I'm working with double integrals, and came about a problem where I had to calculate the following integral:

$$ \int \int_D \left | 3x+4y \right|dx dy$$

where D is the region contained within the circle whose equation is given by $x^2+y^2 \leq 2$.

At first I thought I directly might use polar coordinates, but that seemed to complicate things when we have absolute values in the integrand. I went to check on if there was any hints to the problem, and there was.

The book hinted that a substitution of the form $$\begin{pmatrix} u \\v \end{pmatrix} = \begin{pmatrix}3/5 & 4/5 \\-4/5& 3/5 \\\end{pmatrix} \begin{pmatrix}x \\y \end{pmatrix}$$

And when I made that substitution, I realized that the new region, let's call it $A$ became $u^2+v^2 \leq 2$, the Jacobian became 1, and the double integral became:

$$\int \int_A 5 |u| du dv$$ which then easily can be calculated through polar coordinates.

It was the first time seeing the substitution, so I tried to generalize it for a double integral of the form:

$$I := \int \int_D \left | \alpha x +\beta y \right | dx dy$$ where $D$ is a region given by the circle whose equation is $x^2+y^2 \leq \gamma^2$. I later used the same method. First I let:

$$\begin{pmatrix} u \\v \end{pmatrix} = \frac{1}{||(\alpha, \beta)||}\begin{pmatrix}\alpha & \beta \\- \beta& \alpha \\\end{pmatrix} \begin{pmatrix}x \\y \end{pmatrix}$$

Then, as previously, I get the equivalent region $A$ given by $u^2+v^2\leq \gamma^2$, and the Jacobian $J = 1$, just as in the previous numerical example. We also realize that $\left| \alpha x +\beta y \right | = ||(\alpha, \beta)|| \cdot|u|$, and so our double integral just becomes:

$$||(\alpha, \beta)|| \int \int_A |u| dudv$$

which then, through polar coordinates, gives us that $I$ can be evaluated to $\frac{4 ||(\alpha, \beta)|| \gamma^3}{3}$.

What I realized from this, is that this substitution is really efficient for calculating double integrals with the given integrand. However, I can't really geometrically see how these transformations look like.

I think that if someone could provide any geometrical intution on what these transformations look like between the $xy$ - plane, to the $uv$ - plane and to the $r\theta$ - plane, this would really deepen my understanding on why this substitution is efficient in the way it is.

Thank you in advance for any such contributions.

  • 2
    $\begingroup$ $xy \to uv$ is a rotation (presumably) to a more favorable coordinate system. $\endgroup$
    – Gregory
    Feb 4, 2022 at 14:55

1 Answer 1


This transformation is a rotation of the plane, since $\begin{pmatrix}\alpha & \beta \\- \beta& \alpha \\\end{pmatrix}$ is a rotation matrix for $\alpha,\beta$ satisfying $\alpha^2+\beta^2=1$. This makes it clear why the domain of integration remains a circle and also why the Jacobian determinant is $1$ (rotations don't stretch the plane).

The function you're integrating is $z=|3x+4y|$, and other than the absolute value, the graph of this function is a plane $M$ in $\mathbb{R}^3$. By rotating the $x,y$ plane to become a new plane with basis vectors $u,v$ we can make $M$ be the graph of a function of the form $z=k u$ (or $z=kv$). Geometrically this is possible since we can rotate until the function describing the plane "rises only along one of the coordinates" $u$ or $v$, that is, is constant along the other coordinate.

Concretely, we want a rotation which aligns one of the coordinates with the direction of steepest ascent. Steepest ascent means largest directional derivative and this happens along the direction of the gradient, that is, along $(3,4)$, or when normalized, $(\frac{3}{5},\frac{4}{5})$. The rotation matrix which sends $(1,0)$ to $(\frac{3}{5},\frac{4}{5})$ is \begin{pmatrix}3/5 & -4/5 \\4/5& 3/5 \\\end{pmatrix} so that after the rotation the plane $M$ is going to be the graph of $z=ku$ (here $k=5$).

  • 1
    $\begingroup$ Thank you for that beautiful explanation! I get what's happening now :) $\endgroup$
    – Tanamas
    Feb 5, 2022 at 10:30

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