How to Set Up a Double Integral Over x² + y² ≤ 4 in Polar Coordinates

The region $x^2+y^2\leq 4$ is the disk centered at the origin with radius $2$. It determines the limits of integration, but not the value of a double integral by itself: you also need the integrand. Since no integrand is specified here, the exact integral cannot be evaluated uniquely.

You can still set up any integral over this disk in polar coordinates. The setup and a worked example follow.

Convert the disk to polar coordinates

In polar coordinates, $x=r\cos\theta$, $y=r\sin\theta$, and $x^2+y^2=r^2$. Thus the condition becomes $r^2\leq 4$. Since $r$ is nonnegative, the radius ranges from $0$ to $2$. The disk includes every direction around the origin, so the angle ranges through a full turn.

The bounds are

$$0\leq r\leq 2,\qquad 0\leq\theta\leq 2\pi.$$

For a function $f(x,y)$, the change of variables gives

$$\iint_{x^2+y^2\leq 4} f(x,y)\,dA = \int_0^{2\pi}\int_0^2 f(r\cos\theta,r\sin\theta)\,r\,dr\,d\theta.$$

The factor $r$ is the Jacobian determinant for the polar-coordinate transformation. It accounts for how area changes: a small polar rectangle has area approximately $r\,dr\,d\theta$, not just $dr\,d\theta$.

Worked example: integrate $x^2+y^2$

Suppose the missing integrand were $x^2+y^2$. In polar coordinates it becomes $r^2$, so

$$\begin{aligned} \iint_{x^2+y^2\leq 4}(x^2+y^2)\,dA &=\int_0^{2\pi}\int_0^2 r^2\,r\,dr\,d\theta\\ &=\int_0^{2\pi}\int_0^2 r^3\,dr\,d\theta\\ &=\int_0^{2\pi}4\,d\theta\\ &=8\pi. \end{aligned}$$

Here the integrand contributes $r^2$, and the area element contributes another factor of $r$. The radial integral is $\int_0^2 r^3\,dr=4$.

Different integrands give different answers over the same disk. For instance, if the integrand is $1$, the integral is the disk’s area:

$$\int_0^{2\pi}\int_0^2 r\,dr\,d\theta=4\pi.$$

This confirms why the region alone does not determine a unique value.

A practical setup method

For a specific problem, proceed in this order:

1. Rewrite the region using $r$ and $\theta$.
2. Set the limits so that each point in the region is covered once.
3. Substitute $x=r\cos\theta$ and $y=r\sin\theta$ into the integrand.
4. Multiply by $r$ for the area element, then integrate.

For this centered disk, the bounds are particularly simple. If a problem instead gives part of the disk, such as an upper semicircle, the radial bounds may still be $0$ to $2$, but the angular interval must be restricted to cover just that part.

Common mistakes

• Leaving out the Jacobian: The polar area element is $dA=r\,dr\,d\theta$. Omitting $r$ changes the integral.
• Using the wrong radius: The boundary equation is $r^2=4$, hence the radius is $2$, not $4$.
• Using incomplete angular bounds: For the entire disk, the angle must cover a full revolution, such as $0\leq\theta\leq2\pi$.
• Assuming the region is the integrand: The inequality specifies where to integrate, not what function to integrate. A numerical answer requires the function or a complete integral expression.