Practice: Line Integrals and Path Parametrisation

Direct application

Evaluate $\int_{C} (- y) d x + x d y$ where $C$ runs from $(0, 0)$ to $(1, 0)$ along the $x$ -axis, then from $(1, 0)$ to $(1, 1)$ parallel to the $y$ -axis.

Give the exact value.

Your answer

Enter the value. It is checked against the answer and the precision this task asks for.

2 hints available, least help first.

Hint 1: Retrieval cue

On a leg where one coordinate is constant, that differential is zero and one term drops out.

Hint 2: Next step

Substitute $y = 0$ into the first leg and $x = 1$ into the second before integrating.

Classification · Recognition

For $F = (- y, x)$ , the integral around the unit square $[0, 1] \times [0, 1]$ traversed counterclockwise is $2$ .

What is the integral around the same square traversed clockwise?

Give the exact value.

Your answer

Enter the value. It is checked against the answer and the precision this task asks for.

2 hints available, least help first.

Hint 1: Retrieval cue

Substitute every occurrence of $x$ and $y$ , not only the differentials.

Hint 2: Concept cue

Reduce the whole expression to a single integral in the parameter before evaluating.

Direct application

Evaluate $\int_{C} (2 x y) d x + (x^{2}) d y$ where $C$ is the curve $r (t) = (t, t^{2})$ for $0 \leq t \leq 1$ .

Give the exact value.

Your answer

Enter the value. It is checked against the answer and the precision this task asks for.

2 hints available, least help first.

Hint 1: Retrieval cue

Substitute every occurrence of $x$ and $y$ , not only the differentials.

Hint 2: Concept cue

Reduce the whole expression to a single integral in the parameter before evaluating.

Error diagnosis · Classification

A student evaluates $\int_{C} (x y) d x + (x^{2}) d y$ along $r (t) = (t, t^{2})$ for $0 \leq t \leq 1$ and writes:

" $d x = d t$ and $d y = 2 t d t$ , so the integral is $\int_{0}^{1} [x y + 2 t x^{2}] d t$ ."

What has gone wrong?

2 hints available, least help first.

Hint 1: Retrieval cue

After substituting, how many different letters should remain in the integrand?

Hint 2: Concept cue

The parametrisation gives $x$ and $y$ in terms of $t$ . Use both.

Construction · Direct application · Explanation

Let $F = (y, 2 x)$ .

(a) Evaluate $\int_{C} F \cdot d r$ where $C$ is the straight segment from $(0, 0)$ to $(2, 1)$ , stating the parametrisation and its range.

(b) Evaluate the same integral along the polygonal route $(0, 0) \to (2, 0) \to (2, 1)$ , treating each leg separately and saying which term vanishes on each and why.

(c) Traverse the route in (b) backwards, from $(2, 1)$ to $(0, 0)$ . Give the value, and say what the comparison establishes about which object a line integral is a property of.

Write your answer

Write your answer, then compare it with the worked solution.

3 hints available, least help first.

Hint 1: Retrieval cue

Substitute every occurrence of $x$ and $y$ , not only the differentials.

Hint 2: Concept cue

On a leg where one coordinate is constant, its differential is zero and one term drops.

Hint 3: Strategy cue

In (c), nothing about the field changed. Ask what reversing the range does to $d x$ and $d y$ .

Compare with the worked solution

Comparing does not record a result. Judging your own written answer cannot show that you can do this without help.

(a) The straight segment. Parametrise $r (t) = (2 t, t)$ for $0 \leq t \leq 1$ , so $d x = 2 d t$ and $d y = d t$ . Substituting both the coordinates and the differentials:

\int_{0}^{1} [(t) (2) + 2 (2 t) (1)] d t = \int_{0}^{1} 6 t d t = 3 .

The range must be stated: the same point set traced by $r (t) = (t, t / 2)$ over $0 \leq t \leq 2$ gives the same $3$ , because the substitution introduces a derivative factor that cancels against the change of variable. A parametrisation is a device for computing, not part of the answer.

(b) The polygonal route. Each leg separately.

Leg 1, $(0, 0) \to (2, 0)$ . Here $y = 0$ , so $d y = 0$ and the $2 x d y$ term vanishes; and $P = y = 0$ , so the $P d x$ term vanishes too. Contribution $0$ .

Leg 2, $(2, 0) \to (2, 1)$ . Here $x = 2$ is constant, so $d x = 0$ and the $P d x$ term vanishes. What remains is $\int_{0}^{1} 2 (2) d y = 4$ .

Total: $0 + 4 = 4$ .

Note $3 \neq 4$ : the two routes share endpoints and disagree, so this field's integral depends on the path. Why is one line: $\partial P / \partial y = 1$ against $\partial Q / \partial x = 2$ , which differ, so no potential exists and path dependence was to be expected.

(c) The same route reversed. Reversing direction reverses the sign of every $d x$ and $d y$ while leaving $F$ untouched, so the integral changes sign: $- 4$ . Verified leg by leg: from $(2, 1)$ to $(2, 0)$ gives $\int_{1}^{0} 4 d y = - 4$ , and the final leg from $(2, 0)$ to $(0, 0)$ contributes $0$ as before.

What the comparison establishes. The point set is identical in (b) and (c); only the direction of travel differs, and the value differs by a factor of $- 1$ . So a line integral is a property of the oriented curve, not of the curve as a set of points. A value quoted without a stated direction is ambiguous by a sign, which is why the orientation belongs beside the number rather than being inferred from context.

A complete answer does each of these:

parametrises and evaluates
reports orientation

Practice: Line Integrals and Path Parametrisation

Direct application

Classification · Recognition

Direct application

Error diagnosis · Classification

Construction · Direct application · Explanation

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