Linear Substitutions for Elliptic Regions

Why Transform an Ellipse?

An ellipse can have unequal axis lengths, a shifted center, or axes that are not aligned with the coordinate directions. These features can make a double integral awkward to set up directly. A suitable linear substitution removes the elliptic shape: it maps the region to a disk, where symmetry and polar coordinates are easier to use.

The substitution must do more than change the boundary. It also changes the area element and the integrand. The determinant of the transformation gives the area-scaling factor, while substituting into the integrand expresses the quantity being integrated in the new coordinates.

Key idea: Map the ellipse to a disk, transform the integrand using the same map, and multiply by the absolute value of the Jacobian determinant.

Scaling an Axis-Aligned Ellipse

Consider the ellipse centered at the origin with semiaxis lengths $a>0$ and $b>0$:

E=\left\{(x,y):\frac{x^2}{a^2}+\frac{y^2}{b^2}\leq 1\right\}.

Set $x=au$ and $y=bv$. Then the boundary condition becomes $u^2+v^2\leq 1$, so the transformed region is the unit disk $D$. The Jacobian determinant of the map $(u,v)\mapsto(x,y)$ is $ab$, and therefore $dA=ab\,du\,dv$.

\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}a&0\\0&b\end{pmatrix} \begin{pmatrix}u\\v\end{pmatrix}, \qquad \left|\frac{\partial(x,y)}{\partial(u,v)}\right| = \left|\begin{vmatrix}a&0\\0&b\end{vmatrix}\right| =ab.

Thus, for an integrable function $f$, the change-of-variables theorem gives the integral over the ellipse as an integral over the disk, with both the function and area element transformed.

\iint_E f(x,y)\,dA = ab\iint_D f(au,bv)\,du\,dv.

Worked Example: Finding the Area of an Axis-Aligned Ellipse

Find the area of $E=\{(x,y):x^2/9+y^2/4\leq 1\}$. Here $a=3$ and $b=2$. Use $x=3u$, $y=2v$, so the image is the unit disk and the absolute Jacobian is $3\cdot2=6$.

For the area, the integrand is $1$. Hence

\begin{aligned} \operatorname{Area}(E) &=\iint_E 1\,dA\\ &=6\iint_D 1\,du\,dv\\ &=6\pi\\ &=6\pi. \end{aligned}

The unit disk has area $\pi$, so the transformed area is $6\pi$. As a check, the ellipse has semiaxes $3$ and $2$, and its area is $\pi ab=\pi(3)(2)=6\pi$, agreeing with the substitution.

The Matrix Form of an Ellipse

A centered ellipse can also be specified by a quadratic form. Let $\mathbf{A}$ be a real symmetric positive-definite $2\times2$ matrix. Positive definiteness means that $\mathbf{z}^{\top}\mathbf{A}\mathbf{z}>0$ for every nonzero vector $\mathbf{z}$. For $\rho>0$, define the elliptic region

E=\left\{\mathbf{x}\in\mathbb{R}^2: \mathbf{x}^{\top}\mathbf{A}\mathbf{x}\leq \rho^2\right\}.

Symmetry and positive definiteness ensure that the ellipse has perpendicular principal axes and positive squared axis scales. More precisely, $\mathbf{A}$ has an orthonormal eigenbasis with positive eigenvalues. In that basis its quadratic form is a sum of positive multiples of the squares of the coordinates.

Definition: The positive-definite square root $\mathbf{A}^{1/2}$ is the symmetric matrix formed by keeping the eigenvectors of $\mathbf{A}$ and replacing each positive eigenvalue $\lambda$ with $\sqrt{\lambda}$. It satisfies $\mathbf{A}^{1/2}\mathbf{A}^{1/2}=\mathbf{A}$.

Use the linear substitution $\mathbf{u}=\mathbf{A}^{1/2}\mathbf{x}/\rho$, or equivalently $\mathbf{x}=\rho\mathbf{A}^{-1/2}\mathbf{u}$. Then $\mathbf{x}^{\top}\mathbf{A}\mathbf{x}=\rho^2\mathbf{u}^{\top}\mathbf{u}$. Thus the ellipse maps to the unit disk. The absolute determinant of the inverse map is $\rho^2/\sqrt{\det\mathbf{A}}$.

Theorem: If $\mathbf{A}$ is symmetric positive definite and $\rho>0$, then the substitution $\mathbf{x}=\rho\mathbf{A}^{-1/2}\mathbf{u}$ maps $\mathbf{u}^{\top}\mathbf{u}\leq1$ onto $\mathbf{x}^{\top}\mathbf{A}\mathbf{x}\leq\rho^2$, and $$ dA=\frac{\rho^2}{\sqrt{\det\mathbf{A}}}\,du\,dv. $$ Consequently, the area of the ellipse is $\pi\rho^2/\sqrt{\det\mathbf{A}}$.

Why the Determinant Gives the Area Factor

The map from $\mathbf{u}$ to $\mathbf{x}$ is linear. Its Jacobian matrix is $\rho\mathbf{A}^{-1/2}$. In two dimensions, multiplying a matrix by $\rho$ multiplies its determinant by $\rho^2$. Also, the determinant of $\mathbf{A}^{-1/2}$ is the reciprocal of the positive square root of $\det\mathbf{A}$. Taking the absolute determinant gives the area factor.

Transform the boundary.
For $\mathbf{x}=\rho\mathbf{A}^{-1/2}\mathbf{u}$, substitute into the quadratic form to get $\mathbf{x}^{\top}\mathbf{A}\mathbf{x}=\rho^2\mathbf{u}^{\top}\mathbf{u}$. The ellipse condition is equivalent to $\mathbf{u}^{\top}\mathbf{u}\leq1$.

Compute the Jacobian.
The derivative matrix is $\rho\mathbf{A}^{-1/2}$. Its determinant has absolute value $\rho^2|\det(\mathbf{A}^{-1/2})|$.

Simplify the determinant.
If the eigenvalues of $\mathbf{A}$ are $\lambda_1,\lambda_2>0$, then $\det\mathbf{A}=\lambda_1\lambda_2$ and $\det(\mathbf{A}^{-1/2})=1/\sqrt{\lambda_1\lambda_2}=1/\sqrt{\det\mathbf{A}}$. Hence $dA=\rho^2/\sqrt{\det\mathbf{A}}\,du\,dv$.

Integrate the constant function.
The image is the unit disk, whose area is $\pi$. Multiplying by the area factor gives $\operatorname{Area}(E)=\pi\rho^2/\sqrt{\det\mathbf{A}}$.

This also explains the axis-aligned formula: taking $\mathbf{A}=\operatorname{diag}(1/a^2,1/b^2)$ and $\rho=1$ gives $\sqrt{\det\mathbf{A}}=1/(ab)$, so the area is $\pi ab$.

Integrating Quadratic Forms

The same map is useful when the integrand contains the quadratic form that defines the ellipse. Under $\mathbf{x}=\rho\mathbf{A}^{-1/2}\mathbf{u}$, the form becomes $\rho^2\mathbf{u}^{\top}\mathbf{u}$. On the unit disk, this is $\rho^2 r^2$ in polar coordinates. Since $\int_D r^2\,dA=\int_0^{2\pi}\int_0^1 r^3\,dr\,d\theta=\pi/2$, the corresponding integral has a simple general value.

Useful consequence: For the ellipse above, $$ \iint_E \mathbf{x}^{\top}\mathbf{A}\mathbf{x}\,dA = \frac{\rho^4}{\sqrt{\det\mathbf{A}}} \iint_D \mathbf{u}^{\top}\mathbf{u}\,du\,dv = \frac{\pi\rho^4}{2\sqrt{\det\mathbf{A}}}. $$

Worked Example: Integrating the Defining Quadratic Form

Evaluate $\iint_E (x^2/9+y^2/4)\,dA$ over $E=\{(x,y):x^2/9+y^2/4\leq1\}$. Use $x=3u$, $y=2v$. The transformed region is $D$, the absolute Jacobian is $6$, and the integrand becomes $u^2+v^2$.

In polar coordinates, $u=r\cos\theta$, $v=r\sin\theta$, and $du\,dv=r\,dr\,d\theta$. Therefore,

\begin{aligned} \iint_E \left(\frac{x^2}{9}+\frac{y^2}{4}\right)dA &=6\iint_D (u^2+v^2)\,du\,dv\\ &=6\int_0^{2\pi}\int_0^1 r^2r\,dr\,d\theta\\ &=6\left(\int_0^{2\pi}d\theta\right)\left(\int_0^1r^3\,dr\right)\\ &=6(2\pi)\left(\frac14\right)\\ &=3\pi. \end{aligned}

For a check, the general formula has $\rho=1$ and $\det\mathbf{A}=1/36$, so it gives $\pi/(2\sqrt{1/36})=\pi/(2/6)=3\pi$, as required.

A Rotated Ellipse and a Different Quadratic Integrand

A matrix substitution also handles ellipses whose axes are rotated relative to the $x$- and $y$-axes. The orthonormal coordinates along the principal axes keep lengths and angles unchanged; scaling those coordinates then maps the ellipse to a disk. The Jacobian must include both scales.

Worked Example: Integrating Over a Rotated Ellipse

Evaluate $\iint_E(x^2+y^2)\,dA$, where $E$ is given by $2x^2+2xy+2y^2\leq6$. Introduce orthonormal coordinates $s=(x+y)/\sqrt{2}$ and $t=(x-y)/\sqrt{2}$. Direct expansion gives $x^2+y^2=s^2+t^2$ and $2x^2+2xy+2y^2=3s^2+t^2$. Thus the region is $3s^2+t^2\leq6$.

Set $s=\sqrt{2}\,u$, $t=\sqrt{6}\,v$. The region condition becomes $6u^2+6v^2\leq6$, so $u^2+v^2\leq1$. In terms of the original coordinates, $x=u+\sqrt{3}\,v$ and $y=u-\sqrt{3}\,v$. The Jacobian determinant is

\left|\frac{\partial(x,y)}{\partial(u,v)}\right| = \left|\begin{vmatrix}1&\sqrt{3}\\1&-\sqrt{3}\end{vmatrix}\right| =2\sqrt{3}.

The integrand transforms as $x^2+y^2=2u^2+6v^2$. On the unit disk, symmetry gives $\iint_Du^2\,du\,dv=\iint_Dv^2\,du\,dv=\pi/4$; each value also follows by integrating $r^2\cos^2\theta$ or $r^2\sin^2\theta$ with area element $r\,dr\,d\theta$. Therefore,

\begin{aligned} \iint_E(x^2+y^2)\,dA &=2\sqrt{3}\iint_D(2u^2+6v^2)\,du\,dv\\ &=2\sqrt{3}\left(2\frac{\pi}{4}+6\frac{\pi}{4}\right)\\ &=2\sqrt{3}\left(2\pi\right)\\ &=4\pi\sqrt{3}. \end{aligned}

As a check, the transformed integrand and Jacobian together give $2\sqrt{3}(2u^2+6v^2)$, whose disk integral is $4\pi\sqrt{3}$, matching the computed value. The determinant is nonzero, so this linear map is invertible and the disk covers the ellipse exactly once.

A Common Jacobian Mistake

A frequent error is to use the reciprocal of the area factor in the wrong direction. For the ellipse $x^2/9+y^2/4\leq1$, the map from disk coordinates to ellipse coordinates is $x=3u,\ y=2v$. Its determinant is $6$, so $dA=6\,du\,dv$. A mistaken calculation might write $dA=(1/6)\,du\,dv$ and conclude that the area is $\pi/6$. That reverses the map: $du\,dv=(1/6)\,dA$, but the integral over the ellipse is being rewritten in terms of $du\,dv$. The correct relation gives area $6\iint_D1\,du\,dv=6\pi$.

Check the direction: If you define the original coordinates as functions of the new ones, compute $\left|\partial(x,y)/\partial(u,v)\right|$. This is the factor multiplying $du\,dv$ when replacing $dA$.

When the Method Applies

The matrix form requires a symmetric positive-definite matrix, which ensures that the quadratic inequality describes a bounded ellipse and that the linear map is invertible. A singular matrix has zero determinant and cannot give this disk-to-ellipse substitution. For an ellipse centered at a point $\mathbf{c}$, first include the translation $\mathbf{x}=\mathbf{c}+\rho\mathbf{A}^{-1/2}\mathbf{u}$. Translation does not change the Jacobian determinant. In every case, transform the integrand as well as the region, and use the absolute determinant for area.

Check Your Understanding

Use the disk substitution and keep track of both the transformed integrand and the area factor.

What substitution maps $x^2/16+y^2/25\leq1$ to the unit disk, and what is its absolute Jacobian?
What condition on a symmetric matrix ensures that its quadratic inequality defines a bounded, nondegenerate ellipse?
Why does a translation of the ellipse’s center not change the absolute Jacobian factor?
For a matrix-defined ellipse, what does the quadratic form become after mapping to the unit disk?
Which area element should be used when $x$ and $y$ are specified as functions of $u$ and $v$?