Integrated Polar Analysis of a Lemniscate

One Equation, Two Lobes

The polar equation \(r^2=a^2\cos(2\theta)\), with \(a>0\), describes a lemniscate: a curve with two lobes that meet at the pole. Its geometry is encoded in the condition that \(r^2\) must be nonnegative. Therefore, real points occur only when \(\cos(2\theta)\geq 0\).

For the right lobe, choose \(r\geq 0\) and \(-\pi/4\leq\theta\leq\pi/4\). The radius is zero at the two endpoints and positive between them. At \(\theta=0\), it reaches its maximum \(r=a\), so this interval traces the right lobe once. The other lobe is traced, for example, over \(3\pi/4\leq\theta\leq5\pi/4\).

Definition: In this tutorial, a lobe means one of the two closed loops that meet at the pole. We use \(a>0\) and \(r\geq0\), and take the right lobe to have \(-\pi/4\leq\theta\leq\pi/4\).

The scale parameter \(a\) sets the size of the curve. The maximum radius of either lobe is \(a\), its area scales with \(a^2\), and its arc length scales with \(a\). The same angular interval works for every positive value of \(a\).

The Pole and Its Tangents

The pole is a meeting point of the two lobes, not a smooth point with just one tangent. To find the tangent lines, first write the curve in Cartesian coordinates. For \(r\neq0\), use \(x=r\cos\theta\), \(y=r\sin\theta\), and \(\cos(2\theta)=(x^2-y^2)/r^2\). Substitution gives the equivalent equation

$$ (x^2+y^2)^2=a^2(x^2-y^2). $$

At the pole, both sides vanish, so the equation alone does not immediately show the directions in which the curve arrives. Instead, for points on the curve other than the pole, divide the polar equation by \(a^2\): \(r^2/a^2=\cos(2\theta)\). As a point approaches the pole, \(r\) approaches zero, so \(\cos(2\theta)\) approaches zero. The possible limiting angles satisfy \(\cos(2\theta)=0\), giving directions \(\theta=\pi/4\) and \(\theta=3\pi/4\), considered as lines through the origin. These are the lines \(y=x\) and \(y=-x\).

The right lobe confirms how these directions arise. At its upper endpoint, \(\theta\) approaches \(\pi/4\) and the position vector points in direction \(\pi/4\). At its lower endpoint, \(\theta\) approaches \(-\pi/4\) and the position vector points in direction \(-\pi/4\). Each limiting direction determines a tangent line at the pole. The two lobes together meet there with these two distinct tangent lines.

Result: The lemniscate has two tangent lines at the pole: \(y=x\) and \(y=-x\). They are the lines along which the curve approaches the pole.

Worked Example: Finding the Pole Tangents When \(a=2\)

For \(a=2\), the polar equation is \(r^2=4\cos(2\theta)\). In Cartesian form, it becomes

$$ (x^2+y^2)^2=4(x^2-y^2). $$

Along the curve, \(r^2=4\cos(2\theta)\). As \(r\to0\), this implies \(\cos(2\theta)\to0\), so the limiting directions satisfy \(2\theta=\pi/2+k\pi\), or \(\theta=\pi/4+k\pi/2\), for integers \(k\). Modulo \(\pi\), these give the two line directions \(\pi/4\) and \(3\pi/4\).

A line through the pole with angle \(\pi/4\) has slope \(\tan(\pi/4)=1\), so its equation is \(y=x\). The line with angle \(3\pi/4\) has slope \(\tan(3\pi/4)=-1\), so its equation is \(y=-x\).

As a check, the lowest-degree terms in the Cartesian equation are \(-4(x^2-y^2)\). Their zero set factors as \((x-y)(x+y)=0\), giving the same two lines.

Area of One Lobe

For a polar curve traced once between angles \(\alpha\) and \(\beta\), the area enclosed by that trace is \(A=\frac12\int_\alpha^\beta r^2\,d\theta\), provided the radius is real and the interval describes the region once. For the right lobe, those conditions hold on \([-\pi/4,\pi/4]\). Substituting the lemniscate equation reduces the area calculation to an integral of \(\cos(2\theta)\).

Result: The area of one lobe is \(\frac{a^2}{2}\). The two lobes have equal area, so the total area enclosed by the lemniscate is \(a^2\).

Worked Example: Area of One Lobe When \(a=4\)

Here \(r^2=16\cos(2\theta)\), and the right lobe is traced once from \(-\pi/4\) to \(\pi/4\). Therefore

\begin{aligned} A &=\frac12\int_{-\pi/4}^{\pi/4}16\cos(2\theta)\,d\theta\\ &=8\left[\frac12\sin(2\theta)\right]_{-\pi/4}^{\pi/4}\\ &=4\bigl(\sin(\pi/2)-\sin(-\pi/2)\bigr)\\ &=4(1-(-1))\\ &=8. \end{aligned}

As a check, the general one-lobe formula gives \(a^2/2=4^2/2=8\). The area is positive, as expected for a region.

Arc Length of One Lobe

For a polar curve \(r=r(\theta)\), the arc-length element is \(ds=\sqrt{r^2+(dr/d\theta)^2}\,d\theta\). Apply this on the interior of the right lobe, where \(\cos(2\theta)>0\) and \(r=a\sqrt{\cos(2\theta)}\). The endpoints need separate attention because the radius is zero there and its derivative is unbounded.

Differentiate the positive square-root expression on the interior of the interval:

\frac{dr}{d\theta} = -\frac{a\sin(2\theta)}{\sqrt{\cos(2\theta)}}.

Substituting into the arc-length element gives a useful simplification. On the interior, \(\cos(2\theta)>0\), so the square root below is well-defined:

\begin{aligned} \sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2} &=\sqrt{a^2\cos(2\theta) +\frac{a^2\sin^2(2\theta)}{\cos(2\theta)}}\\ &=a\sqrt{\frac{\cos^2(2\theta)+\sin^2(2\theta)}{\cos(2\theta)}}\\ &=\frac{a}{\sqrt{\cos(2\theta)}}. \end{aligned}

Thus the arc length of one lobe is represented by an improper integral: the integrand is unbounded at both endpoints, so the integral is interpreted as a limit from inside the interval. This setup does not require an elementary antiderivative.

Result: The arc length of the right lobe is

L=a\int_{-\pi/4}^{\pi/4}\frac{d\theta}{\sqrt{\cos(2\theta)}}.

The integral is improper at \(\theta=\pm\pi/4\), but it converges.

Why the Arc-Length Integral Converges

At either endpoint, \(\cos(2\theta)\) approaches zero, so it is important not to treat the displayed arc-length integrand as a bounded function on the closed interval. To check convergence, consider the right endpoint and set \(\delta=\pi/4-\theta\). For \(0<\delta\leq\pi/4\), \(\cos(2\theta)=\sin(2\delta)\). The inequality \(\sin u\geq 2u/\pi\) for \(0\leq u\leq\pi/2\) then gives \(\cos(2\theta)\geq4\delta/\pi\). Consequently, the integrand is bounded above by \(\frac{a\sqrt{\pi}}{2\sqrt{\delta}}\), whose integral over a finite interval next to \(\delta=0\) is finite. The same argument, using \(\delta=\theta+\pi/4\), applies at the left endpoint. Hence both endpoint singularities are integrable.

Worked Example: Setting Up Arc Length When \(a=3\)

For \(r^2=9\cos(2\theta)\), choose the right-lobe interval \([-\pi/4,\pi/4]\). On its interior, \(r=3\sqrt{\cos(2\theta)}\), so

\frac{dr}{d\theta} =-\frac{3\sin(2\theta)}{\sqrt{\cos(2\theta)}}.

Substituting each term into the arc-length formula gives

\begin{aligned} \sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2} &=\sqrt{9\cos(2\theta)+\frac{9\sin^2(2\theta)}{\cos(2\theta)}}\\ &=3\sqrt{\frac{\cos^2(2\theta)+\sin^2(2\theta)}{\cos(2\theta)}}\\ &=\frac{3}{\sqrt{\cos(2\theta)}}. \end{aligned}

The one-lobe length is therefore set up as

L=3\int_{-\pi/4}^{\pi/4}\frac{d\theta}{\sqrt{\cos(2\theta)}}.

For a check, the integrand is positive on the interior, and it has the same square-root endpoint behavior just established, so this is a finite positive length. The factor \(3\) also agrees with the fact that lengths scale linearly with \(a\).

A Common Mistake: Treating \(r^2\) as \(r\)

The equation gives \(r^2=a^2\cos(2\theta)\), not \(r=a^2\cos(2\theta)\). For the chosen lobe, taking the nonnegative square root gives \(r=a\sqrt{\cos(2\theta)}\). Forgetting the square root changes both the derivative and the arc-length setup.

For example, the incorrect assignment \(r=a^2\cos(2\theta)\) would give \(dr/d\theta=-2a^2\sin(2\theta)\), and hence the incorrect integrand \(a^2\sqrt{\cos^2(2\theta)+4\sin^2(2\theta)}\). This does not describe the stated curve: at \(\theta=0\), it predicts radius \(a^2\), whereas the original equation gives \(r=a\). Correctly taking the square root yields \(r=a\) at \(\theta=0\) and the derived length integrand \(a/\sqrt{\cos(2\theta)}\).

A second point worth keeping in view is that the curve has two pole tangents. Treating the crossing at the pole as though it were a single smooth passage would miss one of the two lines. The angular limits show both directions explicitly.

Putting the Analysis Together

The same angular description supports three distinct conclusions. The interval \([-\pi/4,\pi/4]\) selects the right lobe; the polar area formula gives its exact area; and the polar arc-length formula leads to a convergent improper integral for its length. The Cartesian equation and limiting directions reveal the two tangent lines at the pole.

Select one lobe.
Use \(-\pi/4\leq\theta\leq\pi/4\), where \(\cos(2\theta)\geq0\).

Find its area.
Integrate \(\frac12a^2\cos(2\theta)\) over the selected interval to obtain \(a^2/2\).

Find the pole tangents.
As \(r\to0\), the curve equation forces \(\cos(2\theta)\to0\), giving \(y=x\) and \(y=-x\).

Set up arc length.
Use \(r=a\sqrt{\cos(2\theta)}\) to obtain the convergent integral \(a\int_{-\pi/4}^{\pi/4}(\cos(2\theta))^{-1/2}\,d\theta\).

Check Your Understanding

Use the lobe interval and formulas developed above to answer these questions.

Why is \([-\pi/4,\pi/4]\) a suitable interval for tracing the right lobe once?
What two tangent lines does the lemniscate have at the pole, and how do they follow from the limiting angular directions?
Find the area of one lobe when \(a=5\).
For a general \(a>0\), what is \(dr/d\theta\) on the interior of the right-lobe interval?
Why does the unbounded arc-length integrand still give a finite length?