Line integral of F = <3, y> along the segment from (0, 0) to (2, 2) Step by Step

Step by step

The question

Question: find the line integral of F = ⟨3, y⟩ along the segment from (0, 0) to (2, 2).

F = ⟨3, y⟩

along the segment from (0, 0) to (2, 2)

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 = 2t

x = 2t

and y = 2t, with t from 0 to 1.

y = 2t

Then dx is the derivative of x, times dt,

dx = 2 dt

and dy the same for y.

dy = 2 dt

Here is the path on the grid.

It goes from (0, 0) to (2, 2).

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 = 3 and Q = 2t.

F = ⟨3, 2t⟩

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

∫ P dx + Q dy

So P dx + Q dy is 3 times 2, plus 2t times 2, all times dt.

P dx + Q dy = 3 × 2 dt + 2t × 2 dt

Multiply out and add the like terms: 6 + 4t.

P dx + Q dy = (6 + 4t) dt

Careful!

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

They carry the derivatives 2 and 2.

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 (6 + 4t) dt = [6t + 2t²] from 0 to 1

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

∫ from 0 to 1 of (6 + 4t) dt = (6(1) + 2(1)²) − (6(0) + 2(0)²)

The bottom part is 0.

So it is 8.

∫ from 0 to 1 of (6 + 4t) dt = 8

4 Check

Let's check another way.

Here ∂P/∂y equals ∂Q/∂x, so F has a potential: f = 3x + y² over 2.

∂P/∂y = 0 = ∂Q/∂x: f = 3x + y² over 2

Then the answer is f at the end minus f at the start: 8.

The same.

Right!

f(2, 2) − f(0, 0) = 8 − 0 = 8

Let's look back.

First: write the path in t.

Then: put the path in.

Then: integrate.

Last: check.

Answer