Lines and Parabolas as Polar Radial Boundaries

From Cartesian Boundaries to Radial Limits

A line or parabola is usually described in Cartesian coordinates, but polar integration asks a different question: for each angle, how far does a ray travel before it reaches the boundary? Converting the equation of a curve is only part of the work. The angular interval and the inequality describing the region determine which solutions are valid radial limits.

Throughout, use \(x=r\cos\theta\), \(y=r\sin\theta\), and \(r\ge 0\). A radial bound describes points on a ray from the origin, so an algebraic solution for \(r\) is useful only when it is nonnegative and applies on the stated angular interval.

Key idea: Substitute \(x=r\cos\theta\) and \(y=r\sin\theta\) into the Cartesian boundary equation. Then use the region’s inequalities and \(r\ge 0\) to determine which solution is the radial boundary.

Lines in Polar Coordinates

A line \(Ax+By=C\) becomes a polar equation by direct substitution:

r(A\cos\theta+B\sin\theta)=C.

When \(A\cos\theta+B\sin\theta\ne 0\), solving gives

r=\frac{C}{A\cos\theta+B\sin\theta}.

This expression is a radial boundary only on angles where it is defined and nonnegative. For instance, if \(C>0\), the denominator must be positive. If \(C<0\), the denominator must be negative. An equation for the line does not by itself specify which side of the line belongs to a region; that comes from the original inequality.

If \(C=0\), the line passes through the origin. Its polar equation is \(r(A\cos\theta+B\sin\theta)=0\). For points with \(r>0\), this specifies an angle, not a nonzero radial distance. Such a line commonly supplies an angular edge of a polar region.

Worked Example: A First-Quadrant Triangle Under a Line

Find polar bounds for the region \(x\ge 0\), \(y\ge 0\), \(x+y\le 4\). The coordinate inequalities restrict the angle to \(0\le\theta\le\frac{\pi}{2}\). Substituting into the line gives

r\cos\theta+r\sin\theta=4, \qquad r=\frac{4}{\cos\theta+\sin\theta}.

On the chosen interval, both sine and cosine are nonnegative, so their sum is positive. The region’s inequality becomes \(r(\cos\theta+\sin\theta)\le 4\). Since \(r\ge0\), the radial bounds are therefore

0\le r\le\frac{4}{\cos\theta+\sin\theta}, \qquad 0\le\theta\le\frac{\pi}{2}.

Check the boundary at \(\theta=\frac{\pi}{4}\). The upper radial value is \(4/(\sqrt{2})=2\sqrt{2}\), giving \(x=r\cos\theta=2\) and \(y=r\sin\theta=2\). Thus \(x+y=4\), as required. At \(\theta=0\), the upper value is \(4\), giving the vertex \((4,0)\); at \(\theta=\frac{\pi}{2}\), it gives \((0,4)\).

Worked Example: A Vertical Line on a Restricted Angular Interval

Consider the part of the wedge \(-\frac{\pi}{3}\le\theta\le\frac{\pi}{3}\) lying between the origin and the vertical line \(x=2\). Substitution into the line gives

r\cos\theta=2, \qquad r=2\sec\theta.

On this interval, \(\cos\theta\ge\frac12>0\), so the line is reached at a positive radius. The region is described by \(0\le x\le2\); along each ray this means \(0\le r\cos\theta\le2\). Its bounds are

0\le r\le 2\sec\theta, \qquad -\frac{\pi}{3}\le\theta\le\frac{\pi}{3}.

For a check, at \(\theta=\frac{\pi}{3}\), the boundary radius is \(r=2\cdot2=4\). The resulting point has \(x=4\cdot\frac12=2\) and \(y=4\cdot\frac{\sqrt3}{2}=2\sqrt3\), so it lies on \(x=2\). At \(\theta=-\frac{\pi}{3}\), the point has the same \(x\)-coordinate and \(y=-2\sqrt3\), also on the line. The positive cosine on the interval is essential: dividing by a cosine that changes sign would require rechecking the inequality and the admissible angles.

Parabolas in Polar Coordinates

For a vertical parabola \(y=ax^2+bx+c\), substitution gives

r\sin\theta = a r^2\cos^2\theta+b r\cos\theta+c.

Unlike the line equation, this may be quadratic in \(r\). A ray can meet a parabola at more than one point, so any resulting roots must be checked for nonnegativity and against the region being described. For the common parabola \(y=ax^2\), which passes through the origin, the equation factors as

r\sin\theta=a r^2\cos^2\theta, \qquad r\bigl(\sin\theta-a r\cos^2\theta\bigr)=0.

The factor \(r=0\) is the origin. For a nonzero point with \(\cos\theta\ne0\), the other intersection is \(r=\frac{\sin\theta}{a\cos^2\theta}\). When \(a>0\) and \(0<\theta<\frac{\pi}{2}\), this radius is positive. For example, the inequality \(y\ge ax^2\) becomes \(r\sin\theta\ge a r^2\cos^2\theta\). For \(r>0\), division by \(r\) gives \(r\le\frac{\sin\theta}{a\cos^2\theta}\). Thus this side of the parabola gives an upper radial bound on those angles.

Worked Example: The Region Between a Line and a Parabola

Find polar bounds for the first-quadrant region between \(y=2x^2\) and \(y=x\). The inequalities are \(y\ge2x^2\) and \(y\le x\). The line \(y=x\) corresponds to \(\tan\theta=1\), so in the first quadrant the region lies between the rays \(\theta=0\) and \(\theta=\frac{\pi}{4}\).

Convert the parabola boundary:

r\sin\theta=2r^2\cos^2\theta.

For \(r>0\), this gives \(r=\frac{\sin\theta}{2\cos^2\theta}\). The inequality \(y\ge2x^2\) gives \(r\le\frac{\sin\theta}{2\cos^2\theta}\) after division by \(r\). The other inequality, \(y\le x\), becomes \(r\sin\theta\le r\cos\theta\). For \(r>0\), this is \(\tan\theta\le1\), which is exactly the angular restriction already identified. Therefore the bounds are

0\le r\le\frac{\sin\theta}{2\cos^2\theta}, \qquad 0\le\theta\le\frac{\pi}{4}.

Check the boundary at \(\theta=\frac{\pi}{6}\). The upper radius is \(\frac{1/2}{2(3/4)}=\frac13\). The corresponding Cartesian point is \(x=\frac13\cdot\frac{\sqrt3}{2}=\frac{\sqrt3}{6}\), \(y=\frac13\cdot\frac12=\frac16\). Since \(2x^2=2\cdot\frac{3}{36}=\frac16=y\), it lies on the parabola. At \(\theta=\frac{\pi}{4}\), the upper radius is \(\frac{\sqrt2}{2}\big/\bigl(2\cdot\frac12\bigr)=\frac{\sqrt2}{2}\). This gives \(x=y=\frac12\), where \(2x^2=x=y\); the two curves meet there.

Why the Parabola Formula Works

The radial formula for \(y=ax^2\) follows from the geometry of a ray and the algebra of the boundary. A point on a ray has coordinates \((r\cos\theta,r\sin\theta)\). On the parabola, its coordinates must satisfy \(y=ax^2\), so substituting the ray coordinates yields \(r\sin\theta=a r^2\cos^2\theta\). For a point other than the origin, \(r>0\), allowing division by \(r\). If \(a>0\) and the angle is in the first quadrant, division by the positive quantity \(a\cos^2\theta\) then gives the unique nonzero intersection radius.

The side inequality also follows directly rather than from a sketch alone. At a fixed first-quadrant angle, \(y\ge ax^2\) means \(r\sin\theta\ge a r^2\cos^2\theta\). Dividing by \(r>0\) and rearranging gives \(r\le\frac{\sin\theta}{a\cos^2\theta}\). Thus points above the parabola along that ray are closer to the origin than its nonzero intersection. The origin itself is included by continuity and satisfies the original inequality.

Takeaway: The equation of a boundary gives candidate intersections with rays. The region’s inequality, the condition \(r\ge0\), and the selected angular interval determine which candidate is an upper or lower radial limit.

A Common Mistake: Losing a Factor of \(r\)

For \(y=x^2\), an incorrect substitution might claim \(r\sin\theta=r^2\cos^2\theta\) and then write \(\sin\theta=\cos^2\theta\). That step divides by \(r\) on the left but not on the right, and the result incorrectly makes the boundary an angular condition rather than a radial one.

The correct algebra is

r\sin\theta=r^2\cos^2\theta \quad\Longrightarrow\quad r\bigl(\sin\theta-r\cos^2\theta\bigr)=0.

For \(r>0\), divide both sides by \(r\) to obtain \(\sin\theta=r\cos^2\theta\), and then \(r=\frac{\sin\theta}{\cos^2\theta}\) when \(\cos\theta\ne0\). The origin, \(r=0\), is a separate intersection and should not be lost by dividing by \(r\).

Choosing and Checking Radial Bounds

A reliable conversion keeps three decisions distinct: identify the angular interval, solve the boundary equation for nonnegative \(r\), and apply the original inequality to determine which side of that boundary is in the region. When a parabola gives multiple candidate radial solutions, substitute them into the boundary equation and inspect which intersections lie on the ray and border the intended region. A formula that is algebraically correct can still be invalid on angles where it is negative, undefined, or unrelated to the region.

These checks matter before setting up a polar integral: the radial limits describe the actual region, not merely a curve equation. If the radial boundary changes across the angular range, the region must be described using appropriate angular intervals rather than forcing one bound to apply everywhere.

Check Your Understanding

For each question, use the boundary equation together with the relevant inequalities and angular interval.

Convert \(3x+4y=12\) into a polar equation. On which first-quadrant angles is its radial solution positive?
For \(y=3x^2\), derive the nonzero radial intersection and state the first-quadrant bound for the side \(y\ge3x^2\).
Why does a line through the origin usually determine an angular boundary rather than a nonzero radial boundary?
For \(y=x^2\), identify both the origin and nonzero intersection when \(\theta=\frac{\pi}{4}\), and verify the nonzero point in Cartesian coordinates.
What must be checked before using a solution of a quadratic equation in \(r\) as a radial limit?