Stretching grids: the Jacobian factor

Mathematics · Calculus & Analysis · ages 20-21

Name ______________________ Date ____________

The map x = 2u, y = 3v has constant partials. What is its Jacobian determinant?
Answer: ______________
What does the Jacobian of a coordinate map measure?
- How much the map stretches area near each point
- How curved the boundary of the region is
- How many variables the integral has
What is the area element in polar coordinates?
- dr dt
- r dr dt
- r squared dr dt
The map x = 2u, y = 3v has constant partials. What is its Jacobian determinant?
Answer: ______________
The map u = x/2, v = y/3 sends the rectangle 0 <= x <= 4, 0 <= y <= 9 to a rectangle in the uv-plane. What is the area of that uv rectangle?
Answer: ______________
Which is the volume element in cylindrical coordinates?
- dz dr dtheta
- r dz dr dtheta
- r squared dz dr dtheta
For x = u + v and y = u - v, what is the absolute value of the Jacobian determinant?
Answer: ______________
What is the Jacobian determinant of the map x = u + v, y = u - v?
- 2
- -2
- 0
- 1
A student integrates over a disc of radius 2 but writes the area element as dr dt. What is the effect?
- The result is unchanged; r only shifts the limits
- The integral becomes impossible to evaluate
- The result is too small, missing a factor that grows with r
What is the Jacobian determinant of the map x = 3u + v, y = u + 2v?
Answer: ______________

Answer key

For grown-ups. Fold this page away before handing over the rest.

6 · The Jacobian matrix is diagonal with entries 2 and 3, so the determinant is 2 x 3 = 6.
How much the map stretches area near each point · J is the local stretch factor, so it corrects the area element.
r dr dt · The polar Jacobian is r, so area is r dr dt.
6 · The determinant is 2 times 3, which is 6.
6 · u runs from 0 to 2 and v runs from 0 to 3, so the image is a 2 by 3 rectangle of area 6.
r dz dr dtheta · Cylindrical stretching contributes one factor of r, as polar does.
2 · The determinant is minus 2, and area uses its absolute value, 2.
-2 · Partials give the matrix [[1, 1], [1, -1]], whose determinant is (1)(-1) - (1)(1) = -2.
The result is too small, missing a factor that grows with r · Dropping r discards the stretch, so area near the rim counts far too little.
5 · The matrix [[3, 1], [1, 2]] has determinant 3 x 2 - 1 x 1 = 5.