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Cylindrical Triple Integrals · Tutorial 629 of 999

Angular Bounds for Wedges and Sectors

Translate plane boundaries into correct angular intervals, then combine them with radial and vertical limits to describe wedge-shaped and partial-cylinder regions.

Intermediate 10 min read

What You'll Learn

  • Convert vertical planes through the axis into angular boundary rays.
  • Select the correct angle interval by combining the half-space inequalities.
  • Handle angles that cross the negative x-axis without choosing the wrong branch.
  • Combine angular bounds with cylindrical and oblique radial limits.
  • State vertical limits separately and recognize when they do not affect the angle.
  • Check bounds by interpreting the resulting region and verifying its volume.

From Boundary Planes to Angular Bounds

A wedge-shaped solid may be bounded by vertical planes through the \(z\)-axis. In the \(xy\)-plane, each such plane appears as a line through the origin. That line determines possible angular boundary directions, but it does not by itself determine which side of the line belongs to the solid. The inequalities describing the solid select the actual angular interval.

In cylindrical coordinates, \(x=r\cos\theta\), \(y=r\sin\theta\), with \(r\ge 0\). We use angles in \([0,2\pi)\) unless a continuous interval outside that range makes the geometry clearer. Angular endpoints describe the same ray if they differ by \(2\pi\); choose an interval that describes the region once, without duplicating it.

Definition: An angular bound is an interval of values of \(\theta\) describing the directions from the \(z\)-axis occupied by a region. For a solid bounded by vertical planes through that axis, the interval is determined by the rays where those planes meet the \(xy\)-plane, together with the inequalities selecting the region between them.

How a Plane Inequality Selects Angles

A vertical plane through the \(z\)-axis has an equation of the form \(ax+by=0\), where \(a\) and \(b\) are not both zero. In cylindrical coordinates, its equation becomes \(r(a\cos\theta+b\sin\theta)=0\). For points with \(r>0\), the boundary directions therefore satisfy \(a\cos\theta+b\sin\theta=0\). The plane contains two opposite rays, so it gives two possible boundary angles, not just one.

An inequality involving this plane selects one side. For example, \(ax+by\ge 0\) becomes \(a\cos\theta+b\sin\theta\ge 0\) when \(r>0\). The point \(r=0\) lies on the axis and satisfies the original plane inequality regardless of \(\theta\); it does not create an extra angular boundary.

Theorem: Let \(\phi\) be an angle such that \(a\cos\theta+b\sin\theta=\sqrt{a^2+b^2}\cos(\theta-\phi)\). Then \(ax+by\ge 0\), for \(r>0\), holds for angles in the semicircle \(\phi-\frac{\pi}{2}\le\theta\le\phi+\frac{\pi}{2}\), interpreted modulo \(2\pi\). Multiple plane inequalities select the intersection of their angular ranges.

Why the Half-Plane Rule Works

Here is the derivation. Choose \(\phi\) so that \(\cos\phi=a/\sqrt{a^2+b^2}\) and \(\sin\phi=b/\sqrt{a^2+b^2}\). The cosine difference identity gives

$$ a\cos\theta+b\sin\theta =\sqrt{a^2+b^2}\bigl(\cos\phi\cos\theta+\sin\phi\sin\theta\bigr) =\sqrt{a^2+b^2}\cos(\theta-\phi). $$

The factor \(\sqrt{a^2+b^2}\) is positive, so the inequality is equivalent to \(\cos(\theta-\phi)\ge 0\). Cosine is nonnegative on \([-\pi/2,\pi/2]\), modulo \(2\pi\). Thus the selected directions form the stated semicircle. If a region is defined by several plane inequalities, a direction is allowed precisely when it satisfies every inequality, which is why their angular ranges must be intersected.

1
Find the boundary directions.
Set each vertical plane equation through the axis equal to zero and identify both rays in the \(xy\)-plane.
2
Use the inequalities.
For each boundary plane, determine which side is included; do not choose an interval from the boundary line alone.
3
Intersect the allowed angles.
Keep only the angles satisfying all plane constraints, choosing a continuous interval if it crosses \(0\).
4
Attach the other bounds.
State the radial and vertical limits that describe the rest of the solid. Check whether any additional boundary changes those limits as \(\theta\) varies.

Worked Example: A First-Quadrant Partial Cylinder

Worked Example: A First-Quadrant Partial Cylinder

Describe and find the volume of the solid inside \(x^2+y^2\le 4\), in the first quadrant, between \(z=-1\) and \(z=3\). The first-quadrant constraints are \(x\ge 0\) and \(y\ge 0\). Substituting \(x=r\cos\theta\) and \(y=r\sin\theta\), these become \(r\cos\theta\ge 0\) and \(r\sin\theta\ge 0\). For \(r>0\), both hold in the first quadrant, so \(0\le\theta\le\pi/2\). The cylinder gives \(0\le r\le 2\), and the horizontal planes give \(-1\le z\le 3\).

The volume element is \(r\,dz\,dr\,d\theta\), so the volume is

$$ \begin{aligned} V &=\int_0^{\pi/2}\int_0^2\int_{-1}^3 r\,dz\,dr\,d\theta\\ &=\int_0^{\pi/2}\int_0^2 r\bigl(3-(-1)\bigr)\,dr\,d\theta\\ &=\int_0^{\pi/2}\int_0^2 4r\,dr\,d\theta\\ &=\int_0^{\pi/2}\left[2r^2\right]_0^2\,d\theta\\ &=\int_0^{\pi/2}8\,d\theta =4\pi. \end{aligned} $$

As a check, the base is a quarter-disk of area \(\frac14\pi(2)^2=\pi\), and the height is \(3-(-1)=4\). Its volume is \(4\pi\), matching the integral.

Angular Bounds Do Not Determine Every Bound

Finding the angle interval is one part of describing the solid, not a replacement for the radial and vertical limits. A cylindrical side \(x^2+y^2=R^2\) gives the constant radial bound \(r=R\), while a plane such as \(x+y=c\) gives \(r(\cos\theta+\sin\theta)=c\). The latter can create a radial limit that depends on \(\theta\). Horizontal planes give vertical limits directly; they do not, by themselves, restrict \(\theta\).

For a proposed radial upper bound obtained by dividing by a trigonometric expression, check its sign throughout the chosen angular interval. Also check that the boundary describes the intended side of the plane. The same plane equation can be paired with either inequality, giving different parts of space.

Worked Example: A Wedge Cut by an Oblique Plane

Worked Example: A Wedge Cut by an Oblique Plane

Find the cylindrical bounds and volume for the region \(x\ge 0\), \(y\ge 0\), \(x+y\le 2\), and \(0\le z\le 1\). The first two inequalities give \(0\le\theta\le\pi/2\). The third becomes \(r\cos\theta+r\sin\theta\le 2\). On this angular interval, \(\cos\theta+\sin\theta>0\), so it gives \(0\le r\le 2/(\cos\theta+\sin\theta)\). The complete bounds are therefore \(0\le\theta\le\pi/2\), \(0\le r\le 2/(\cos\theta+\sin\theta)\), and \(0\le z\le 1\).

Using the cylindrical volume element,

$$ \begin{aligned} V &=\int_0^{\pi/2}\int_0^{2/(\cos\theta+\sin\theta)}\int_0^1 r\,dz\,dr\,d\theta\\ &=\int_0^{\pi/2}\int_0^{2/(\cos\theta+\sin\theta)}r\,dr\,d\theta\\ &=\int_0^{\pi/2}\frac12\left(\frac{2}{\cos\theta+\sin\theta}\right)^2d\theta\\ &=\int_0^{\pi/2}\frac{2}{(\cos\theta+\sin\theta)^2}\,d\theta. \end{aligned} $$

Since \(\cos\theta+\sin\theta=\sqrt{2}\cos(\theta-\pi/4)\), the last integrand is \(\sec^2(\theta-\pi/4)\). Hence

$$ \begin{aligned} V &=\int_0^{\pi/2}\sec^2(\theta-\pi/4)\,d\theta\\ &=\left[\tan(\theta-\pi/4)\right]_0^{\pi/2}\\ &=\tan(\pi/4)-\tan(-\pi/4)=1-(-1)=2. \end{aligned} $$

To check, the base in the \(xy\)-plane is the right triangle with intercepts \(2\) on both axes. Its area is \(\frac12(2)(2)=2\), and the height is \(1\), so its volume is \(2\). The check also confirms that the angle-dependent radial bound covers the triangular base, rather than the whole quarter-disk.

Angles That Cross the Negative \(x\)-Axis

An angular interval need not start at \(0\), and it can be more convenient to let its upper endpoint exceed \(2\pi\) than to split the region into two intervals. For instance, \([3\pi/4,5\pi/4]\) describes a sector centered on the negative \(x\)-axis. This interval is continuous and does not double-count any interior directions.

Worked Example: A Sector on the Left Side

Consider the region defined by \(x\le 0\), \(y\ge x\), \(y\le -x\), \(1\le\sqrt{x^2+y^2}\le 2\), and \(-1\le z\le 1\). The radial bounds are \(1\le r\le 2\). To find the angles, substitute cylindrical coordinates into the plane inequalities. They become \(\cos\theta\le 0\), \(\sin\theta-\cos\theta\ge 0\), and \(\sin\theta+\cos\theta\le 0\), respectively. All three hold on \(3\pi/4\le\theta\le5\pi/4\). The vertical bounds remain \(-1\le z\le1\).

The volume is

$$ \begin{aligned} V &=\int_{3\pi/4}^{5\pi/4}\int_1^2\int_{-1}^1 r\,dz\,dr\,d\theta\\ &=\int_{3\pi/4}^{5\pi/4}\int_1^2 2r\,dr\,d\theta\\ &=\int_{3\pi/4}^{5\pi/4}\left[r^2\right]_1^2\,d\theta\\ &=\int_{3\pi/4}^{5\pi/4}3\,d\theta =3\left(\frac{5\pi}{4}-\frac{3\pi}{4}\right) =\frac{3\pi}{2}. \end{aligned} $$

For a check, the sector angle is \(\pi/2\), so its annular base area is \(\frac12(2^2-1^2)(\pi/2)=3\pi/4\). The height is \(2\), giving volume \(3\pi/2\), as calculated.

Common Mistake: Treating a Boundary Line as One Ray

A plane through the \(z\)-axis meets the \(xy\)-plane in a full line, with two opposite rays. For example, \(y=x\) has directions \(\theta=\pi/4\) and \(\theta=5\pi/4\). The equation alone does not say which ray bounds a sector or which side of the plane is included.

A related error is using the inverse tangent without checking the quadrant. For \(x\le0\) and \(y\ge0\), the calculation \(y/x\le0\) does not justify choosing a negative angle or an angle in the fourth quadrant. The point is in the second quadrant, so its angles lie between \(\pi/2\) and \(\pi\). Correcting the quadrant and applying every inequality gives the appropriate angular interval. Always test a direction known to lie inside the proposed region; if it fails one of the original inequalities, the interval is wrong.

Why These Bounds Matter

Every angular interval describes a set of directions, not a complete three-dimensional solid. A usable cylindrical description must also specify how far the solid extends from the axis and how high it extends vertically. Some constraints affect only the angle, some affect only \(z\), and others couple \(r\) to \(\theta\). Separating those roles helps prevent missing bounds and makes the resulting iterated integral match the actual region.

Takeaway: Convert each plane inequality into a condition on \(\theta\) when it passes through the \(z\)-axis, intersect the allowed angular ranges, and then state the radial and vertical limits separately. Verify the final description against the original inequalities and the geometry.

Check Your Understanding

For each question, identify the relevant plane inequalities and explain how they determine the angular interval.

  1. What angular interval corresponds to \(x\ge0\) and \(y\le0\), using angles in \([0,2\pi)\)?
  2. Why does the plane \(y=x\) correspond to two opposite boundary directions?
  3. For \(x\ge0\), \(y\ge0\), and \(x+y\le5\), which bounds depend on \(\theta\), and which do not?
  4. How can an interval such as \([7\pi/4,9\pi/4]\) represent a sector crossing angle \(0\)?
  5. What check can detect an angular interval that selects the wrong side of a boundary plane?

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