From Cartesian Monomials to Polar Integrals
A Cartesian monomial \(x^m y^n\) may look less convenient to integrate over a circular region than a radial function. In polar coordinates, however, its powers separate into a radial factor and an angular factor. This makes the calculation systematic: evaluate one power integral in \(r\), one trigonometric integral in \(\theta\), and multiply the results.
For a disk or annulus centered at the origin, the polar area element is \(dA=r\,dr\,d\theta\). The extra factor \(r\) is essential: it changes the radial power by one.
The General Formula for a Disk or Annulus
Let the region be a disk of radius \(R>0\), or a concentric annulus with inner radius \(a\) and outer radius \(b\), where \(0\leq a<b\). In either case the angular coordinate covers a full revolution, \(0\leq\theta\leq2\pi\). Write the radial interval as \(\alpha\leq r\leq\beta\): for a disk, \(\alpha=0\) and \(\beta=R\); for an annulus, \(\alpha=a\) and \(\beta=b\).
Because the transformed monomial is a product of a function of \(r\) and a function of \(\theta\), the double integral separates into two single integrals. The radial factor is a power integral. The angular factor depends on the exponents and on the full-circle bounds.
The angular factor has a useful parity rule. If either exponent is odd, its full-circle integral is zero. If both exponents are even, write \(m=2p\) and \(n=2q\), where \(p,q\) are nonnegative integers. Then the angular factor is
Here \(k!\) denotes the factorial of a nonnegative integer \(k\), with \(0!=1\). Together, these formulas give an explicit answer for every Cartesian monomial with nonnegative integer exponents on a disk or concentric annulus.
Why the Angular Formula Works
First consider the zero case. If \(m\) is odd, replacing \(\theta\) by \(\pi-\theta\) reverses the sign of \(\cos\theta\) and leaves \(\sin\theta\) unchanged. The full-circle integral cancels between corresponding parts of the circle. If \(n\) is odd, replacing \(\theta\) by \(-\theta\) reverses the sign of \(\sin\theta\) and leaves \(\cos\theta\) unchanged; periodicity again gives cancellation. Thus the angular integral is zero whenever at least one exponent is odd.
For even exponents, define the first-quadrant integral \(I_{pq}=\int_0^{\pi/2}\cos^{2p}\theta\sin^{2q}\theta\,d\theta\). The integrand has the same integral on all four quadrants, so the full-circle integral is \(4I_{pq}\). The following recurrence explains the factorial formula.
For \(p\geq1\), integrate the derivative of \(\cos^{2p-1}\theta\sin^{2q+1}\theta\) from \(0\) to \(\pi/2\). The expression is zero at both endpoints, so its derivative has integral zero. Using \(\sin^2\theta=1-\cos^2\theta\) gives $$ (2p-1)I_{p-1,q}=(2p+2q)I_{pq}. $$ Thus \(I_{pq}=\frac{2p-1}{2p+2q}I_{p-1,q}\).
Interchanging sine and cosine, or making the substitution \(\theta\mapsto\frac{\pi}{2}-\theta\), gives \(I_{pq}=\frac{2q-1}{2p+2q}I_{p,q-1}\) for \(q\geq1\).
Repeatedly applying these recurrences reduces the integral to \(I_{00}=\int_0^{\pi/2}1\,d\theta=\pi/2\). The resulting expression is $$ I_{pq}=\frac{\pi}{2}\frac{(2p)!(2q)!}{4^{p+q}p!q!(p+q)!}. $$ Multiplying by four gives the stated full-circle formula.
The separation of the double integral is justified because the polar region is a rectangle in the \((r,\theta)\) bounds and the transformed integrand is a product of a radial function and an angular function. Equivalently, integrating the product over these independent bounds gives the product of the two single integrals.
Worked Examples
Worked Example: \(x^2y^2\) over a Disk of Radius 2
Evaluate \(\iint_D x^2y^2\,dA\), where \(D\) is the disk \(0\leq r\leq2\).
Here \(m=n=2\). The polar integrand, including the area element, is $$ r^{2+2+1}\cos^2\theta\sin^2\theta = r^5\cos^2\theta\sin^2\theta. $$ The angular exponents are \(2p=2\) and \(2q=2\), so \(p=q=1\). The angular factor is $$ 2\pi\frac{2!\,2!}{4^2\,1!\,1!\,2!} = 2\pi\frac{4}{32} = \frac{\pi}{4}. $$ The radial factor is $$ \int_0^2r^5\,dr = \left[\frac{r^6}{6}\right]_0^2 = \frac{64}{6} = \frac{32}{3}. $$ Therefore, $$ \iint_Dx^2y^2\,dA = \frac{32}{3}\cdot\frac{\pi}{4} = \frac{8\pi}{3}. $$ As a check, the antiderivative \(r^6/6\) differentiates to \(r^5\), and the angular formula gives \(\pi/4\) for these exponents. Their product is exactly the value above.
Worked Example: \(x^2y^2\) over an Annulus
Evaluate \(\iint_A x^2y^2\,dA\) over the annulus \(1\leq r\leq2\).
The angular factor is unchanged from the disk example, since the annulus also covers \(0\leq\theta\leq2\pi\): $$ \int_0^{2\pi}\cos^2\theta\sin^2\theta\,d\theta = \frac{\pi}{4}. $$ The radial factor now has lower limit \(1\): $$ \int_1^2r^5\,dr = \left[\frac{r^6}{6}\right]_1^2 = \frac{2^6-1^6}{6} = \frac{63}{6} = \frac{21}{2}. $$ Thus $$ \iint_Ax^2y^2\,dA = \frac{21}{2}\cdot\frac{\pi}{4} = \frac{21\pi}{8}. $$ For a check, the radial antiderivative still differentiates to \(r^5\), and its endpoint difference is \(64/6-1/6=63/6\). The annulus excludes the inner disk, as reflected in the subtraction of \(1^6/6\).
Worked Example: \(x^4y^2\) over the Unit Disk
Evaluate \(\iint_Dx^4y^2\,dA\) for the unit disk \(0\leq r\leq1\).
Here \(m=4\), \(n=2\), so \(m+n+1=7\). The angular powers are \(2p=4\), \(2q=2\), or \(p=2\), \(q=1\). Therefore, $$ \int_0^{2\pi}\cos^4\theta\sin^2\theta\,d\theta = 2\pi\frac{4!\,2!}{4^3\,2!\,1!\,3!} = 2\pi\frac{48}{64\cdot12} = \frac{\pi}{8}. $$ The radial integral is $$ \int_0^1r^7\,dr = \left[\frac{r^8}{8}\right]_0^1 = \frac{1}{8}. $$ Multiplying gives $$ \iint_Dx^4y^2\,dA = \frac{\pi}{8}\cdot\frac{1}{8} = \frac{\pi}{64}. $$ The check is direct: differentiating \(r^8/8\) gives \(r^7\), and the angular formula at \(p=2,q=1\) evaluates to \(\pi/8\).
Worked Example: \(x^3y^2\) over a Disk of Radius 3
Evaluate \(\iint_Dx^3y^2\,dA\) over the disk \(0\leq r\leq3\).
The transformed integrand is $$ r^{3+2+1}\cos^3\theta\sin^2\theta = r^6\cos^3\theta\sin^2\theta. $$ The cosine exponent is odd, so the angular integral over a full revolution is zero. In particular, the angular integrand at \(\pi-\theta\) is the negative of its value at \(\theta\), and the contributions cancel. The radial factor is finite: $$ \int_0^3r^6\,dr = \left[\frac{r^7}{7}\right]_0^3 = \frac{3^7}{7} = \frac{2187}{7}. $$ Consequently, $$ \iint_Dx^3y^2\,dA = \frac{2187}{7}\cdot0 = 0. $$ The radial antiderivative differentiates to \(r^6\); the zero result comes from the angular factor, not from the radial calculation.
A Common Pitfall: Forgetting the Area Element
A frequent error is to substitute \(x=r\cos\theta\) and \(y=r\sin\theta\) but omit the additional \(r\) in \(dA=r\,dr\,d\theta\). For example, that mistake would turn \(x^2y^2\,dA\) into \(r^4\cos^2\theta\sin^2\theta\,dr\,d\theta\), instead of \(r^5\cos^2\theta\sin^2\theta\,dr\,d\theta\).
On the unit disk, the incorrect calculation would give $$ \frac{\pi}{4}\int_0^1r^4\,dr = \frac{\pi}{4}\cdot\frac{1}{5} = \frac{\pi}{20}. $$ The correct calculation includes the Jacobian factor: $$ \frac{\pi}{4}\int_0^1r^5\,dr = \frac{\pi}{4}\cdot\frac{1}{6} = \frac{\pi}{24}. $$ The radial exponent must be \(m+n+1\), not merely \(m+n\).
Using the Formula Reliably
For any monomial \(x^m y^n\) on a centered disk or annulus, first add the exponents and then add one more power of \(r\) for the area element. Next check whether either angular exponent is odd. If so, the integral is zero. Otherwise, use the even-power formula for the angular factor and evaluate the radial power between its actual inner and outer radii.
This separation is useful because the two factors record different information: the angular integral describes how the monomial varies around a circle, while the radial integral accounts for the range of distances included in the region. Changing a disk to an annulus changes the radial factor but leaves the full-circle angular factor the same.
Check Your Understanding
Use the polar form and the full-circle bounds to answer these questions.
- What radial power occurs after converting \(x^3y^4\,dA\) to polar coordinates?
- Why is the integral of \(x^2y^3\) over a disk centered at the origin zero?
- What is the angular integral of \(\cos^2\theta\sin^2\theta\) from \(0\) to \(2\pi\)?
- For \(x^2y^2\) over \(2\leq r\leq4\), what endpoint difference appears in the radial factor?
- What factor is lost if the polar area element is incorrectly written as \(dr\,d\theta\)?