5.3 Double Integrals in Polar Coordinates

Big idea

Rectangular coordinates make circles awkward. A disk of radius $a$ becomes $-\sqrt{a^2-x^2} \le y \le \sqrt{a^2-x^2}$, and after the inner integral you are usually staring at a square root that has to be handled by a trigonometric substitution. Meanwhile the integrand itself often carries the combination $x^2 + y^2$, which is just the squared distance from the origin dressed in the wrong notation. Both problems have the same cure: describe the plane by distance and direction instead of by two perpendicular distances.

Setting $x = r\cos\theta$ and $y = r\sin\theta$ turns the disk into $0 \le r \le a$, $0 \le \theta \le 2\pi$, which is a rectangle in the $(r,\theta)$ plane. It turns $x^2 + y^2$ into $r^2$. What it does not do is preserve area, and that is the one thing you have to pay attention to. A grid in $r$ and $\theta$ does not cut the plane into congruent cells; near the origin the cells are slivers and far out they are wide. The correction factor that accounts for this is $r$, and it must ride along inside every polar double integral.

The correction is not a fudge. Take the piece of the plane between radii $r_1$ and $r_2$ and between angles $\theta_1$ and $\theta_2$. It is the difference of two circular sectors, so its area is exactly $\tfrac{1}{2}(r_2^2 - r_1^2)(\theta_2 - \theta_1)$. Factoring the difference of squares rewrites that as $\tfrac{1}{2}(r_2 + r_1)(r_2 - r_1)(\theta_2 - \theta_1)$, which is the average radius times $\Delta r$ times $\Delta\theta$. So each cell contributes $r^{*}\Delta r\, \Delta\theta$ with $r^{*}$ the midpoint radius, and the Riemann sum that results has the factor $r$ built into it from the start.

The payoff shows up twice. Some integrals are merely easier in polar form, and some are impossible without it, because the antiderivative that rectangular coordinates demand does not exist in elementary terms while the polar version is a routine substitution. Learning to spot $x^2 + y^2$ and circular boundaries is most of the skill.

Decoder

Substituting $x = r\cos\theta$ and $y = r\sin\theta$ converts a double integral over a region into an integral in $r$ and $\theta$ over the corresponding polar description, with the area element $dA$ replaced by $r\, dr\, d\theta$.

The whole content of the statement is the extra $r$. Every other piece is bookkeeping: rewrite the integrand in terms of $r$ and $\theta$, rewrite the region as a range of angles and a range of radii. Drop the $r$ and you are computing a weighted average with the wrong weights, which typically shows up as an answer with the wrong power of the radius in it.

The usual convention is $r \ge 0$, with $\theta$ sweeping an interval of length at most $2\pi$. Under that convention $r$ is never negative, so $|r| = r$ and no absolute value is needed on the area element. If you ever allow negative $r$ to trace a curve, the factor becomes $|r|$, and it is simpler to avoid the situation by choosing the angle range so that $r$ stays nonnegative.

Two signals tell you to switch. The first is a boundary that is a circle, a ring, a wedge, or any curve whose equation is short in polar form. The second is an integrand built from $x^2 + y^2$, including its square root $\sqrt{x^2+y^2} = r$ and anything composed with it. When both signals fire, the polar version is usually a one-line computation.

Definitions and results

Polar coordinates. With $x = r\cos\theta$ and $y = r\sin\theta$, the quantity $r \ge 0$ is the distance from the origin and $\theta$ is the angle from the positive $x$-axis. Consequently $x^2 + y^2 = r^2$ and $\tan\theta = y/x$ away from the $y$-axis.

Area of a polar rectangle. The set $r_1 \le r \le r_2$, $\theta_1 \le \theta \le \theta_2$ has area

$$ \frac{1}{2}\big( r_2^2 - r_1^2 \big)\big( \theta_2 - \theta_1 \big) = \bar{r}\, \Delta r\, \Delta\theta, \qquad \bar{r} = \frac{r_1+r_2}{2} $$

The area element. In the limit of fine grids the exact statement above becomes

$$ dA = r\, dr\, d\theta $$

Integrals over a polar rectangle. If $D$ is given by $a \le r \le b$ and $\alpha \le \theta \le \beta$, then

$$ \iint_D f(x,y)\, dA = \int_{\alpha}^{\beta} \int_a^b f(r\cos\theta,\, r\sin\theta)\, r\, dr\, d\theta $$

Regions bounded by polar curves. If the region is swept by angles $\alpha \le \theta \le \beta$ and, for each such angle, radii from $r = h_1(\theta)$ out to $r = h_2(\theta)$, then

$$ \iint_D f\, dA = \int_{\alpha}^{\beta} \int_{h_1(\theta)}^{h_2(\theta)} f(r\cos\theta,\, r\sin\theta)\, r\, dr\, d\theta $$

The angle limits are the constants here, playing the role that $x$ played in a type I region.

Area in polar form. Taking $f = 1$ in the previous formula and doing the inner integral gives

$$ A(D) = \frac{1}{2} \int_{\alpha}^{\beta} \Big( h_2(\theta)^2 - h_1(\theta)^2 \Big) d\theta $$

which is the area formula for polar curves from single-variable calculus, now derived rather than assumed.

Separation happens often. On a polar rectangle, integrands of the form $g(r)h(\theta)$ split into a product exactly as in rectangular coordinates. Since $r$ itself is a function of $r$ alone, a radially symmetric integrand contributes a factor of $\beta - \alpha$ from the angle.

Worked examples

An integral rectangular coordinates cannot finish

Evaluate $\iint_D e^{-(x^2+y^2)}\, dA$ where $D$ is the disk $x^2 + y^2 \le 4$.

In rectangular coordinates the inner integral would be $\int e^{-x^2}e^{-y^2}\, dy$ with the exponential in $y$ integrated against no helpful factor, and $e^{-y^2}$ has no elementary antiderivative. In polar coordinates the region is $0 \le r \le 2$, $0 \le \theta \le 2\pi$, and the $r$ from the area element is exactly the factor the substitution $u = r^2$ wants:

$$ \int_0^{2\pi} \int_0^2 e^{-r^2} r\, dr\, d\theta = \int_0^{2\pi} \left[ -\frac{e^{-r^2}}{2} \right]_0^2 d\theta = 2\pi \cdot \frac{1 - e^{-4}}{2} = \pi\big( 1 - e^{-4} \big) $$

That is about $3.08$. The disk has area $4\pi \approx 12.57$ and the integrand never exceeds $1$, so the answer had to be well under $12.57$, and it is positive as required.

Volume under a paraboloid

Find the volume of the solid between the disk $x^2 + y^2 \le 9$ and the surface $z = 9 - x^2 - y^2$. The surface is $z = 9 - r^2$, which is $0$ exactly on the boundary circle, so the solid is a dome sitting on the disk.

$$ \int_0^{2\pi} \int_0^3 \big( 9 - r^2 \big) r\, dr\, d\theta = \int_0^{2\pi} \left[ \frac{9r^2}{2} - \frac{r^4}{4} \right]_0^3 d\theta = 2\pi \left( \frac{81}{2} - \frac{81}{4} \right) = \frac{81\pi}{2} $$

Check it against a known fact: a paraboloid dome fills exactly half of the cylinder that circumscribes it. That cylinder has radius $3$ and height $9$, so its volume is $81\pi$, and half of that is $\tfrac{81\pi}{2}$. The two agree.

Integrating over a ring

Evaluate $\iint_D \sqrt{x^2+y^2}\, dA$ where $D$ is the annulus $1 \le x^2 + y^2 \le 4$. The integrand is $r$ and the region is the polar rectangle $1 \le r \le 2$, $0 \le \theta \le 2\pi$:

$$ \int_0^{2\pi} \int_1^2 r \cdot r\, dr\, d\theta = 2\pi \left[ \frac{r^3}{3} \right]_1^2 = 2\pi \cdot \frac{7}{3} = \frac{14\pi}{3} $$

About $14.66$. The annulus has area $\pi(4-1) = 3\pi \approx 9.42$, so the average value of $r$ over the ring is about $1.56$. That is between $1$ and $2$ and sits above the midpoint $1.5$, which is right, because the outer part of a ring carries more area than the inner part.

A region whose radius depends on the angle

Find the area enclosed by the cardioid $r = 1 + \cos\theta$. Here the inner limits are not both constants: for each angle the radius runs from the origin out to the curve.

$$ \int_0^{2\pi} \int_0^{1+\cos\theta} r\, dr\, d\theta = \int_0^{2\pi} \frac{(1+\cos\theta)^2}{2}\, d\theta = \frac{1}{2} \int_0^{2\pi} \Big( 1 + 2\cos\theta + \cos^2\theta \Big) d\theta $$

Over a full period $\int_0^{2\pi} \cos\theta\, d\theta = 0$ and $\int_0^{2\pi} \cos^2\theta\, d\theta = \pi$, since the average of $\cos^2$ is $\tfrac{1}{2}$. So the area is $\tfrac{1}{2}(2\pi + 0 + \pi) = \tfrac{3\pi}{2}$, about $4.71$.

Sanity check the size. The curve reaches out to $r = 2$ at $\theta = 0$ and pinches to the origin at $\theta = \pi$, so the enclosed area should fall between the area of the unit disk, $\pi \approx 3.14$, and that of the disk of radius $2$, $4\pi \approx 12.57$. It does.

Practice

Convert, then integrate. Write the region as a range of angles and a range of radii, replace $x^2+y^2$ by $r^2$ wherever it appears, and write the area element as $r\, dr\, d\theta$ before doing anything else. Several of these give you a rectangular integral and expect you to recognize that polar form is the way to finish it.

Quiz

Five items on converting regions and integrands to polar form, the area element, and volumes and areas over circular regions.