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Triple Integrals Over Wedge-Shaped Solids

Translate wedge boundaries into angular limits, include the cylindrical volume factor, and evaluate triple integrals over restricted sectors.

Intermediate 10 min read

What You'll Learn

  • Identify angular bounds from the vertical planes that form a wedge.
  • Describe a wedge-shaped solid using angular, radial, and vertical bounds.
  • Set up cylindrical triple integrals with restricted angular intervals.
  • Evaluate wedge integrals for constant integrands and variable heights.
  • Handle angle-dependent integrands after converting them to cylindrical form.
  • Check an answer using geometry, antiderivatives, and the bounds.

Why Wedge-Shaped Solids Need Angular Bounds

A cylindrical description is especially natural when a solid is bounded by circular cylinders or has a vertical axis of symmetry. A wedge-shaped solid adds another feature: its projection onto the \(xy\)-plane occupies only part of a circle. The two sides of that wedge are often vertical half-planes passing through the \(z\)-axis.

In cylindrical coordinates, those sides are described by restricting the angle \(\theta\). The radial and vertical bounds then describe how far the solid extends within that angular sector and how high it reaches. The central task is to keep these three kinds of bounds distinct.

Key idea: For a wedge between the rays \(\theta=\alpha\) and \(\theta=\beta\), the angular bounds are \(\alpha\leq\theta\leq\beta\). The radial and vertical bounds describe the solid inside that sector. In a cylindrical triple integral, the volume element is \(r\,dz\,dr\,d\theta\).

Describing a Wedge-Shaped Solid

At a point away from the \(z\)-axis, cylindrical coordinates describe the horizontal position by a radius \(r\) and angle \(\theta\), with \(x=r\cos\theta\) and \(y=r\sin\theta\). A vertical half-plane through the \(z\)-axis corresponds to a fixed angle. The region between two such half-planes is an angular sector.

For a single angular interval, a common description of a solid is

$$ \alpha\leq\theta\leq\beta,\qquad a(\theta)\leq r\leq b(\theta),\qquad L(r,\theta)\leq z\leq U(r,\theta). $$

Here \(a(\theta)\) and \(b(\theta)\) are the inner and outer radial bounds, while \(L(r,\theta)\) and \(U(r,\theta)\) are the lower and upper vertical bounds. The bounds must describe an actual solid: \(0\leq a(\theta)\leq b(\theta)\), and \(L(r,\theta)\leq U(r,\theta)\) throughout the indicated ranges. The angles must use a consistent interval representing the intended sector.

When these bounds describe the solid without needing to split it into separate pieces, the triple integral of a function \(f\) is

$$ \iiint_S f(x,y,z)\,dV = \int_{\alpha}^{\beta} \int_{a(\theta)}^{b(\theta)} \int_{L(r,\theta)}^{U(r,\theta)} f(r\cos\theta,r\sin\theta,z)\,r\,dz\,dr\,d\theta. $$

The factor \(r\) is essential: it is part of the cylindrical volume element. The integrand must also be rewritten in terms of \(r\), \(\theta\), and \(z\) before evaluation.

Why the Integral Has This Form

The angular interval selects the directions inside the wedge. For each selected direction, the radial bounds select the points in the horizontal projection. At each such point, the vertical bounds select a segment parallel to the \(z\)-axis. Together, these bounds describe the solid through its vertical slices.

The cylindrical volume factor can be seen from the coordinate map \(\Phi(r,\theta,z)=(r\cos\theta,r\sin\theta,z)\). Its derivative matrix, with columns corresponding to \(r\), \(\theta\), and \(z\), is

$$ D\Phi= \begin{pmatrix} \cos\theta & -r\sin\theta & 0\\ \sin\theta & r\cos\theta & 0\\ 0 & 0 & 1 \end{pmatrix}, \qquad \det(D\Phi)=r\cos^2\theta+r\sin^2\theta=r. $$

Since \(r\geq0\), the absolute value of this determinant is \(r\). Thus a small cylindrical coordinate box has volume approximately \(r\,dz\,dr\,d\theta\). Integrating over the stated bounds gives the formula: the bounds select every point of \(S\), and the factor \(r\) accounts for the volume represented by each coordinate element.

Practical setup rule: First identify the wedge’s angular interval. Then determine the radial and vertical bounds within that interval. Finally, substitute cylindrical expressions into the integrand and include the factor \(r\).

Worked Example: Constant Integrand in a Vertical Wedge

Worked Example: Constant Integrand in a Vertical Wedge

Find the volume of the solid bounded by \(r=2\), \(z=0\), \(z=3\), and the vertical half-planes \(\theta=\frac{\pi}{6}\) and \(\theta=\frac{\pi}{3}\).

For volume, the integrand is \(1\). The bounds are \(\frac{\pi}{6}\leq\theta\leq\frac{\pi}{3}\), \(0\leq r\leq2\), and \(0\leq z\leq3\). Therefore

$$ V=\int_{\pi/6}^{\pi/3}\int_0^2\int_0^3 r\,dz\,dr\,d\theta. $$

Integrate first with respect to \(z\):

$$ \int_0^3 r\,dz=3r. $$

Then integrate with respect to \(r\):

$$ \int_0^2 3r\,dr = 3\left[\frac{r^2}{2}\right]_0^2 = 3\left(\frac{4}{2}-0\right) = 6. $$

The angular width is \(\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}\). Hence

$$ V=\int_{\pi/6}^{\pi/3}6\,d\theta = 6\left[\theta\right]_{\pi/6}^{\pi/3} = 6\left(\frac{\pi}{3}-\frac{\pi}{6}\right) = \pi. $$

Check geometrically: the horizontal cross-section is a circular sector of radius \(2\) and angle \(\frac{\pi}{6}\), so its area is \(\frac{1}{2}(2^2)(\frac{\pi}{6})=\frac{\pi}{3}\). Multiplying by the height \(3\) gives \(3\cdot\frac{\pi}{3}=\pi\), matching the integral.

Worked Example: A Wedge Under a Paraboloid

Worked Example: A Wedge Under a Paraboloid

Find the volume of the solid with \(\frac{\pi}{4}\leq\theta\leq\frac{\pi}{2}\), \(0\leq r\leq1\), and \(0\leq z\leq4-r^2\).

The integrand is \(1\), and the upper surface is \(z=4-r^2\). Using the cylindrical volume factor,

$$ V=\int_{\pi/4}^{\pi/2}\int_0^1\int_0^{4-r^2}r\,dz\,dr\,d\theta. $$

The inner integral is

$$ \int_0^{4-r^2}r\,dz = r\bigl((4-r^2)-0\bigr) = 4r-r^3. $$

Integrating radially gives

$$ \int_0^1(4r-r^3)\,dr = \left[2r^2-\frac{r^4}{4}\right]_0^1 = 2-\frac14 = \frac74. $$

Finally, integrate over the angular interval:

$$ V=\int_{\pi/4}^{\pi/2}\frac74\,d\theta = \frac74\left[\theta\right]_{\pi/4}^{\pi/2} = \frac74\left(\frac{\pi}{2}-\frac{\pi}{4}\right) = \frac{7\pi}{16}. $$

Check the radial antiderivative: differentiating \(2r^2-\frac{r^4}{4}\) gives \(4r-r^3\), the radial integrand. The evaluated radial factor is \(\frac74\), and the angular width is \(\frac{\pi}{4}\); their product is \(\frac{7\pi}{16}\). The result is positive, as required for a volume.

Worked Example: An Angle-Dependent Integrand

Worked Example: An Angle-Dependent Integrand

Evaluate \(\iiint_S x\,dV\), where \(S\) is given by \(\frac{\pi}{6}\leq\theta\leq\frac{\pi}{3}\), \(0\leq r\leq2\), and \(0\leq z\leq1\).

Because \(x=r\cos\theta\), the integrand in cylindrical coordinates is \(r\cos\theta\). Multiplying by the volume factor \(r\) gives the integrand \(r^2\cos\theta\) in the iterated integral:

$$ \iiint_S x\,dV = \int_{\pi/6}^{\pi/3}\int_0^2\int_0^1 r^2\cos\theta\,dz\,dr\,d\theta. $$

The \(z\)-integral contributes a factor of \(1\), since

$$ \int_0^1 r^2\cos\theta\,dz = r^2\cos\theta(1-0) = r^2\cos\theta. $$

For the radial integral,

$$ \int_0^2r^2\,dr = \left[\frac{r^3}{3}\right]_0^2 = \frac83. $$

For the angular integral,

$$ \int_{\pi/6}^{\pi/3}\cos\theta\,d\theta = \left[\sin\theta\right]_{\pi/6}^{\pi/3} = \frac{\sqrt3}{2}-\frac12 = \frac{\sqrt3-1}{2}. $$

Multiplying the factors gives

$$ \iiint_S x\,dV = \frac83\cdot\frac{\sqrt3-1}{2} = \frac{4(\sqrt3-1)}{3}. $$

Check: the antiderivatives used are correct because \(\frac{d}{dr}(r^3/3)=r^2\) and \(\frac{d}{d\theta}(\sin\theta)=\cos\theta\). Also \(x=r\cos\theta\geq0\) throughout this wedge, so a positive integral is consistent with the sign of the integrand.

A Common Mistake: Using the Full Angular Range

A wedge occupies only the angles specified by its boundary half-planes. Replacing that interval with a full revolution changes the region. For the first example, the correct angular interval has width \(\frac{\pi}{6}\), not \(2\pi\).

If one incorrectly uses \(0\leq\theta\leq2\pi\), the calculation becomes

$$ \int_0^{2\pi}\int_0^2\int_0^3r\,dz\,dr\,d\theta = \int_0^{2\pi}6\,d\theta = 12\pi. $$

That answer is twelve times the correct volume because \(2\pi\) is twelve times the wedge’s angular width \(\frac{\pi}{6}\). The corrected setup retains the actual wedge bounds, \(\frac{\pi}{6}\leq\theta\leq\frac{\pi}{3}\), and gives volume \(\pi\). Before integrating, check that the angular interval represents the intended sector rather than a full circle.

Checking a Wedge Integral Before Evaluating

A short bounds check often catches setup errors before any integration is needed. Verify that the angle starts and ends at the half-planes forming the wedge, that the radial bounds reach from the inner boundary to the outer boundary, and that the vertical bounds run from the bottom surface to the top surface. Then check that the integrand is expressed in cylindrical coordinates and that the extra factor \(r\) is present.

For a volume, the integrand before applying the volume element is \(1\), so the cylindrical integrand is \(r\). For another quantity, convert the specified function first and then multiply by \(r\). For example, if the function is \(x\), its cylindrical expression is \(r\cos\theta\), making the full integrand \(r^2\cos\theta\).

Check Your Understanding

Use the bounds and volume element carefully when answering these questions.

  1. Which cylindrical coordinate describes the boundary formed by a vertical half-plane through the \(z\)-axis?
  2. For a wedge between the half-planes \(\theta=\frac{\pi}{3}\) and \(\theta=\frac{2\pi}{3}\), what is the angular interval?
  3. Write the triple integral for the volume of the region \(0\leq\theta\leq\frac{\pi}{4}\), \(0\leq r\leq3\), and \(0\leq z\leq2\).
  4. If the integrand is \(y\), what expression replaces it in cylindrical coordinates? What is the full integrand after including the volume factor?
  5. Why does using \(0\leq\theta\leq2\pi\) generally give the wrong region for a wedge?

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