals.app/practice/physics-i/line-integrals/you-must-compute-the-work-c-f-dr-along-a-complicated-spiral-path

### 6. Sort each property: does it hold for a **conservative** vector field or a **non-conservative** one?

**Answer:**

- Conservative: Work around every closed loop is zero, A potential function $f$ with $F = \nabla f$ exists
- Non-conservative: Work can differ along two paths with the same endpoints, Friction-like: always drains energy along any real path

**Why:** The three faces of conservativeness (potential exists ⇔ path-independent ⇔ zero loop integrals) stand or fall together. Friction fails all three at once, which is exactly why "potential energy of friction" is not a thing.

Page: https://tryals.app/practice/physics-i/line-integrals/sort-each-property-does-it-hold-for-a-conservative-vector-field-or-a