Integrals over Rose-Curve Regions

Rose Curves as Polar Integration Regions

A rose curve can enclose several separate petals, so an integral over the whole region must account for each petal without counting any part twice. The useful strategy is to describe one petal with its own angular interval, then sum over the petals. Rotational symmetry can shorten that sum, but only when the integrand has the same symmetry.

Consider the standard rose curve \(r=a\cos(n\theta)\), where \(a>0\) and \(n\in\mathbb{N}\). The curve is traced using a signed radius: when \(r<0\), the point is plotted in the direction opposite to the angle \(\theta\). That feature affects how many distinct petals the curve has.

Definition: A petal of a rose curve is one of its enclosed lobes, including the region between the curve and the origin. For \(r=a\cos(n\theta)\), each petal has maximum distance \(a\) from the origin.

Petal Count and Angular Bounds

For \(r=a\cos(n\theta)\), the zeros occur when \(\cos(n\theta)=0\). Around a direction where the curve reaches its maximum radius, consecutive zeros bound one petal. In coordinates centered on that petal, its angular interval has half-width \(\pi/(2n)\), and its radial boundary is \(a\cos(n(\theta-\alpha))\), where \(\alpha\) is the direction of the petal’s center.

The total number of distinct petals depends on whether \(n\) is odd or even. The reason is the signed radius: a negative radius at angle \(\theta\) represents a point at angle \(\theta+\pi\) with positive radius. When \(n\) is odd, this maps each negative-radius lobe onto a lobe already traced. When \(n\) is even, it maps the negative-radius lobes to additional petals.

Petal-count rule: The rose \(r=a\cos(n\theta)\) has \(n\) petals when \(n\) is odd and \(2n\) petals when \(n\) is even. If the number of petals is \(N\), their center directions are \(\alpha_j=2\pi j/n\) for odd \(n\), and \(\alpha_j=\pi j/n\) for even \(n\), with \(j=0,\ldots,N-1\).

For a petal centered at \(\theta=\alpha_j\), the radial boundary is positive on the interval from \(\alpha_j-\pi/(2n)\) to \(\alpha_j+\pi/(2n)\). The complete petal region therefore has \(0\le r\le a\cos(n(\theta-\alpha_j))\) on that interval. These intervals describe individual petals in the plane, including those that arise from negative values of the original signed radius.

\alpha_j-\frac{\pi}{2n}\le\theta\le\alpha_j+\frac{\pi}{2n}, \qquad 0\le r\le a\cos\bigl(n(\theta-\alpha_j)\bigr).

Setting Up Integrals Over One Petal or the Whole Rose

In polar coordinates, the area element is \(dA=r\,dr\,d\theta\). Thus, if \(h(r,\theta)\) is an integrand written in polar coordinates, its integral over one petal centered at \(\alpha_j\) is

\int_{\alpha_j-\pi/(2n)}^{\alpha_j+\pi/(2n)} \int_0^{a\cos(n(\theta-\alpha_j))} h(r,\theta)\,r\,dr\,d\theta.

For a Cartesian integrand \(f(x,y)\), substitute \(x=r\cos\theta\) and \(y=r\sin\theta\), so that \(h(r,\theta)=f(r\cos\theta,r\sin\theta)\). The integral over the complete rose is the sum of the integrals over its \(N\) petals. Their boundaries meet only at the origin, which does not affect a double integral.

\iint_{\text{complete rose}} h(r,\theta)\,dA = \sum_{j=0}^{N-1} \int_{\alpha_j-\pi/(2n)}^{\alpha_j+\pi/(2n)} \int_0^{a\cos(n(\theta-\alpha_j))} h(r,\theta)\,r\,dr\,d\theta.

If rotating the rose through \(2\pi/N\) leaves the integrand unchanged, each petal contributes the same value and the sum is \(N\) times one petal’s integral. A radial integrand, which depends only on \(r\), always has this property. An integrand that depends on \(\theta\) may not.

Why the Petal Bounds Work

For a petal centered at \(\alpha_j\), the angular distance from its center is at most \(\pi/(2n)\). Therefore \(-\pi/2\le n(\theta-\alpha_j)\le\pi/2\), so \(\cos(n(\theta-\alpha_j))\ge0\). The boundary radius is consequently nonnegative throughout the interval, and it vanishes at both endpoints.

At the center, \(\theta=\alpha_j\), the cosine equals \(1\), so the radial boundary reaches \(a\). Between the center and either endpoint, the boundary traces one side of the petal. At each angle in that interval, the values of \(r\) from \(0\) to the boundary fill the region between that side and the origin. This gives exactly the petal’s interior, not just its curved boundary.

Choose a petal center.
Use the petal-count rule to identify the direction \(\alpha_j\).

Set the angular interval.
Use the zeros on either side of the center, at \(\alpha_j\pm\pi/(2n)\).

Set the radial interval.
At each angle, \(r\) runs from \(0\) to \(a\cos(n(\theta-\alpha_j))\).

Sum the petal contributions.
Multiply one petal’s integral by \(N\) only if rotation through \(2\pi/N\) preserves the integrand.

Worked Examples

Worked Example: Area of One Petal and the Complete Rose

Find the area of one petal and the complete region enclosed by \(r=3\cos(4\theta)\).

Here \(n=4\), so the rose has \(N=2n=8\) petals. Consider the petal centered at \(\alpha=0\). Its bounds are \(-\pi/8\le\theta\le\pi/8\) and \(0\le r\le3\cos(4\theta)\). Using \(dA=r\,dr\,d\theta\), its area is

\begin{aligned} A_{\text{petal}} &=\int_{-\pi/8}^{\pi/8}\int_0^{3\cos(4\theta)}r\,dr\,d\theta\\ &=\frac12\int_{-\pi/8}^{\pi/8}9\cos^2(4\theta)\,d\theta\\ &=\frac92\cdot\frac14\int_{-\pi/2}^{\pi/2}\cos^2u\,du\\ &=\frac92\cdot\frac14\cdot\frac{\pi}{2} =\frac{9\pi}{16}. \end{aligned}

The substitution is \(u=4\theta\), so \(d\theta=du/4\). The identity \(\int_{-\pi/2}^{\pi/2}\cos^2u\,du=\pi/2\) follows from \(\cos^2u=(1+\cos(2u))/2\). All eight petals have equal area, so

A_{\text{complete rose}}=8\left(\frac{9\pi}{16}\right)=\frac{9\pi}{2}.

As a check, the one-petal integral is positive, and multiplying by the eight congruent petals gives the complete area. The result has units of square units when the length parameter \(3\) is expressed in length units.

Worked Example: A Radial Moment Over a Three-Petal Rose

Evaluate \(\iint_R(x^2+y^2)\,dA\), where \(R\) is the complete region enclosed by \(r=2\cos(3\theta)\).

Since \(x^2+y^2=r^2\), the integrand together with the polar area element is \(r^3\,dr\,d\theta\). The curve has three petals. For the petal centered at zero, the bounds are \(-\pi/6\le\theta\le\pi/6\) and \(0\le r\le2\cos(3\theta)\). Its contribution is

\begin{aligned} I_{\text{petal}} &=\int_{-\pi/6}^{\pi/6}\int_0^{2\cos(3\theta)}r^3\,dr\,d\theta\\ &=\frac14\int_{-\pi/6}^{\pi/6}\bigl(2\cos(3\theta)\bigr)^4\,d\theta\\ &=4\int_{-\pi/6}^{\pi/6}\cos^4(3\theta)\,d\theta\\ &=\frac43\int_{-\pi/2}^{\pi/2}\cos^4u\,du\\ &=\frac43\cdot\frac{3\pi}{8} =\frac{\pi}{2}. \end{aligned}

For the last integral, \(\cos^4u=(3+4\cos(2u)+\cos(4u))/8\); the cosine terms integrate to zero over \([-\pi/2,\pi/2]\), leaving \(3\pi/8\). Because \(x^2+y^2\) is radial, rotation preserves the integrand, so all three petals contribute equally:

\iint_R(x^2+y^2)\,dA=3\left(\frac{\pi}{2}\right)=\frac{3\pi}{2}.

The result is positive, as expected for a nonnegative integrand. It also has units of length to the fourth power if \(x\) and \(y\) are measured in length units.

Worked Example: A Cartesian Second Moment Over a Four-Petal Rose

Evaluate \(\iint_R x^2\,dA\) for the complete rose \(r=2\cos(2\theta)\).

This rose has four petals. First evaluate the radial moment \(\iint_R(x^2+y^2)\,dA=\iint_R r^2\,dA\) by integrating over one petal and multiplying by four. For the petal centered at zero, the angular bounds are \(-\pi/4\le\theta\le\pi/4\), and the radial boundary is \(2\cos(2\theta)\):

\begin{aligned} \iint_{\text{one petal}}r^2\,dA &=\int_{-\pi/4}^{\pi/4}\int_0^{2\cos(2\theta)}r^3\,dr\,d\theta\\ &=\frac14\int_{-\pi/4}^{\pi/4}16\cos^4(2\theta)\,d\theta\\ &=4\cdot\frac12\int_{-\pi/2}^{\pi/2}\cos^4u\,du\\ &=2\cdot\frac{3\pi}{8} =\frac{3\pi}{4}. \end{aligned}

Thus the full radial moment is \(4(3\pi/4)=3\pi\). The region is unchanged by a rotation through \(\pi/2\), which exchanges \(x^2\) and \(y^2\). Their integrals are therefore equal. Since their sum is the radial moment, each is half of \(3\pi\):

\iint_R x^2\,dA = \frac12\iint_R(x^2+y^2)\,dA = \frac{3\pi}{2}.

As a check, the two equal second moments add to \(3\pi\), the radial moment computed directly from the four petals.

When Symmetry Does Not Permit Multiplication

A complete rose may be rotationally symmetric while the integrand is not. In that case, the integral over each petal need not be equal. The symmetry shortcut applies to the pair consisting of the region and the integrand, not to the region alone.

Common pitfall: It is incorrect to multiply one petal’s integral by the number of petals without checking that rotation leaves the integrand unchanged.

For a concrete example, take the four-petal rose \(r=\cos(2\theta)\) and the integrand \(x=r\cos\theta\). The petal centered at zero contributes

I_0=\frac13\int_{-\pi/4}^{\pi/4}\cos\theta\cos^3(2\theta)\,d\theta =\frac{16\sqrt{2}}{105}>0.

Multiplying this value by four would give \(64\sqrt{2}/105\), but the complete integral is zero. Reflection across the \(y\)-axis leaves the rose unchanged and changes \(x\) to \(-x\), so contributions from opposite sides cancel. Equivalently, the petal centered at \(\pi\) contributes \(-I_0\), and the two petals centered on the \(y\)-axis have zero \(x\)-integral by their left-right symmetry. The correction is to sum the actual petal contributions, or to use a symmetry that preserves the whole integrand and region.

Check Your Understanding

Use the petal bounds and symmetry conditions to answer these questions.

How many petals does \(r=5\cos(7\theta)\) have, and what are the angular bounds for the petal centered at zero?
Write the double-integral bounds for the area of one petal of \(r=4\cos(2\theta)\).
What condition on an integrand allows one petal’s integral to be multiplied by the total number of petals?
Why can a negative radius not be treated as a point in the same direction as its angle?
For the four-petal rose \(r=2\cos(2\theta)\), explain why the complete integral of \(x\) is zero even though the integral over the petal centered at zero is positive.