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Line integral of F = <x, y> along the segment from (0, 0) to (3, 4)

Step by step, with the reason for every step · Line integrals and work · Calculus 3

Step by step

The question

Question: find the line integral of F = ⟨x, y⟩ along the segment from (0, 0) to (3, 4).

F = ⟨x, y⟩
along the segment from (0, 0) to (3, 4)

1 Write the path in t

A segment: start, plus t times the step from start to end.

t goes from 0 to 1.

start + t × (end − start)

So x = 3t

x = 3t

and y = 4t, with t from 0 to 1.

y = 4t

Then dx is the derivative of x, times dt,

dx = 3 dt

and dy the same for y.

dy = 4 dt

Here is the path on the grid.

It goes from (0, 0) to (3, 4).

The blue arrows show F at a few points on the path.

2 Put the path in

Put x and y in t into F: P = 3t and Q = 4t.

F = ⟨3t, 4t⟩

The line integral is P dx plus Q dy, added along the path.

∫ P dx + Q dy

So P dx + Q dy is 3t times 3, plus 4t times 4, all times dt.

P dx + Q dy = 3t × 3 dt + 4t × 4 dt

Multiply out and add the like terms: 25t.

P dx + Q dy = 25t dt

Careful!

Many people put t into F but forget dx and dy.

They carry the derivatives 3 and 4.

Careful: dx and dy become dt pieces too

3 Integrate

Integrate each term: the power goes up by 1, and we divide by the new power.

∫ from 0 to 1 of 25t dt = 25t22 from 0 to 1

Put in the top t = 1, then take away the bottom t = 0.

∫ from 0 to 1 of 25t dt = 25(1)22 − 25(0)22

The bottom part is 0.

So it is 25/2.

∫ from 0 to 1 of 25t dt = 252

4 Check

Let's check another way.

Here ∂P/∂y equals ∂Q/∂x, so F has a potential: f = x² over 2 + y² over 2.

∂P/∂y = 0 = ∂Q/∂x: f = x² over 2 + y² over 2

Then the answer is f at the end minus f at the start: 25/2.

The same.

Right!

f(3, 4) − f(0, 0) = 25/2 − 0 = 25/2

Let's look back.

First: write the path in t.

Then: put the path in.

Then: integrate.

Last: check.

Answer
∫ from 0 to 1 of 25t dt = 252

Your turn

Try this one yourself, then check:

F = ⟨3x, 2y⟩, along the segment from (0, 0) to (2, 3)
Show the answer
15

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