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How to compute a line integral along a straight segment

The recipe: parametrise, substitute, integrate

A line integral $\int_{L} P d x + Q d y$ along a straight segment is turned into an ordinary definite integral in three moves. Write both coordinates through one parameter, $x = x (t)$ , $y = y (t)$ ; replace $d x$ and $d y$ by $x^{'} (t) d t$ and $y^{'} (t) d t$ ; integrate from the value of $t$ at the starting point to the value at the finishing point. Two details decide the sign and the value. The limits follow the direction of travel - swapping the endpoints flips the sign. And when $\frac{\partial P}{\partial y} \neq = \frac{\partial Q}{\partial x}$ the field is not conservative, so the answer genuinely depends on the path: the fact that the path is a straight segment is part of the problem, not a convenience.

Problem

Find $\int_{B A} 2 (x^{2} + y^{2}) d x + (x + y)^{2} d y$ , where $A (1; 1)$ and $B (2; 2)$ , along the straight segment.

Answer

\int_{B A} 2 (x^{2} + y^{2}) d x + (x + y)^{2} d y = - \frac{56}{3} \approx - 18.67

Checked by hand: the substitution and the limits are correct. Note the direction: from A to B the answer would be +56/3.

Step-by-step solution

1. Parametrise the segment

Both points lie on the line $y = x$ , so the whole segment is described by one parameter:

x = t, y = t, d x = d t, d y = d t

The path runs from $B (2; 2)$ to $A (1; 1)$ , so $t$ runs from $2$ down to $1$ - the limits keep that order.

2. Substitute and simplify

\int_{B A} 2 (x^{2} + y^{2}) d x + (x + y)^{2} d y = \int_{2}^{1} [2 (t^{2} + t^{2}) + (t + t)^{2}] d t

Inside the brackets:

2 \cdot 2 t^{2} + (2 t)^{2} = 4 t^{2} + 4 t^{2} = 8 t^{2}

so the integral becomes

\int_{2}^{1} 8 t^{2} d t

3. Integrate

\int_{2}^{1} 8 t^{2} d t = [\frac{8 t ^{3}}{3}]_{2}^{1} = \frac{8}{3} - \frac{64}{3} = - \frac{56}{3} \approx - 18.67

The minus sign is not an accident: we walked from $B$ to $A$ , that is, in the direction of decreasing $t$ . Reversing the direction would give $+ \frac{56}{3}$ .

4. Why the path had to be given

Check the potential test for this field:

\frac{\partial P}{\partial y} = \frac{\partial}{\partial y} [2 (x^{2} + y^{2})] = 4 y, \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x} (x + y)^{2} = 2 (x + y)

These are not equal, so the field is not conservative and the integral really does depend on the route. Along a different curve joining the same two points the answer would be different.

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