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16.5 Triple Integrals in Cylindrical Coordinates

cylindrical: hybrid of polar and cartesian

  • plane of "floor" → polar
  • height → Cartesian

point in cylindrical: \( (r, \theta, z) \)

Note: \( r \) and \( \theta \) are polar components, while \( z \) is Cartesian.

3D Cartesian axes showing point (r, theta, z) with its vertical height z and polar projection on the xy-plane.

The diagram illustrates a point in 3D space. The distance from the origin to the projection on the \( xy \)-plane is \( r \), the angle from the \( x \)-axis is \( \theta \), and the vertical distance is \( z \).

shadow of point on \( xy \)-plane

Conversion Formulas

\( (x, y, z) \to (r, \theta, z) \)

\[ r^2 = x^2 + y^2 \]
\[ \tan \theta = \frac{y}{x} \]

(polar components)

\[ z = z \]

\( (r, \theta, z) \to (x, y, z) \)

\[ x = r \cos \theta \]
\[ y = r \sin \theta \]

(polar components)

\[ z = z \]
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cylindrical is good when volume is cylinder-like

cylinder-like volume looks simple in cylindrical

\( \{ (r, \theta, z) : 1 \le r \le 2, \quad 0 \le \theta \le \pi/2, \quad 3 \le z \le 4 \} \)

"floor" in polar

A 2D graph showing a shaded quarter-annulus in the first quadrant between radii 1 and 2.

The "floor" region in the \( xy \)-plane.

A 3D sketch of a wedge-shaped volume element floating above the xy-plane, defined by cylindrical bounds.

The resulting 3D volume in cylindrical coordinates.

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Example: Triple Integral in Cartesian Coordinates

\[ \int_{-3}^{3} \int_{0}^{\sqrt{9-x^2}} \int_{0}^{9-x^2-y^2} \sqrt{x^2+y^2} \, dz \, dy \, dx \]

terrible in Cartesian

let's examine the volume we are integrating over

\(-3 \leq x \leq 3\)

\(0 \leq y \leq \sqrt{9-x^2}\)

"floor"

upper half circle radius 3

A 2D graph showing a shaded semi-circle on the xy-plane with radius 3, centered at the origin, above the x-axis.

this is a region in which polar is good

\(0 \leq \theta \leq \pi\)

\(0 \leq r \leq 3\)

now we look at \(z\)

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\[ 0 \leq z \leq 9-x^2-y^2 \]
  • xy-plane
  • paraboloid opening down vertex at \(z=9\)

in polar/cylindrical

\[ 9-x^2-y^2 = 9-(x^2+y^2) \]\[ = 9-r^2 \]
A 3D coordinate system showing a shaded paraboloid volume opening downwards from z=9 to the xy-plane.

now the bounds in cylindrical:

\(0 \leq r \leq 3\), \(0 \leq \theta \leq \pi\), \(0 \leq z \leq 9-r^2\)

original integral:

\[ \int_{-3}^{3} \int_{0}^{\sqrt{9-x^2}} \int_{0}^{9-x^2-y^2} \sqrt{x^2+y^2} \, dz \, dy \, dx \]

Conversion notes: bounds go into \(r, \theta\) bounds; integrand becomes \(r\); differential becomes \(r \, dz \, dr \, d\theta\).

becomes

\[ \int_{0}^{\pi} \int_{0}^{3} \int_{0}^{9-r^2} r \cdot r \, dz \, dr \, d\theta = \int_{0}^{\pi} \int_{0}^{3} \int_{0}^{9-r^2} r^2 \, dz \, dr \, d\theta \]\[ = \dots = \boxed{\frac{162\pi}{5}} \]
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Example: Triple Integral in Cylindrical Coordinates

\[ \int_{0}^{4} \int_{0}^{1/\sqrt{2}} \int_{x}^{\sqrt{1-x^2}} e^{-x^2-y^2} \, dy \, dx \, dz \]

Can't even start in Cartesian. Convert to cylindrical.

Limits of Integration

\( 0 \le x \le \frac{1}{\sqrt{2}} \)

\( x \le y \le \sqrt{1-x^2} \) (line \( y=x \) to upper circle radius 1)

\( 0 \le z \le 4 \)

2D region in xy-plane bounded by y=x, the y-axis, and a circular arc y=sqrt(1-x^2).

The volume looks like

3D sketch of a cylindrical wedge with height 4 and a base defined by the xy-region.
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Convert the "floor" to polar

\( y=x \) slope 1

bisect first quadrant in two halves

so angle is \( 45^\circ = \frac{\pi}{4} \)

Polar coordinate sketch showing a sector from pi/4 to pi/2 with radius r=1.

New Limits in Polar

\( 0 \le r \le 1 \)

\( \frac{\pi}{4} \le \theta \le \frac{\pi}{2} \)

Convert the integral:

\[ \int_{0}^{4} \int_{0}^{1/\sqrt{2}} \int_{x}^{\sqrt{1-x^2}} e^{-x^2-y^2} \, dy \, dx \, dz \]

Using substitutions: \( e^{-x^2-y^2} \to e^{-r^2} \) and \( dy \, dx \to r \, dr \, d\theta \)

\[ = \int_{0}^{4} \int_{\pi/4}^{\pi/2} \int_{0}^{1} r e^{-r^2} \, dr \, d\theta \, dz = \dots = \frac{\pi}{2}(1 - e^{-1}) \]
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Example: Calculating Mass of a Solid

Find mass of solid bounded above by sphere radius 2:

\[ x^2 + y^2 + z^2 = 4 \]
and bounded below by cone:
\[ z = \sqrt{x^2 + y^2} \]
with density \( \rho(x, y, z) = z \).

3D diagram showing a cone inside a sphere, with the intersection forming an ice cream cone shape.

The solid is made of material with density \( z \).

A separate sketch of the cone portion of the solid, resembling an ice cream cone.

Finding the Region of Integration

The shadow on the \( xy \)-plane is the "floor". It's a circle with a certain radius.

Substitute \( z = \sqrt{x^2 + y^2} \) into \( x^2 + y^2 + z^2 = 4 \):

\[ x^2 + y^2 + x^2 + y^2 = 4 \rightarrow 2x^2 + 2y^2 = 4 \rightarrow x^2 + y^2 = 2 \]

This is a circle of radius \( \sqrt{2} \).

2D graph of a circle on the xy-plane with radius square root of 2, shaded to show the region of integration.

\[ 0 \le r \le \sqrt{2} \]\[ 0 \le \theta \le 2\pi \]

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Setting up the Triple Integral

Determining the bounds for \( z \):

  • Above cone: \( z = \sqrt{x^2 + y^2} \rightarrow z = r \)
  • Below sphere: \( z = \sqrt{4 - x^2 - y^2} \rightarrow z = \sqrt{4 - r^2} \)
\[ r \le z \le \sqrt{4 - r^2} \]

Calculation

\[ \text{mass} = \int_{0}^{2\pi} \int_{0}^{\sqrt{2}} \int_{r}^{\sqrt{4-r^2}} z \cdot r \, dz \, dr \, d\theta = \dots = 2\pi \]

We accumulate this (density \( z \)) inside the ice cream cone. Note the Jacobian factor \( r \) is included in the differential \( r \, dz \, dr \, d\theta \).