Worked examples
The question
Question: find the line integral of F = ⟨y, x⟩ along the segment from (0, 0) to (1, 2).
F = ⟨y, x⟩
along the segment from (0, 0) to (1, 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 = t
x = t
and y = 2t, with t from 0 to 1.
y = 2t
Then dx is the derivative of x, times dt,
dx = 1 dt
and dy the same for y.
dy = 2 dt
Here is the path on the grid.
It goes from (0, 0) to (1, 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 = 2t and Q = t.
F = ⟨2t, t⟩
The line integral is P dx plus Q dy, added along the path.
∫ P dx + Q dy
So P dx + Q dy is 2t times 1, plus t times 2, all times dt.
P dx + Q dy = 2t × 1 dt + t × 2 dt
Multiply out and add the like terms: 4t.
P dx + Q dy = 4t dt
Careful!
Many people put t into F but forget dx and dy.
They carry the derivatives 1 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 4t dt = 2t2 from 0 to 1
Put in the top t = 1, then take away the bottom t = 0.
∫ from 0 to 1 of 4t dt = 2(1)2 − 2(0)2
The bottom part is 0.
So it is 2.
∫ from 0 to 1 of 4t dt = 2
4 Check
Let's check another way.
Here ∂P/∂y equals ∂Q/∂x, so F has a potential: f = xy.
∂P/∂y = 1 = ∂Q/∂x: f = xy
Then the answer is f at the end minus f at the start: 2.
The same.
Right!
f(1, 2) − f(0, 0) = 2 − 0 = 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 4t dt = 2
▶ Watch it solved
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 ∫ from 0 to 1 of (6 + 4t) dt = 8
▶ Watch it solved
The question
Question: find the work done by F = ⟨xy, x⟩ along y = x² from (0, 0) to (2, 4).
F = ⟨xy, x⟩
along y = x² from (0, 0) to (2, 4)
1 Write the path in t
On the curve, let x be t.
Then y is t².
t goes from 0 to 2.
x = t, y = t²
So x = t
x = t
and y = t², with t from 0 to 2.
y = t2
Then dx is the derivative of x, times dt,
dx = 1 dt
and dy the same for y.
dy = 2t dt
Here is the path on the grid.
It goes from (0, 0) to (2, 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 = t³ and Q = t.
F = ⟨t3, t⟩
The line integral is P dx plus Q dy, added along the path.
∫ P dx + Q dy
So P dx + Q dy is t³ times 1, plus t times 2t, all times dt.
P dx + Q dy = t3 × 1 dt + t × 2t dt
Multiply out and add the like terms: t³ + 2t².
P dx + Q dy = (t3 + 2t2) dt
Careful!
Many people put t into F but forget dx and dy.
They carry the derivatives 1 and 2t.
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 2 of (t3 + 2t2) dt = [t44 + 2t33] from 0 to 2
Put in the top t = 2, then take away the bottom t = 0.
∫ from 0 to 2 of (t3 + 2t2) dt = ((2)44 + 2(2)33) − ((0)44 + 2(0)33)
The bottom part is 0.
So it is 28/3.
∫ from 0 to 2 of (t3 + 2t2) dt = 283
4 Check
Let's check: cut the path into 200 tiny steps and add F dot each step, by computer.
About 9.33.
200 small steps: sum of F • step ≈ 9.333
Our answer is 9.333.
Almost the same.
Right!
28/3 = 9.333
Let's look back.
First: write the path in t.
Then: put the path in.
Then: integrate.
Last: check.
Answer ∫ from 0 to 2 of (t3 + 2t2) dt = 283
▶ Watch it solved
