Recipe

Evaluate a line integral of a vector field

Use when: you need \int_C \mathbf{F}\cdot d\mathbf{r} for a specific curve C and field \mathbf{F}

Goal: the numerical value of the work integral \int_C \mathbf{F}\cdot d\mathbf{r}

Taught in The Fundamental Theorem for line integrals

1. 1

   Check whether \mathbf{F} is already known, or stated, to be a gradient field \nabla f for some specific f.

   Example

   For \mathbf{F}=(2xy,\ x^2+3y^2), Section 3 already found f(x,y)=x^2y+y^3 with \nabla f=\mathbf{F}, so this field qualifies for the fast route.

   • \mathbf{F}=\nabla f for a known f — skip directly to Step 2.
   • no potential function is known or given — continue to Step 3 and parametrize C directly.

   Why: recognizing which method applies before starting avoids parametrizing a curve whose shape is about to become irrelevant.

2. 2

   Find C's true starting point A and ending point B, ignoring everything about the curve's interior shape, and compute f(B)-f(A).

   Example

   Let C be any piecewise-smooth curve from A=(0,0) to B=(1,2). Then f(B)-f(A) = (1^2\cdot2+2^3) - (0) = 10.

   Why: the Fundamental Theorem guarantees \int_C\mathbf{F}\cdot d\mathbf{r}=f(B)-f(A) exactly whenever \mathbf{F} is conservative, with no parametrization of C involved at all. This is the final answer for this branch.

3. 3

   Write C as \mathbf{r}(t) for t\in[a,b], then substitute \mathbf{r}(t)'s components into \mathbf{F}.

   Example

   Let \mathbf{F}=(y,\,-x) and \mathbf{r}(t)=(\cos t,\,\sin t), 0\le t\le \pi/2. Then \mathbf{F}(\mathbf{r}(t)) = (\sin t,\,-\cos t).

   Why: this is the same substitution every line integral needs; there is no shortcut available once no potential function is known.

4. 4

   Compute \mathbf{r}'(t).

   Example

   \mathbf{r}'(t) = (-\sin t,\,\cos t).

   Why: d\mathbf{r} = \mathbf{r}'(t)\,dt is what turns the line integral into an ordinary integral in t.

5. 5

   Compute \mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t), then integrate that scalar function from t=a to t=b.

   Example

   \mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t) = \sin t(-\sin t) + (-\cos t)(\cos t) = -1. So \displaystyle\int_0^{\pi/2}(-1)\,dt = -\frac{\pi}{2}.

   Why: this dot-and-integrate step is the definition of the line integral itself, and it works for any \mathbf{F} and any piecewise-smooth C, conservative or not. This is the final answer for this branch.