Page 1

Original handwritten page 1
Original handwritten page 1

Lesson 22: Triple Integrals in Cylindrical Coordinates (16.5)

Warmup
Evaluate \(\iint_D (4 - x^2 - y^2) \, dA\), where \(D\) is the region below:
A graph shows a shaded circle centered at the origin on a Cartesian coordinate system. The circle is labeled with the equation \(x^2 + y^2 = 4\) and its radius \(r = 2\). A blue ray is drawn from the origin into the first quadrant, with an arc indicating the angle \(\theta\) from the positive x-axis and the radius \(r\).
Review: Polar Coordinates

\(x = r \cos \theta\)
\(y = r \sin \theta\)

\(r^2 = x^2 + y^2\)
\(\tan \theta = \frac{y}{x}\)

\[\iint_D f(x, y) \, dA = \int_{\theta \text{ Bounds}} \int_{r \text{ Bounds}} f(r \cos \theta, r \sin \theta) \, r \, dr \, d\theta\]
Solution
Bounds: \(0 \le r \le 2\), \(0 \le \theta \le 2\pi\) \[\iint_D (4 - x^2 - y^2) \, dA = \int_0^{2\pi} \int_0^2 (4 - r^2) r \, dr \, d\theta\] \[= \int_0^{2\pi} \int_0^2 (4r - r^3) \, dr \, d\theta\] \[= \int_0^{2\pi} \left[ 2r^2 - \frac{r^4}{4} \right]_0^2 \, d\theta\] \[= \int_0^{2\pi} (8 - 4) \, d\theta = \int_0^{2\pi} 4 \, d\theta\] \[= [4\theta]_0^{2\pi} = 8\pi\]

Page 2

Original handwritten page 2
Original handwritten page 2

Example: Finding Volume with Triple Integrals

Example: Find the volume of the region \( D \) between \( z = x^2 + y^2 \) & \( z = 4 \)

The diagram depicts a three-dimensional region \( D \) bounded below by a paraboloid opening upwards from the origin and above by a horizontal plane. The paraboloid is labeled \( z = x^2 + y^2 \) and the horizontal plane is labeled \( z = 4 \). The region between these two surfaces is shaded and labeled \( D \). Below the paraboloid, on the \( xy \)-plane, a circular "shadow" region is labeled \( R \). Dashed lines project the circular intersection of the plane and the paraboloid down onto this region \( R \) on the \( xy \)-plane. A vertical red line segment inside the paraboloid illustrates the vertical distance between the lower surface and the upper plane for a fixed point \( (x, y) \).

\[ V = \iiint_D 1 \, dV \]

Fix \( (x, y) \) in the shadow \( R \):

\[ x^2 + y^2 \leq z \leq 4 \]

Boundary of shadow = intersection \( x^2 + y^2 = 4 \).

\[ \iiint_D 1 \, dV = \iint_R \left[ \int_{x^2 + y^2}^4 1 \, dz \right] dA = \iint_R (4 - x^2 - y^2) \, dA \]

Use polar coordinates to compute:

Page 3

Original handwritten page 3
Original handwritten page 3

Cylindrical Coordinates

Combination of Cartesian (in 1 variable) & polar (in 2 variables)

Visual Description: A 2D Cartesian coordinate system showing the relationship between Cartesian and polar coordinates. An arrow extends from the origin to a point labeled \((x, y)\). The radial distance of this arrow is labeled \(r\), and the angle it makes with the positive \(x\)-axis is labeled \(\theta\).

\[ \begin{aligned} x &= r \cos \theta \\ y &= r \sin \theta \\ z &= z \end{aligned} \longleftrightarrow \begin{aligned} r &= \sqrt{x^2 + y^2} \\ \theta &= \tan^{-1}\left(\frac{y}{x}\right) \\ z &= z \end{aligned} \]

Visual Description: A 3D coordinate system with \(x\), \(y\), and \(z\) axes. A point in 3D space is labeled \((x, y, z)\). A vertical dashed blue line drops from this point down to the \(xy\)-plane, meeting it at the point \((x, y, 0)\); the vertical height is labeled \(z\). In the \(xy\)-plane, a solid blue line connects the origin to the point \((x, y, 0)\), labeled as \(r\). The angle between the positive \(x\)-axis and this radial line is labeled \(\theta\).

\[ \iiint_{D} f(x, y, z) \, dV = \int_{\text{bounds of } \theta} \int_{\text{bounds of } r} \int_{\text{bounds of } z} f(r \cos \theta, r \sin \theta, z) \cdot r \, dz \, dr \, d\theta \]

Page 4

Original handwritten page 4
Original handwritten page 4

Graphs in Cylindrical Coordinates

Example 1: \(\theta = \frac{\pi}{4}\) (where \(r\) and \(z\) are free)
Visual Description of 2D Plot: A Cartesian coordinate system showing the \(xy\)-plane. A blue ray originates from the origin and extends into the first quadrant. The angle between the ray and the positive \(x\)-axis is labeled \(\frac{\pi}{4}\). Two points are marked on this ray with the coordinates \((1, \frac{\pi}{4})\) and \((r, \frac{\pi}{4})\).
Visual Description of 3D Plot: A 3D Cartesian coordinate system with \(x\), \(y\), and \(z\) axes. A vertical plane is depicted that contains the \(z\)-axis and extends into the first octant along the line where \(y = x\) in the \(xy\)-plane. Parallel blue lines are drawn vertically on the surface of this plane to illustrate its extension in the \(z\)-direction.
Solution: Represents a plane: \(\theta = \frac{\pi}{4} \Rightarrow \tan\left(\frac{\pi}{4}\right) = \frac{y}{x} \Rightarrow 1 = \frac{y}{x} \Rightarrow y = x\).

Page 5

Original handwritten page 5
Original handwritten page 5

Example 2

Example 2: \(r = 1\) in 2D circle of Radius 1

A diagram showing a 2D Cartesian coordinate system with x and y axes. A blue circle is centered at the origin, representing a circle with a radius of 1.

\(r = 1\)
\(0 \leq \theta \leq 2\pi\)

In 3D, \(z\) is "free"

A 3D coordinate system with x, y, and z axes. A blue vertical cylinder is centered along the vertical z-axis. The cylinder is defined by its vertical edges and two circular cross-sections, illustrating that the radius \(r = 1\) is constant for all values of \(z\).

Circular Cylinder.

Page 6

Original handwritten page 6
Original handwritten page 6

Example 3

eg(3): \( z = r \) or \( z = \sqrt{x^2 + y^2} \)

Single Cone.

A 3D coordinate system graph illustrating a single cone opening upwards. The vertex of the cone is positioned at the origin (0, 0, 0). The cone extends upwards along the vertical z-axis. Horizontal circular cross-sections are drawn at regular intervals, using solid lines for the front and dashed lines for the back to create a three-dimensional perspective.

Similarly: \( z = -r = -\sqrt{x^2 + y^2} \) is a single cone

A 3D coordinate system graph illustrating a single cone opening downwards. The vertex of the cone is positioned at the origin (0, 0, 0). The cone extends downwards into the negative z-region. Parallel horizontal cross-sections are indicated with dashed lines for the back edges to depict the three-dimensional volume of the downward-facing cone.

Page 7

Original handwritten page 7
Original handwritten page 7

eg 4:

\[ z = 6 - r^2 = 6 - x^2 - y^2 \]

\( z = 6 \leadsto 6 = 6 - x^2 - y^2 \leadsto x^2 + y^2 = 0 \)

\( z > 6 \leadsto z = 6 - x^2 - y^2 \leadsto x^2 + y^2 = 6 - z \quad \text{X} \)

\( z < 6 \leadsto x^2 + y^2 = 6 - z \leadsto \text{Circles} \)

Visual Description: A hand-drawn sketch of a 3D coordinate system showing the x, y, and z axes. A paraboloid is drawn opening downwards, starting from a peak on the positive z-axis (representing \( z = 6 \)). The sketch includes horizontal traces represented by elliptical lines (dashed for the background parts and solid for the foreground) at two different levels below the peak to illustrate the circular cross-sections of the surface.

Paraboloid Facing down.

Page 8

Original handwritten page 8
Original handwritten page 8

Evaluating Triple Integrals using Cylindrical Coordinates

eg â‘ : Find volume of the region \(D = \{ (r, \theta, z) : 1 \leq r \leq 5; \frac{\pi}{4} \leq \theta \leq \frac{3\pi}{4}; 2 \leq z \leq 7 \}\)
\[ \iiint_{D} 1 \, dV = \int_{\pi/4}^{3\pi/4} \int_{1}^{5} \int_{2}^{7} 1 \cdot r \, dz \, dr \, d\theta \] \[ = \int_{\pi/4}^{3\pi/4} \int_{1}^{5} 5r \, dr \, d\theta = \int_{\pi/4}^{3\pi/4} \left[ \frac{5r^2}{2} \right]_{1}^{5} \, d\theta \] \[ = \int_{\pi/4}^{3\pi/4} 60 \, d\theta = 60 \left( \frac{3\pi}{4} - \frac{\pi}{4} \right) = 60 \cdot \frac{\pi}{2} = 30\pi \]

Page 9

Original handwritten page 9
Original handwritten page 9

Defining a Region in Cylindrical Coordinates

\[ D = \{ (r, \theta, z) : 1 \leq r \leq 5; \frac{\pi}{4} \leq \theta \leq \frac{3\pi}{4}; 2 \leq z \leq 7 \} \]

The constraints \( 1 \leq r \leq 5 \) and \( \frac{\pi}{4} \leq \theta \leq \frac{3\pi}{4} \) are bracketed together and labeled as the \( xy \)-plane projection.

Visual Description: Geometric Representation of Region \( D \)

The notes provide two diagrams illustrating the mathematical definition above:

Left Diagram (2D Projection): This diagram shows the region \( R \) in the \( xy \)-plane using polar coordinates. It features two concentric circles centered at the origin: an inner circle with radius \( r = 1 \) and an outer circle with radius \( r = 5 \). Two boundary lines are drawn from the origin at angles \( \theta = \frac{\pi}{4} \) and \( \theta = \frac{3\pi}{4} \). The area captured between these two radii and two angles is shaded in blue and labeled with a red \( R \).

Right Diagram (3D Visualization): This diagram illustrates the full volume \( D \) in a 3D coordinate system with \( x, y, \) and \( z \) axes.

  • The region \( R \) from the first diagram is projected onto the \( xy \)-plane and shaded light blue.
  • Extending upward from region \( R \) along the \( z \)-axis is a 3D volume shaded yellow and labeled with a red \( D \).
  • The volume is a segment of a hollow cylinder (a cylindrical shell slice).
  • The vertical extent of the yellow volume is marked on the \( z \)-axis, beginning at \( z = 2 \) and ending at \( z = 7 \).

Page 10

Original handwritten page 10
Original handwritten page 10

Example 2

Find the volume of the region between \(z = \sqrt{x^2+y^2}\) and \(z = 6 - x^2 - y^2\)

In cylindrical coordinates:

  • \(z = \sqrt{x^2+y^2}\) becomes \(z = r\)
  • \(z = 6 - x^2 - y^2\) becomes \(z = 6 - r^2\)

A 3D coordinate graph illustrates the region of integration \(D\). A cone opening upwards from the origin, defined by \(z = \sqrt{x^2+y^2}\), intersects with a paraboloid opening downwards from \(z=6\), defined by \(z = 6 - x^2 - y^2\). The bounded volume between these two surfaces is labeled \(D\). A vertical red line segment within the volume indicates the range of \(z\) from the cone surface to the paraboloid surface. Below the surfaces, on the \(xy\)-plane, a circular "shadow" region is shaded with purple vertical lines and labeled \(R\). Dotted lines project the boundaries of the intersection down to the boundary of region \(R\).

Volume Calculation

\[V = \iiint_D 1 \, dV\]

Boundary of shadow

The boundary is determined by the intersection of the two surfaces:

\[6 - r^2 = r \implies r^2 + r - 6 = 0\] \[(r + 3)(r - 2) = 0\] \[r = 2\]

Integral Setup

To set up the triple integral, fix a point in the region \(R\) on the \(xy\)-plane. The vertical range is: \(r \le z \le 6 - r^2\)

\[\iiint_D 1 \, dV = \iint_R \left( \int_r^{6-r^2} 1 \, dz \right) dA\] \[= \int_0^{2\pi} \int_0^2 \int_r^{6-r^2} r \, dz \, dr \, d\theta\]

Page 11

Original handwritten page 11
Original handwritten page 11

Example 3

Find the volume of the region between \(y = \sqrt{x^2 + z^2}\) & \(y = 2 - \sqrt{x^2 + z^2}\)

A 3D coordinate system showing the \(x\), \(y\), and \(z\) axes. The volume \(D\) is bounded by two cones oriented along the \(y\)-axis. One cone, labeled \(y = r\), has its vertex at the origin and opens towards the positive \(y\) direction. The second cone, labeled \(y = 2 - r\), has its vertex at \(y = 2\) and opens towards the origin. The two cones intersect, forming a double-cone shape. A red shaded region on the \(xz\)-plane represents the "shadow" or projection \(R\), which is a circular disk of radius 1.

Cartesian in \(y\), polar in \(xz\):

\[x^2 + z^2 = r^2\] \[y = \sqrt{x^2 + z^2} = r\] \[y = 2 - \sqrt{x^2 + z^2} = 2 - r\]

Boundary of shadow = intersection:

\[r = 2 - r \implies 2r = 2 \implies r = 1\]

\[V = \iiint_D 1 \, dV = \iint\limits_{R} \left( \int_{r}^{2-r} 1 \, dy \right) dA = \int_{0}^{2\pi} \int_{0}^{1} \int_{r}^{2-r} r \, dy \, dr \, d\theta\]

Where \(R = \text{Disk of radius 1}\)

Page 12

Original handwritten page 12
Original handwritten page 12

Example

\[ \int_0^{\frac{\sqrt{2}}{2}} \int_y^{\sqrt{1-y^2}} \int_0^{1-x^2-y^2} 1 \, dz \, dx \, dy \] evaluate by converting to Cylindrical Coordinates

Solution

Convert to Bounds in polar:

The differential elements \(dx \, dy\) are converted to \(r \, dr \, d\theta\).

Analyzing the limits of integration for \(x\):

\[ y \leq x \leq \sqrt{1-y^2} \]

Left Bound: \(x = y\)
Right Bound: \(x = \sqrt{1-y^2} \implies x^2 + y^2 = 1\)

A graph shows the projection of the region of integration onto the \(xy\)-plane. The axes are labeled \(x\) and \(y\). A purple line \(y=x\) passes through the origin into the first quadrant. A green circular arc representing \(x^2 + y^2 = 1\) connects the \(x\)-axis to the \(y\)-axis. The intersection of the line and the arc is marked with a point at \((\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})\). The region between the \(x\)-axis, the line \(y=x\), and the circular arc is shaded with blue horizontal lines. The angle \(\theta\) is indicated starting from the positive \(x\)-axis toward the line \(y=x\).

Bounds in polar:

\[ 0 \leq r \leq 1 \] \[ 0 \leq \theta \leq \frac{\pi}{4} \]

The conversion to cylindrical coordinates yields: \[ \int_0^{\frac{\sqrt{2}}{2}} \int_y^{\sqrt{1-y^2}} \int_0^{1-x^2-y^2} 1 \, dz \, dx \, dy = \int_0^{\frac{\pi}{4}} \int_0^1 \int_0^{1-r^2} r \, dz \, dr \, d\theta \]