Line integral of F = <x^2, y> along the segment from (0, 0) to (1, 1) Step by Step

Step by step

The question

Question: find the line integral of F = ⟨x², y⟩ along the segment from (0, 0) to (1, 1).

F = ⟨x², y⟩

along the segment from (0, 0) to (1, 1)

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 = t

x = t

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

y = t

Then dx is the derivative of x, times dt,

dx = 1 dt

and dy the same for y.

dy = 1 dt

Here is the path on the grid.

It goes from (0, 0) to (1, 1).

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 = t² and Q = t.

F = ⟨t², t⟩

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

∫ P dx + Q dy

So P dx + Q dy is t² times 1, plus t times 1, all times dt.

P dx + Q dy = t² × 1 dt + t × 1 dt

Multiply out and add the like terms: t² + t.

P dx + Q dy = (t² + t) dt

Careful!

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

They carry the derivatives 1 and 1.

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 (t² + t) dt = [t³3 + t²2] from 0 to 1

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

∫ from 0 to 1 of (t² + t) dt = ((1)³3 + (1)²2) − ((0)³3 + (0)²2)

The bottom part is 0.

So it is 5/6.

∫ from 0 to 1 of (t² + t) dt = 56

4 Check

Let's check another way.

Here ∂P/∂y equals ∂Q/∂x, so F has a potential: f = x³ over 3 + y² over 2.

∂P/∂y = 0 = ∂Q/∂x: f = x³ over 3 + y² over 2

Then the answer is f at the end minus f at the start: 5/6.

The same.

Right!

f(1, 1) − f(0, 0) = 5/6 − 0 = 5/6

Let's look back.

First: write the path in t.

Then: put the path in.

Then: integrate.

Last: check.

Answer