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A transformation of the plane, the Jacobian determinant as the local area-stretching factor, why the formula carries an absolute value, how polar and spherical coordinates fall out as special cases, and choosing a substitution that straightens a region.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to build the Jacobian matrix of a transformation, compute its determinant, and use the change-of-variables formula to rewrite a double integral over a transformed region. You will also be able to recover the polar and spherical volume elements as special cases, choose a substitution that straightens an awkward region, and say what a negative Jacobian determinant means.
Substitution in a one-variable integral, where $dx$ became $\frac{dx}{du}\,du$; the extra factor of $r$ in polar coordinates and of $\rho^{2}\sin\phi$ in spherical ones; and the Jacobian matrix from the chain rule lesson. This lesson is the single rule all of those turn out to be instances of.
A transformation $T$ sends a point $(u,v)$ to a point $(x,y) = (x(u,v), y(u,v))$. It is one-to-one on a region when no two points of that region go to the same image. The Jacobian matrix is the matrix of the four partial derivatives; the Jacobian determinant, written $\partial(x,y)/\partial(u,v)$ or $J$, is its determinant. The region $S$ in the $uv$ plane is the preimage of the region $R$ that $T$ carries it to.
Let $T$ carry a region $S$ in the $uv$ plane one-to-one onto a region $R$ in the $xy$ plane, with continuous partial derivatives. Then
$$\int\!\!\int_R f(x,y)\,dA = \int\!\!\int_S f\big(x(u,v), y(u,v)\big)\;\left|\frac{\partial(x,y)}{\partial(u,v)}\right|\,du\,dv,$$
where the Jacobian determinant is
$$\frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} \\[6pt] \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v} \end{vmatrix} = x_u y_v - x_v y_u.$$
Why a determinant. Near a point, $T$ is well approximated by its linear part, and a linear map sends the unit square spanned by $\mathbf{i}$ and $\mathbf{j}$ to the parallelogram spanned by the two columns of the Jacobian matrix. The area of that parallelogram is the absolute value of the determinant. So $|J|$ is the local area-stretching factor, and $du\,dv$ times $|J|$ is the area of the image cell.
Why the absolute value. An area is positive. A negative determinant means $T$ reverses orientation — it turns the plane over — which is real information about the map and no information at all about the size of an area.
The old cases, recovered. For polar coordinates $x = r\cos\theta$, $y = r\sin\theta$,
$$J = \begin{vmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{vmatrix} = r\cos^2\theta + r\sin^2\theta = r,$$
which is the factor that was simply asserted two lessons ago. The spherical factor $\rho^{2}\sin\phi$ is the three-variable determinant of the same kind. And in one variable the determinant is the single number $dx/du$, which is ordinary substitution — the only case where the absolute value can be dodged, by letting the limits swap instead.
Choosing a substitution. Look at the region. If its boundary is $x + y = 1$ and $x - y = 3$, put $u = x + y$ and $v = x - y$ and the region becomes a rectangle. Simplifying the integrand is a bonus; simplifying the region is the point.
Another way: picture
Draw a fine square grid on the $uv$ plane and watch what $T$ does to it. The grid arrives in the $xy$ plane bent and stretched: each little square has become a little parallelogram, and $|J|$ at that point is how many times bigger it is. Where $J = 0$ the parallelogram has collapsed to a segment, and the map has folded the plane onto itself there.
Another way: steps
To change variables in a double integral:
Evaluate $\int\int_R (x + y)\,dA$ over the parallelogram with vertices $(0,0)$, $(2,0)$, $(3,1)$ and $(1,1)$. In $xy$ coordinates this needs the region split or the limits written as awkward functions.
The edges lie on $y = 0$, $y = 1$, $x - y = 0$ and $x - y = 2$. So put $u = x - y$ and $v = y$. Then $x = u + v$ and $y = v$, and the region becomes the rectangle $0 \le u \le 2$, $0 \le v \le 1$.
The Jacobian matrix is $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$, whose determinant is $1$ — this transformation is a shear, and a shear preserves area exactly, which is a fact worth knowing independently. The integrand $x + y$ becomes $u + 2v$, so
$$\int_0^1\!\!\int_0^2 (u + 2v)\,du\,dv = \int_0^1 (2 + 4v)\,dv = 4.$$
Notice the order of decisions. The substitution came from the edges of the region; the integrand was rewritten afterwards and happened to stay simple. Choosing $u = x + y$ to simplify the integrand instead would have left a region no easier than the one we started with.
Dropping the absolute value. It produces negative areas and negative masses, which are not small errors but impossible answers.
Inverting the wrong way. The formula needs $x$ and $y$ as functions of $u$ and $v$. If the substitution is given the other way round, either solve it or use the fact that $\partial(x,y)/\partial(u,v)$ is the reciprocal of $\partial(u,v)/\partial(x,y)$ — a genuine shortcut, and one that must be applied to the determinant rather than entry by entry.
Forgetting to change the region. The new limits describe $S$, not $R$. An integral with the new integrand, the new factor and the old limits is over a region that exists in neither plane.
Using a map that is not one-to-one. If $T$ covers the region twice the integral comes out doubled. Polar coordinates on a full disc are only one-to-one once $r > 0$ and $\theta$ is restricted to one turn, and that is why those conventions are stated so carefully.
It is tempting to file $|J|$ alongside the $r$ of polar coordinates as a thing that has to be remembered, a tax on changing variables. It is the opposite: it is the only content of the change-of-variables formula. Everything else — rewriting the integrand, relabelling the limits — is bookkeeping that changes no quantity. The determinant is where the geometry lives.
Read that way, the formula says something simple and memorable. A transformation takes small cells to small parallelograms; the determinant of the derivative is the area of the parallelogram a unit cell becomes; so integrating over the image is integrating over the source with each cell weighted by how much it grew. Every special case in this course, from ordinary substitution to the volume element of a sphere, is that one sentence with the derivative written out.
For $x = r\cos\theta$ and $y = r\sin\theta$, the four partial derivatives are $x_r = \cos\theta$, $x_{\theta} = -r\sin\theta$, $y_r = \sin\theta$ and $y_{\theta} = r\cos\theta$.
The Jacobian matrix is built from these four.
Its determinant is $\cos\theta \cdot r\cos\theta - (-r\sin\theta)\cdot\sin\theta = r(\cos^2\theta + \sin^2\theta) = r$.
One Pythagorean identity finishes it.
So $dA = r\,dr\,d\theta$, which two lessons ago was justified by a picture of an arc. The picture was right, and this is the general theorem it was a special case of.
A formula that was assumed is now proved.
Evaluate $\int\int_R e^{(y-x)/(y+x)}\,dA$ over the triangle with vertices $(0,0)$, $(1,0)$ and $(0,1)$. The integrand demands $u = y - x$ and $v = y + x$.
Here the integrand chooses the substitution, unusually.
Then $x = (v-u)/2$ and $y = (u+v)/2$, so the Jacobian matrix is $\begin{pmatrix} -1/2 & 1/2 \\ 1/2 & 1/2 \end{pmatrix}$ with determinant $-1/2$, and $|J| = 1/2$.
The sign says the map reflects; the size says it halves areas.
The triangle becomes $0 \le v \le 1$, $-v \le u \le v$, so the integral is $\tfrac12\int_0^1\!\int_{-v}^{v} e^{u/v}\,du\,dv = \tfrac12\int_0^1 v(e - e^{-1})\,dv = \tfrac14(e - e^{-1})$.
The inner integral is now elementary, which it was not before.
The partial derivatives are $x_u = 2u$, $x_v = -2v$, $y_u = 2v$ and $y_v = 2u$.
Four derivatives, one per entry.
So $J = (2u)(2u) - (-2v)(2v) = 4u^2 + 4v^2$.
This is never negative, and it is zero only at the origin — so the map preserves orientation everywhere and folds only at that single point. It is the squaring map of complex numbers, and $|J| = 4|w|^2$ is exactly the $|f'|^2$ that the complex analysis course will derive by a completely different route.
Write the Jacobian matrix of the transformation $x = 2u + v$, $y = -4u + 2v$.
This task has no paper form; do it on a device.
Put the stages of changing variables in a double integral over a region of area $5$ into order.
Number the steps in order (write the number in the box):
For $x = 4u - 5v$ and $y = 4u + 2v$, what is the Jacobian determinant $\partial(x,y)/\partial(u,v)$?
Answer:
The map $x = 9u$, $y = 4v$ carries the unit disc to an ellipse. The ellipse's area is a multiple of $\pi$. What is the multiplier?
Answer:
Write the Jacobian matrix of the transformation $x = u + 4v$, $y = 2u + 2v$.
This task has no paper form; do it on a device.
A student computes a Jacobian determinant of $-6$ and reports the transformed area as $-6$ times the original. What has gone wrong?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Put the stages of changing variables in a double integral over a region of area $3$ into order.
Number the steps in order (write the number in the box):
You can build a Jacobian, read its determinant as an area-stretching factor, and change variables in a double integral by choosing a substitution that straightens the region. Next: fields that assign a vector to every point, and the two derivatives that describe them.
10. Your turn: the Jacobian of $x = u^2 - v^2$, $y = 2uv$, step 3