Double integrals in polar coordinates

The last week of the course. Last week you took double integrals in Cartesian coordinates. But if the region is a disk, annulus or sector, the limits turn into radical expressions ($y = \sqrt{1-x^2}$) and the integral often gets ugly, even impossible to compute. Polar coordinates make these regions like rectangles: “radius from $0$ to $2$, angle from $0$ to $2\pi$”. If there is $x^2+y^2$ in the integrand, it also becomes a single $r^2$. The exam has at least one polar question, and often a “convert to polar and evaluate” question.

Golden rules:

1. $dA = r\,dr\,d\theta$. The extra factor $r$ is never forgotten.
2. If the region is bounded by a disk, annulus, sector or a curve $r = f(\theta)$, and/or the integrand has $x^2+y^2$, switch to polar.
3. First sketch the region, read off the $\theta$ range and the $r$ range for each $\theta$.
4. When converting from Cartesian, put $r\cos\theta$, $r\sin\theta$ in place of $x$, $y$; set up the limits again from the region.

Polar coordinates reminder

A point is given by its distance $r$ from the origin and the angle $\theta$ it makes with the positive $x$ axis:

$x = r\cos\theta,\qquad y = r\sin\theta,\qquad x^2+y^2 = r^2,\qquad\tan\theta = \frac yx.$

Why $r\,dr\,d\theta$?

In a Cartesian grid the small piece was a rectangle, with area $\Delta x\,\Delta y$. In a polar grid the small piece is an annular sector: inner radius $r$, thickness $\Delta r$, angular width $\Delta\theta$. The length of its outer edge is about $r\Delta\theta$ (arc length = radius $\times$ angle). So the area is about

$\Delta A\approx(r\,\Delta\theta)(\Delta r) = r\,\Delta r\,\Delta\theta.$

As you move away from the origin, the same $\Delta\theta$ gives a wider arc; the factor $r$ accounts for this.

Theorem. If $D = \{\alpha\le\theta\le\beta,\ r_1(\theta)\le r\le r_2(\theta)\}$ then

$\iint_Df(x,y)\,dA = \int_\alpha^\beta\!\!\int_{r_1(\theta)}^{r_2(\theta)}f(r\cos\theta,\,r\sin\theta)\,r\,dr\,d\theta.$

Writing regions in polar form

Template: draw a ray from the origin into the region. Where does the ray enter the region (lower limit $r_1$) and where does it exit (upper limit $r_2$)? From which angle to which angle must you rotate the ray to sweep the region ($\alpha$, $\beta$)?

Example 1. The area of a disk of radius $3$: $\displaystyle\int_0^{2\pi}\!\!\int_0^3r\,dr\,d\theta = 2\pi\cdot\frac92 = 9\pi$. Check: $\pi\cdot3^2$ ✓.

Example 2. $\displaystyle\iint_D\sqrt{x^2+y^2}\,dA$, $D$ the unit disk. $\sqrt{x^2+y^2} = r$:

$\int_0^{2\pi}\!\!\int_0^1r\cdot r\,dr\,d\theta = 2\pi\cdot\frac13 = \frac{2\pi}{3}.$

Example 3 (annulus, quarter). $\iint x\,dA$ on the part of the annulus $1\le x^2+y^2\le4$ in the first quadrant:

r: 1 → 2,   θ: 0 → π/2,   x = r cos θ
∫_0^{π/2} ∫_1^2 (r cos θ) r dr dθ = (∫_0^{π/2} cos θ dθ)(∫_1^2 r² dr)
                                  = 1 · (8 - 1)/3 = 7/3

When the limits are constant and the integrand has the form $g(\theta)h(r)$, the integral separates into a product of two single integrals. This happens very often in polar questions.

Example 4 (bounded by a polar curve). The area inside the circle $r = 4\sin\theta$. The circle is traced once on the interval $\theta\in[0,\pi]$.

$\int_0^\pi\!\!\int_0^{4\sin\theta}r\,dr\,d\theta = \int_0^\pi8\sin^2\theta\,d\theta = 8\cdot\frac\pi2 = 4\pi.$

Check: $x^2+y^2 = 4y$, that is, $x^2+(y-2)^2 = 4$, radius $2$, area $4\pi$ ✓.

PITFALL · the $\theta$ range. If you write $0\le\theta\le2\pi$ for the circle $r = 2\cos\theta$, you go around the circle twice and find twice the area. Use the interval in which the polar curve is traced once: the angles where $r\ge0$. For rose curves ($r = \cos2\theta$), find the $r\ge0$ interval for one petal as well.

Converting from Cartesian to polar

Example 5. $\displaystyle\int_{-1}^1\!\!\int_0^{\sqrt{1-x^2}}(x^2+y^2)\,dy\,dx$.

1) Read the region: -1 ≤ x ≤ 1,  0 ≤ y ≤ √(1-x²)  → the UPPER half of the unit disk
2) Polar limits: 0 ≤ r ≤ 1,  0 ≤ θ ≤ π
3) Integrand: x² + y² = r²;   dy dx → r dr dθ
4) ∫_0^π ∫_0^1 r² · r dr dθ = π · (1/4) = π/4

In Cartesian this integral would need a long computation with terms like $y^3/3$ and $(1-x^2)^{3/2}$.

TIP · a radical in the limit is a signal. If you see $\sqrt{a^2-x^2}$ or $\sqrt{a^2-y^2}$ in a limit, the region is most likely a piece of a disk. If the integrand also has $x^2+y^2$, the question is calling you to polar.

The Gaussian integral

$\int e^{-x^2}dx$ has no elementary antiderivative; but its value over the whole line is found with the polar trick. If $I = \int_{-\infty}^\infty e^{-x^2}dx$ then

$I^2 = \int_{-\infty}^\infty e^{-x^2}dx\int_{-\infty}^\infty e^{-y^2}dy = \iint_{\mathbb R^2}e^{-(x^2+y^2)}dA = \int_0^{2\pi}\!\!\int_0^\infty e^{-r^2}r\,dr\,d\theta = 2\pi\cdot\frac12 = \pi.$

$I = \sqrt\pi$. The $\sqrt{2\pi}$ in the formula of the normal distribution in probability comes from here. The heart of the trick is again the factor $r$: the expression $e^{-r^2}r$ has an antiderivative, while $e^{-r^2}$ does not.

Volume

For solids whose base region is a disk or annulus, the volume $\iint(\text{top}-\text{bottom})\,dA$ is still valid; you just write it in polar form. To find the base, equate the two surfaces or look at where the surface meets $z = 0$. In Q7 the paraboloid gives the circle $r = 3$ at $z = 0$; in Q8 the cylinder determines the base directly as $r\le1$.

Common mistakes