Integrals over Cardioids and Limacons

Reading the Shape from Its Polar Equation

A cardioid or limaçon often has a polar equation of the form \(r=a+b\cos\theta\), where \(a\) and \(b\) are positive constants. The cosine makes the curve symmetric about the \(x\)-axis: replacing \(\theta\) with \(-\theta\) leaves \(r\) unchanged. An equation involving \(\sin\theta\) gives the corresponding shape with its axis of symmetry rotated.

To integrate over the region bounded by one of these curves, the main task is to identify which angles trace the boundary and whether the radial coordinate stays nonnegative. Once a region is described by \(0\le r\le R(\theta)\), a double integral uses the polar area element \(dA=r\,dr\,d\theta\). The extra factor \(r\) is essential: it accounts for the way area scales with distance from the origin.

Definition: A cardioid is the special limaçon \(r=a+a\cos\theta\) (or an equivalent rotated form). A limaçon has the more general form \(r=a+b\cos\theta\), with positive constants \(a\) and \(b\). When \(a=b\), the curve is a cardioid.

Bounds When the Radius Is Nonnegative

If \(a>b>0\), then \(a+b\cos\theta\ge a-b>0\) for every angle. The curve makes one closed loop without passing through the origin, and its enclosed region is described in one sweep by \(0\le\theta\le2\pi\) and \(0\le r\le a+b\cos\theta\). If \(a=b\), the radius is nonnegative and equals zero at \(\theta=\pi\); the same bounds describe the cardioid’s interior.

More generally, the upper radial bound must be the actual nonnegative distance from the origin along the chosen ray. It is not enough to copy a formula for \(r\): first determine where that formula is nonnegative. If it becomes negative, it represents points in the opposite direction, not points at a negative distance along the ray.

Key setup: For \(r=a+b\cos\theta\) with \(a\ge b>0\), the region inside the curve is \(0\le\theta\le2\pi,\;0\le r\le a+b\cos\theta\). For \(a=b\), this is the cardioid case.

Why the Radial Integral Gives Area

In polar coordinates, a small area element at radius \(r\) has radial thickness \(dr\) and angular width \(d\theta\). Its width along the arc is approximately \(r\,d\theta\), so its area is \(r\,dr\,d\theta\). Consequently, for a region described by \(0\le r\le R(\theta)\), its area is found by integrating \(r\) through the radial interval and then over the allowed angles.

\operatorname{Area}(D) =\int_{\alpha}^{\beta}\int_0^{R(\theta)}r\,dr\,d\theta =\frac12\int_{\alpha}^{\beta}R(\theta)^2\,d\theta.

The second equality follows by evaluating the inner integral: \(\int_0^{R(\theta)}r\,dr=[r^2/2]_0^{R(\theta)}=R(\theta)^2/2\). This area formula applies when the bounds trace the region once with nonnegative radius.

Worked Example: Area Inside a Cardioid

Find the area inside \(r=2(1+\cos\theta)\). Here \(a=b=2\), so the cardioid is traced with \(0\le\theta\le2\pi\), and \(0\le r\le2(1+\cos\theta)\).

Use the polar area element and integrate first with respect to \(r\):

\begin{aligned} A &=\int_0^{2\pi}\int_0^{2(1+\cos\theta)}r\,dr\,d\theta\\ &=\frac12\int_0^{2\pi}\bigl(2(1+\cos\theta)\bigr)^2\,d\theta\\ &=2\int_0^{2\pi}\left(1+2\cos\theta+\cos^2\theta\right)\,d\theta. \end{aligned}

Over a full period, \(\int_0^{2\pi}1\,d\theta=2\pi\), \(\int_0^{2\pi}\cos\theta\,d\theta=0\), and \(\int_0^{2\pi}\cos^2\theta\,d\theta=\pi\). Thus

A=2(2\pi+2\cdot0+\pi)=6\pi.

Check: the squared boundary radius is \(4(1+2\cos\theta+\cos^2\theta)\), and the area formula contributes a factor of \(1/2\). Its constant and squared-cosine terms therefore contribute \(2\pi\) and \(\pi\) inside the final outer factor of \(2\), respectively, while the cosine term integrates to zero. The result is \(6\pi\), as computed.

Using Symmetry in a Polar Integral

For a curve \(r=a+b\cos\theta\), reflection across the \(x\)-axis sends \((r,\theta)\) to \((r,-\theta)\). The region is unchanged by this reflection. In particular, an integrand that changes sign under \(y\mapsto-y\) has integral zero over this region, provided the integral exists. For instance, \(y=r\sin\theta\) changes sign under that reflection. By contrast, a radial integrand such as \(x^2+y^2=r^2\) is unchanged and need not have zero integral.

Worked Example: A Radial Moment over a Limaçon

Evaluate \(\iint_D(x^2+y^2)\,dA\), where \(D\) is inside \(r=2+\cos\theta\). Since \(2+\cos\theta\ge1\), the radius is positive for every angle, so \(0\le r\le2+\cos\theta\) for \(0\le\theta\le2\pi\). In polar coordinates \(x^2+y^2=r^2\), and \(dA=r\,dr\,d\theta\).

\begin{aligned} \iint_D(x^2+y^2)\,dA &=\int_0^{2\pi}\int_0^{2+\cos\theta}r^3\,dr\,d\theta\\ &=\frac14\int_0^{2\pi}(2+\cos\theta)^4\,d\theta. \end{aligned}

Expand the fourth power:

(2+\cos\theta)^4 =16+32\cos\theta+24\cos^2\theta+8\cos^3\theta+\cos^4\theta.

Over a full period, the integrals of \(\cos\theta\) and \(\cos^3\theta\) are zero. Also, \(\int_0^{2\pi}\cos^2\theta\,d\theta=\pi\) and \(\int_0^{2\pi}\cos^4\theta\,d\theta=3\pi/4\). Therefore,

\begin{aligned} \iint_D(x^2+y^2)\,dA &=\frac14\left(16(2\pi)+24\pi+\frac{3\pi}{4}\right)\\ &=\frac14\left(\frac{227\pi}{4}\right) =\frac{227\pi}{16}. \end{aligned}

Check: integrating \(r^3\) radially gives \([r^4/4]_0^{2+\cos\theta}=(2+\cos\theta)^4/4\), which matches the angular integral used. The constant term contributes \(32\pi\), the squared-cosine term contributes \(24\pi\), and the fourth-power term contributes \(3\pi/4\) before the factor \(1/4\); their sum is \(227\pi/4\). Hence the stated value follows.

Limaçons with an Inner Loop

When \(b>a>0\), the radius \(a+b\cos\theta\) is negative for some angles. It is zero when \(\cos\theta=-a/b\), and negative on the interval around \(\theta=\pi\) between those zeros. The negative values trace the inner loop in the opposite direction. To write ordinary polar bounds for that loop, shift the angle by \(\pi\), which reverses the direction and makes the radius positive.

Set \(\alpha=\arccos(a/b)\), so \(0<\alpha<\pi/2\). The part with nonnegative radius traces the outer loop for \(-(\pi-\alpha)\le\theta\le\pi-\alpha\), with \(0\le r\le a+b\cos\theta\). For the inner loop, write the original angle as \(\theta=\phi+\pi\). Then \(-\alpha\le\phi\le\alpha\), and the positive radius is \(-(a+b\cos(\phi+\pi))=b\cos\phi-a\). These are separate bounds for the two loops; the inner-loop bounds do not use a negative radial coordinate.

Worked Example: Area of an Inner Loop

Find the area of the inner loop of \(r=1+2\cos\theta\). Here \(a=1\), \(b=2\), and \(\alpha=\arccos(1/2)=\pi/3\). After shifting the angle by \(\pi\), the inner loop has the positive-radius description \(-\pi/3\le\phi\le\pi/3\), \(0\le r\le2\cos\phi-1\).

\begin{aligned} A_{\mathrm{inner}} &=\frac12\int_{-\pi/3}^{\pi/3}(2\cos\phi-1)^2\,d\phi\\ &=\frac12\int_{-\pi/3}^{\pi/3} \left(4\cos^2\phi-4\cos\phi+1\right)\,d\phi. \end{aligned}

The three terms integrate as follows:

\begin{aligned} \int_{-\pi/3}^{\pi/3}4\cos^2\phi\,d\phi &=4\left(\frac{\pi}{3}+\frac{\sin(2\pi/3)}{2}\right) =\frac{4\pi}{3}+\sqrt3,\\ \int_{-\pi/3}^{\pi/3}-4\cos\phi\,d\phi &=-8\sin(\pi/3)=-4\sqrt3,\\ \int_{-\pi/3}^{\pi/3}1\,d\phi &=\frac{2\pi}{3}. \end{aligned}

Combining them and multiplying by \(1/2\) gives

A_{\mathrm{inner}} =\frac12\left(2\pi-3\sqrt3\right) =\pi-\frac{3\sqrt3}{2}.

Check: the positive radial bound \(2\cos\phi-1\) is zero at \(\phi=\pm\pi/3\) and equals \(1\) at \(\phi=0\), so it describes a nonnegative radius throughout the interval. The integrated expression is approximately \(0.543\), which is positive, as an area must be.

A Common Mistake: Omitting the Jacobian Factor

For the cardioid \(r=2(1+\cos\theta)\), a tempting but incorrect area calculation is \(\frac12\int_0^{2\pi}r\,d\theta\). It gives \(\frac12\int_0^{2\pi}2(1+\cos\theta)\,d\theta=2\pi\), but area depends on the square of the boundary radius, not the radius itself. The correct calculation integrates \(r\,dr\,d\theta\); after the radial integration, this produces \(\frac12\int_0^{2\pi}r(\theta)^2\,d\theta=6\pi\).

For an inner loop, another error is to treat negative values of the curve equation as ordinary upper radial bounds. Instead, shift the angle by \(\pi\) and express the loop with a nonnegative radius, as in the worked example. This keeps the integration bounds aligned with the usual polar area element.

Practical check: Before integrating, verify that the radial upper bound is nonnegative on the stated angular interval and that the chosen interval traces the intended loop exactly once. Then include the factor \(r\) from \(dA=r\,dr\,d\theta\).

Check Your Understanding

For each question, first identify the angular interval and nonnegative radial bound before evaluating an integral.

Set up an integral for the area inside \(r=3+2\cos\theta\).
Why does the area integral for a region \(0\le r\le R(\theta)\) contain \(r\) before the radial integration?
For \(r=1+2\cos\theta\), what positive-radius bounds describe the inner loop?
Which of the functions \(x\) and \(y\) has integral zero over the region inside \(r=2+\cos\theta\), and what symmetry explains the result?
Find the angular values at which \(r=1+2\cos\theta\) is zero.