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Calculus IV Unit 14 Review

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14.2 Evaluation of triple integrals in cylindrical coordinates

14.2 Evaluation of triple integrals in cylindrical coordinates

Written by the Fiveable Content Team • Last updated August 2025
Written by the Fiveable Content Team • Last updated August 2025
Calculus IV
Unit & Topic Study Guides

Integration Setup

Setting Up Triple Integrals and Integration Limits

A triple integral evaluates a function f(x,y,z)f(x, y, z) over a three-dimensional region EE, written as Ef(x,y,z)dV\iiint_E f(x, y, z) \, dV, where dVdV is the volume element.

To set one up, you need to determine the integration limits that describe the boundaries of EE. These limits come from the geometry of the region (a rectangular box, cylinder, cone, sphere, etc.) and often depend on the other variables.

The order of integration specifies which variable you integrate first, second, and third. Common orders include dzdydxdz \, dy \, dx, dzdrdθdz \, dr \, d\theta, and so on. Choosing the right order can make a problem dramatically simpler. Pick the order that gives you the cleanest limits.

Setting Up Triple Integrals and Integration Limits, HartleyMath - Triple Integrals

Iterated Integrals and Evaluation

Iterated integrals break the triple integral into three nested single integrals. Here's how you evaluate one:

  1. Start with the innermost integral. Treat all other variables as constants and integrate with respect to the innermost variable.
  2. Substitute that result into the middle integral. Now integrate with respect to the middle variable, still treating the outermost variable as a constant.
  3. Evaluate the outermost integral to get your final answer.

For example, with the integral abcdeff(x,y,z)dzdydx\int_a^b \int_c^d \int_e^f f(x, y, z) \, dz \, dy \, dx:

  • First compute eff(x,y,z)dz\int_e^f f(x, y, z) \, dz, holding xx and yy constant.
  • Plug that result into cd[]dy\int_c^d [\cdots] \, dy and evaluate, holding xx constant.
  • Finally, evaluate ab[]dx\int_a^b [\cdots] \, dx.

The key thing to watch: when limits depend on other variables (e.g., zz goes from 00 to r2x2\sqrt{r^2 - x^2}), you must respect that dependency at each stage.

Setting Up Triple Integrals and Integration Limits, Triple Integrals in Cylindrical and Spherical Coordinates · Calculus

Cylindrical Coordinates

Cylindrical Volume Element and Change of Variables

Cylindrical coordinates (r,θ,z)(r, \theta, z) are the natural choice whenever the region has symmetry about the zz-axis. The three coordinates are:

  • rr: distance from the zz-axis (always 0\geq 0)
  • θ\theta: angle measured in the xyxy-plane from the positive xx-axis
  • zz: height, same as in Cartesian coordinates

The conversion formulas from cylindrical to Cartesian are:

  • x=rcosθx = r\cos\theta
  • y=rsinθy = r\sin\theta
  • z=zz = z

When you switch to cylindrical coordinates, the volume element changes:

dV=rdrdθdzdV = r \, dr \, d\theta \, dz

That extra factor of rr is critical. It comes from the Jacobian determinant (x,y,z)(r,θ,z)=r\left|\frac{\partial(x,y,z)}{\partial(r,\theta,z)}\right| = r. Forgetting this rr is one of the most common mistakes on exams. Every integrand f(x,y,z)dVf(x,y,z)\,dV becomes f(rcosθ,rsinθ,z)  rdrdθdzf(r\cos\theta,\, r\sin\theta,\, z)\; r\, dr\, d\theta\, dz.

Setting up limits in cylindrical coordinates typically follows this pattern:

  1. Determine the range of θ\theta (often 00 to 2π2\pi for full revolution, or a smaller interval for a wedge).
  2. Determine the range of rr (often 00 to some function of θ\theta, or just a constant for a full cylinder).
  3. Determine the range of zz (often between two surfaces, expressed in terms of rr and θ\theta).

Cylindrical Shells and Integration

The cylindrical shell method connects single-variable techniques to the triple integral framework. A cylindrical shell is a thin hollow cylinder at radius rr with infinitesimal thickness drdr.

For a solid of revolution, the volume using shells is:

V=ab2πrh(r)drV = \int_a^b 2\pi r \, h(r) \, dr

where h(r)h(r) is the height of the shell at radius rr, and rr ranges from aa to bb.

Example: Volume of a cone with base radius RR and height HH.

The cone's surface slopes linearly, so at radius rr, the height is h(r)=HR(Rr)h(r) = \frac{H}{R}(R - r). Setting up the integral:

V=0R2πrHR(Rr)dr=2πHR0R(Rrr2)drV = \int_0^R 2\pi r \cdot \frac{H}{R}(R - r)\, dr = \frac{2\pi H}{R}\int_0^R (Rr - r^2)\, dr

Evaluating:

=2πHR[Rr22r33]0R=2πHR(R32R33)=2πHRR36=13πR2H= \frac{2\pi H}{R}\left[\frac{Rr^2}{2} - \frac{r^3}{3}\right]_0^R = \frac{2\pi H}{R}\left(\frac{R^3}{2} - \frac{R^3}{3}\right) = \frac{2\pi H}{R}\cdot\frac{R^3}{6} = \frac{1}{3}\pi R^2 H

This matches the well-known cone volume formula, which is a good sanity check. Whenever you can verify your triple integral result against a known formula, do it.