MATH-283 (Calculus III) Curriculum

Session 15 - Double Integrals in Polar Coordinates

Handout Worksheet

Fall Semester 2026

Session Objectives & Overview

By the end of this session, you will be able to:

Formulate, sketch, and bound 2-dimensional planar domains using polar coordinates $(r, \theta)$ , establishing exact boundary limits for polar rectangles (annular sectors $a \le r \le b, \, \alpha \le \theta \le \beta$ ) and general polar regions with angle-dependent radial limits $g_1(\theta) \le r \le g_2(\theta)$ and offset circles $r = 2a\cos\theta$ .
Transform double integrals from rectangular $(x, y)$ coordinates to polar $(r, \theta)$ coordinates by applying the coordinate transformations $x = r\cos\theta, \, y = r\sin\theta, \, x^2 + y^2 = r^2$ , mapping boundary curves, and incorporating the Jacobian determinant area scaling factor $dA = r \, dr \, d\theta$ .
Strategically evaluate iterated polar double integrals over radially symmetric domains, leveraging Fubini product separability $\iint_R g(r)h(\theta) \, dA = \left(\int_\alpha^\beta h(\theta) \, d\theta\right)\left(\int_a^b r g(r) \, dr\right)$ for constant boundaries, executing sequential nested integration for non-separable curves (cardioids), and calculating enclosed 3D volumes under surfaces $z = f(r, \theta)$ including paraboloids and hemispheres.

Micro-Lecture

1. Radial Geometry & Curvilinear Grids

Mathematical Foundation: In Cartesian coordinates, planar regions are gridded by orthogonal lines of constant $x$ and constant $y$ . However, when domains exhibit circular, radial, or rotational symmetry, Cartesian boundary equations degenerate into algebraic radicals $y = \pm\sqrt{a^2 - x^2}$ . As established in Double Integrals in Polar Coordinates, transitioning to polar coordinates parameterizes points by radial distance $r \ge 0$ and azimuth angle $\theta \in [0, 2\pi)$ , where $x = r\cos\theta, y = r\sin\theta$ , and $x^2 + y^2 = r^2$ . On this curvilinear grid, circular disks and concentric rings transform into simple constant intervals known as polar rectangles.
Engineering Context: Mechanical and aerospace engineers utilize polar grids to analyze rotational stress in spinning flywheels, turbine impeller disks, and helicopter rotor hubs, where material boundaries naturally conform to constant radial contours.

2. The Polar Area Element & The Jacobian Scaling Factor

Mathematical Foundation: When transforming integration coordinates, the differential area element does not map directly ( $dx \, dy \ne dr \, d\theta$ ). Because the circumferential arc length of an infinitesimal sector expands proportionally with distance from the origin ( $s = r\theta$ ), a polar grid patch with radial width $\Delta r$ and angular width $\Delta \theta$ occupies an area $\Delta A \approx \bar{r} \, \Delta r \, \Delta \theta$ . In the continuous limit, the area element requires the Jacobian scaling factor $r$ , yielding $dA = r \, dr \, d\theta$ . Omitting this factor $r$ is the single most common failure mode in multivariable integration.
Engineering Context: Civil and environmental engineers compute volumetric fluid discharge $Q = \iint v \, dA$ in circular stormwater culverts; inserting the Jacobian factor $r$ correctly weights fluid velocity across expanding concentric pipe flow streamlines.

3. Strategic Evaluation & Rotational Volume Enclosures

Mathematical Foundation: Once transformed, polar integrals frequently exhibit algebraic separability. If the integration boundaries are constants and the integrand factors into $g(r) \cdot h(\theta)$ , Fubini's Theorem permits splitting the double integral into the product of two independent single integrals. When surfaces in three dimensions possess rotational symmetry, such as paraboloids $z = c - r^2$ or hemispheres $z = \sqrt{R^2 - r^2}$ , polar double integrals calculate the enclosed spatial volume $V = \iint_D z(r, \theta) \, r \, dr \, d\theta$ with immediate single-variable $u$ -substitution.
Engineering Context: Optical and structural engineers calculate the storage volume and light-gathering focus of parabolic reflector antennas, astronomical telescope mirrors, and dome pressure hulls using polar volume integration.

Skill Block 1

Worked Example A: Bounding an Annular Sector (Concentric Circles & Rays)

Describe the planar region $R$ in polar coordinates that lies in the first quadrant, bounded between the concentric circles $x^2 + y^2 = 4$ and $x^2 + y^2 = 16$ , and between the ray lines $y = \frac{1}{\sqrt{3}}x$ and $y = x$ . Formulate and evaluate the double integral representing its surface area.

Worked Solution:

Determine Radial Bounds:
State the Circle Boundary Equation in Polar Form (defined in Polar Rectangles and Annular Sectors):
$x^2 + y^2 = r^2$ For the inner circle: $x^2 + y^2 = 4 \implies r_{\text{inner}} = 2$ .
For the outer circle: $x^2 + y^2 = 16 \implies r_{\text{outer}} = 4$ .
Thus, the radial limits are: $2 \le r \le 4$
Determine Angular Bounds:
State the Polar Angle Tangent Formula:
$\tan\theta = \frac{y}{x}$ For the lower bounding line $y = \frac{1}{\sqrt{3}}x$ : $\tan\theta_1 = \frac{1}{\sqrt{3}} \implies \theta_1 = \frac{\pi}{6}$ For the upper bounding line $y = x$ : $\tan\theta_2 = 1 \implies \theta_2 = \frac{\pi}{4}$ Thus, the angular sweep is: $\frac{\pi}{6} \le \theta \le \frac{\pi}{4}$
Formulate and Evaluate the Area Integral:
State the Polar Area Double Integral:
$\text{Area}(R) = \int_{\theta_1}^{\theta_2} \int_{r_1}^{r_2} r \, dr \, d\theta = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \int_{2}^{4} r \, dr \, d\theta$ Evaluate the inner integral with respect to $r$ : $\int_{2}^{4} r \, dr = \left[ \frac{r^2}{2} \right]_{2}^{4} = \frac{16}{2} - \frac{4}{2} = 8 - 2 = 6$ Evaluate the outer integral with respect to $\theta$ : $\text{Area}(R) = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} 6 \, d\theta = [6\theta]_{\frac{\pi}{6}}^{\frac{\pi}{4}} = 6\left(\frac{\pi}{4} - \frac{\pi}{6}\right) = 6\left(\frac{\pi}{12}\right) = \frac{\pi}{2}$

Worked Example B: Region Bounded by Circle and Slanted Line

Find the polar coordinates formulation and evaluate the area of the region $R$ in the first quadrant that lies inside the circle $x^2 + y^2 = 25$ and below the slanted line $y = \sqrt{3}x$ .

Worked Solution:

Extract Radial Limits:
The boundary curve $x^2 + y^2 = 25$ transforms to $r^2 = 25 \implies r = 5$ . Since the region begins at the origin, the radial distance sweeps from $r = 0$ out to the circular perimeter:
$0 \le r \le 5$
Extract Angular Limits:
The region is bounded below by the positive $x$ -axis, where $\theta_1 = 0$ .
The upper boundary line is $y = \sqrt{3}x$ . Apply the Polar Angle Tangent Formula:
$\tan\theta = \frac{y}{x} = \sqrt{3} \implies \theta_2 = \frac{\pi}{3}$ Thus, the angular interval is: $0 \le \theta \le \frac{\pi}{3}$
Evaluate the Area Integral:
State the Polar Area Double Integral (defined in Double Integrals in Polar Coordinates):
$\text{Area} = \int_{0}^{\frac{\pi}{3}} \int_{0}^{5} r \, dr \, d\theta$ Evaluate the inner integral: $\int_{0}^{5} r \, dr = \left[ \frac{r^2}{2} \right]_{0}^{5} = \frac{25}{2}$ Evaluate the outer integral: $\text{Area} = \int_{0}^{\frac{\pi}{3}} \frac{25}{2} \, d\theta = \frac{25}{2}\left(\frac{\pi}{3} - 0\right) = \frac{25\pi}{6}$

Practice Problem 1: Concept & Reading Check

Open this Skill Block on the Memex to locate the information needed to answer the following:

(a) Explain why an infinitesimal area element $\Delta A$ in a polar grid depends on radial distance $r$ , unlike a Cartesian area element $\Delta x \, \Delta y$ .
(b) In Worked Example B, deconstruct how the Cartesian equation $y = \sqrt{3}x$ was converted into the constant upper angular bound $\theta = \frac{\pi}{3}$ .
(c) In the linked engineering context Mechanical Clutch Plate Contact Zone Friction Torque (Mechanical Engineering), identify why powertrain engineers bound clutch friction pads into discrete annular sectors rather than covering the entire disc face.

Solution Key:

(a) In Cartesian coordinates, grid lines are parallel, so sub-rectangles maintain constant width $\Delta x$ and height $\Delta y$ everywhere. In polar coordinates, radial rays diverge as they move outward from the origin. The arc length along the angular direction is $s = r\theta$ , meaning the tangential width of a polar patch is $\Delta s \approx r \Delta \theta$ . Multiplying by the radial increment $\Delta r$ yields the area $\Delta A \approx r \, \Delta r \, \Delta \theta$ , introducing the mandatory Jacobian factor $r$ .
(b) The Cartesian line passes through the origin with slope $m = \sqrt{3}$ . Using the fundamental polar transformation equations $x = r\cos\theta$ and $y = r\sin\theta$ , dividing gives $\frac{y}{x} = \frac{r\sin\theta}{r\cos\theta} = \tan\theta$ . Setting $\tan\theta = \sqrt{3}$ on the first quadrant interval $[0, \frac{\pi}{2}]$ produces the unique angle $\theta = \frac{\pi}{3}\text{ rad}$ ( $60^\circ$ ).
(c) As detailed in Mechanical Clutch Plate Contact Zone Friction Torque (Mechanical Engineering), wet-friction clutch plates must dissipate severe thermal energy during gear engagement. Bounding the friction material into discrete annular sectors between $r = 6\text{ cm}$ and $r = 10\text{ cm}$ leaves angular channels for transmission oil flow, preventing thermal glazing while maintaining sufficient normal contact area for torque transmission.

Show your steps and calculations here.

Practice Problem 2

Describe the region $R$ in polar coordinates and compute its area: the planar set of points between the concentric circles $r = 1$ and $r = 3$ and between the angles $\theta = \frac{\pi}{6}$ and $\theta = \frac{\pi}{2}$ .

Solution Key:

State the Governing Formula:
State the Polar Rectangle Area Formula (defined in Polar Rectangles and Annular Sectors):
$\text{Area}(R) = \int_{\alpha}^{\beta} \int_{a}^{b} r \, dr \, d\theta$ Substitute the given boundaries $a = 1, b = 3, \alpha = \frac{\pi}{6}, \beta = \frac{\pi}{2}$ : $\text{Area}(R) = \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \int_{1}^{3} r \, dr \, d\theta$
Evaluate the Inner Radial Integral:
$\int_{1}^{3} r \, dr = \left[ \frac{r^2}{2} \right]_{1}^{3} = \frac{3^2}{2} - \frac{1^2}{2} = \frac{9 - 1}{2} = 4$
Evaluate the Outer Angular Integral:
$\text{Area}(R) = \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} 4 \, d\theta = [4\theta]_{\frac{\pi}{6}}^{\frac{\pi}{2}} = 4\left(\frac{\pi}{2} - \frac{\pi}{6}\right) = 4\left(\frac{3\pi - \pi}{6}\right) = 4\left(\frac{2\pi}{6}\right) = \frac{4\pi}{3}$

Show your steps and calculations here.

Practice Problem 3

Determine the polar coordinate bounds and set up the double integral for the area of the region $R$ lying inside the circle $x^2 + y^2 \le 9$ and above the line $y = x$ .

Solution Key:

Identify the Radial Boundary:
The boundary circle $x^2 + y^2 = 9$ converts via $x^2 + y^2 = r^2$ to $r^2 = 9 \implies r = 3$ . The region extends from the origin out to the perimeter:
$0 \le r \le 3$
Identify the Angular Sweep:
The line $y = x$ satisfies $\tan\theta = 1$ . In the $xy$ -plane, this line extends through the origin at $\theta = \frac{\pi}{4}$ (Quadrant 1) and $\theta = \frac{5\pi}{4}$ (Quadrant 3). The region lying above the line $y = x$ encompasses the upper-left semicircle, sweeping counter-clockwise:
$\frac{\pi}{4} \le \theta \le \frac{5\pi}{4}$
Formulate and Evaluate the Integral:
State the Polar Area Double Integral (defined in Double Integrals in Polar Coordinates):
$\text{Area} = \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \int_{0}^{3} r \, dr \, d\theta$ Evaluate the inner integral: $\int_{0}^{3} r \, dr = \left[ \frac{r^2}{2} \right]_{0}^{3} = \frac{9}{2}$ Evaluate the outer integral: $\text{Area} = \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \frac{9}{2} \, d\theta = \frac{9}{2}\left(\frac{5\pi}{4} - \frac{\pi}{4}\right) = \frac{9}{2}(\pi) = \frac{9\pi}{2}$ (Note: This represents exactly half of the full circle area $\frac{1}{2}\pi (3)^2 = \frac{9\pi}{2}$ .)

Show your steps and calculations here.

Practice Problem 4

Find the polar coordinate equation and integration bounds for the region enclosed by the offset circle $(x - 2)^2 + y^2 \le 4$ . Formulate the double integral representing its area.

Solution Key:

Convert Cartesian Boundary to Polar Form:
State the Offset Circle Expansion:
$(x - 2)^2 + y^2 = 4 \implies x^2 - 4x + 4 + y^2 = 4 \implies x^2 + y^2 = 4x$ Substitute the polar identities $x^2 + y^2 = r^2$ and $x = r\cos\theta$ : $r^2 = 4r\cos\theta \implies r = 4\cos\theta$
Determine Angular and Radial Limits:
Since the radius must be non-negative ( $r \ge 0$ ), we require $4\cos\theta \ge 0$ , which restricts $\theta$ to Quadrants 1 and 4:
$-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$ For each fixed $\theta$ , the radius extends from the origin ( $r = 0$ ) to the circular boundary: $0 \le r \le 4\cos\theta$
Formulate and Evaluate the Area Integral:
State the General Polar Area Integral (defined in Polar Rectangles and Annular Sectors):
$\text{Area} = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_{0}^{4\cos\theta} r \, dr \, d\theta$ Evaluate the inner integral: $\int_{0}^{4\cos\theta} r \, dr = \left[ \frac{r^2}{2} \right]_{0}^{4\cos\theta} = \frac{16\cos^2\theta}{2} = 8\cos^2\theta$ Substitute the half-angle identity $\cos^2\theta = \frac{1 + \cos(2\theta)}{2}$ : $\text{Area} = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 8\left(\frac{1 + \cos(2\theta)}{2}\right) d\theta = 4 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (1 + \cos(2\theta)) \, d\theta$ Evaluate the outer integral: $\text{Area} = 4 \left[ \theta + \frac{\sin(2\theta)}{2} \right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = 4\left[ \left(\frac{\pi}{2} + 0\right) - \left(-\frac{\pi}{2} + 0\right) \right] = 4(\pi) = 4\pi$

Show your steps and calculations here.

Regroup 1

Clarify why boundaries of the form $x^2 + y^2 = 2ax$ transform into single polar curves $r = 2a\cos\theta$ and why their angular sweep is $[-\pi/2, \pi/2]$ rather than $[0, 2\pi]$ .
Emphasize the geometric difference between a Cartesian rectangle $\Delta x \Delta y$ and a polar sector $\bar{r} \Delta r \Delta \theta$ .

Skill Block 2

Worked Example A: Converting a Non-Elementary Gaussian Radial Integrand

Convert the following Cartesian double integral into polar coordinates and evaluate its exact value:

\int_{0}^{2} \int_{0}^{\sqrt{4 - x^2}} e^{-(x^2 + y^2)} \, dy \, dx

Worked Solution:

Analyze and Map the Cartesian Boundaries:
The inner limits indicate $y$ ranges from $0$ up to $y = \sqrt{4 - x^2}$ . Squaring yields $x^2 + y^2 = 4$ , which is a circle of radius $R = 2$ .
The outer limits indicate $x$ ranges from $0$ to $2$ .
Together, this defines a quarter-disk in the first quadrant. In polar coordinates:
$0 \le r \le 2, \quad 0 \le \theta \le \frac{\pi}{2}$
Transform Integrand and Differential Area Element:
Apply the Polar Transformation Equations (defined in Double Integrals in Polar Coordinates):
$x^2 + y^2 = r^2 \implies e^{-(x^2 + y^2)} = e^{-r^2}$ $dy \, dx = r \, dr \, d\theta$ Assemble the polar integral: $I = \int_{0}^{\frac{\pi}{2}} \int_{0}^{2} e^{-r^2} \cdot r \, dr \, d\theta$
Evaluate the Iterated Integral:
Apply $u$ -substitution for the radial integral: let $u = -r^2$ , so $du = -2r \, dr \implies r \, dr = -\frac{1}{2}du$ .
When $r = 0 \implies u = 0$ ; when $r = 2 \implies u = -4$ .
Evaluate the inner integral:
$\int_{0}^{2} e^{-r^2} r \, dr = \int_{0}^{-4} -\frac{1}{2}e^u \, du = \left[ -\frac{1}{2}e^u \right]_{0}^{-4} = -\frac{1}{2}e^{-4} - \left(-\frac{1}{2}e^0\right) = \frac{1}{2}\left(1 - e^{-4}\right)$ Evaluate the outer integral: $I = \int_{0}^{\frac{\pi}{2}} \frac{1}{2}\left(1 - e^{-4}\right) d\theta = \frac{1}{2}\left(1 - e^{-4}\right) [\theta]_0^{\frac{\pi}{2}} = \frac{\pi}{4}\left(1 - e^{-4}\right)$

Worked Example B: Converting a Higher-Degree Polynomial Radical Integrand

Convert the following rectangular double integral into polar coordinates and evaluate it:

\int_{0}^{1} \int_{0}^{\sqrt{1 - x^2}} (x^2 + y^2)^2 \, dy \, dx

Worked Solution:

Determine Polar Domain Limits:
The region is bounded by $y = 0$ , $y = \sqrt{1 - x^2}$ , and $x \in [0, 1]$ , representing the first-quadrant unit disk:
$0 \le r \le 1, \quad 0 \le \theta \le \frac{\pi}{2}$
Substitute Coordinates and Jacobian:
State the Pythagorean Polar Identity:
$x^2 + y^2 = r^2 \implies (x^2 + y^2)^2 = (r^2)^2 = r^4$ Include the Jacobian factor $dA = r \, dr \, d\theta$ : $I = \int_{0}^{\frac{\pi}{2}} \int_{0}^{1} (r^4) \cdot r \, dr \, d\theta = \int_{0}^{\frac{\pi}{2}} \int_{0}^{1} r^5 \, dr \, d\theta$
Evaluate via Separability:
Apply Fubini's Separability Theorem (defined in Separability in Polar Integrals):
$I = \left( \int_{0}^{\frac{\pi}{2}} d\theta \right) \left( \int_{0}^{1} r^5 \, dr \right) = \left( \frac{\pi}{2} \right) \left[ \frac{r^6}{6} \right]_{0}^{1} = \left(\frac{\pi}{2}\right) \left(\frac{1}{6}\right) = \frac{\pi}{12}$

Practice Problem 5: Concept & Reading Check

Open this Skill Block on the Memex to locate the information needed to answer the following:

(a) State the algebraic substitution used to convert the Cartesian differential element $dy \, dx$ into polar coordinates, and explain the mathematical purpose of the factor $r$ .
(b) In Worked Example A, explain why the original Cartesian integral could not be integrated directly with respect to $y$ , and how inserting $r$ enabled analytical integration.
(c) In the linked engineering application Hydraulic Conduit Laminar Volumetric Flow Rate (Civil & Environmental Engineering), describe how the parabolic flow velocity profile $v(x, y) = 4(0.25 - (x^2+y^2))$ transforms when mapped to polar coordinates.

Solution Key:

(a) The differential area element transforms as $dy \, dx = r \, dr \, d\theta$ . The factor $r$ is the absolute value of the determinant of the Jacobian matrix $\frac{\partial(x,y)}{\partial(r,\theta)}$ . Geometrically, it accounts for the fact that polar grid patches expand in tangential width as the radius increases from the origin ( $\Delta s = r \Delta \theta$ ), ensuring conservation of area across coordinate transformations.
(b) In Cartesian coordinates, the inner integral requires integrating $e^{-y^2}$ with respect to $y$ , which is a non-elementary Gaussian integral with no closed-form algebraic antiderivative. By converting the entire domain into polar coordinates, the integrand becomes $e^{-r^2} \cdot r$ . The extra factor of $r$ matches the derivative of the exponent ( $\frac{d}{dr}(-r^2) = -2r$ ), unlocking standard single-variable $u$ -substitution.
(c) As detailed in Hydraulic Conduit Laminar Volumetric Flow Rate (Civil & Environmental Engineering), fluid flow through a circular pipe is axisymmetric. Substituting $x^2 + y^2 = r^2$ collapses the two-variable Cartesian velocity function $v(x, y) = 4(0.25 - (x^2+y^2))$ into the single radial variable function $v(r) = 1 - 4r^2\text{ m/s}$ , simplifying the volumetric flow rate integral into a straightforward polynomial evaluation.

Show your steps and calculations here.

Practice Problem 6

Convert and evaluate the double integral:

\int_{-2}^{2} \int_{0}^{\sqrt{4 - x^2}} y \, dy \, dx

Solution Key:

Identify the Domain in Polar Coordinates:
The boundary $y = \sqrt{4 - x^2}$ describes the upper semicircle of radius $R = 2$ centered at the origin, with $-2 \le x \le 2$ and $y \ge 0$ .
In polar coordinates:
$0 \le r \le 2, \quad 0 \le \theta \le \pi$
Transform Integrand and Differential:
Substitute $y = r\sin\theta$ and $dy \, dx = r \, dr \, d\theta$ :
$I = \int_{0}^{\pi} \int_{0}^{2} (r\sin\theta) \, r \, dr \, d\theta = \int_{0}^{\pi} \int_{0}^{2} r^2 \sin\theta \, dr \, d\theta$
Evaluate via Separability:
Apply Fubini's Separability Theorem in Polar Coordinates (defined in Separability in Polar Integrals):
$I = \left( \int_{0}^{\pi} \sin\theta \, d\theta \right) \left( \int_{0}^{2} r^2 \, dr \right)$ Evaluate the angular factor: $\int_{0}^{\pi} \sin\theta \, d\theta = [-\cos\theta]_{0}^{\pi} = (-\cos\pi) - (-\cos 0) = -(-1) - (-1) = 2$ Evaluate the radial factor: $\int_{0}^{2} r^2 \, dr = \left[ \frac{r^3}{3} \right]_{0}^{2} = \frac{8}{3} - 0 = \frac{8}{3}$ Multiply: $I = 2 \times \frac{8}{3} = \frac{16}{3}$

Show your steps and calculations here.

Practice Problem 7

Convert the following Cartesian integral into polar coordinates and evaluate its exact closed-form value:

\int_{0}^{1} \int_{0}^{\sqrt{1 - x^2}} \frac{1}{1 + x^2 + y^2} \, dy \, dx

Solution Key:

Map Cartesian Limits to Polar Bounds:
The domain $0 \le x \le 1$ and $0 \le y \le \sqrt{1 - x^2}$ is the first quadrant of the unit disk:
$0 \le r \le 1, \quad 0 \le \theta \le \frac{\pi}{2}$
Transform Integrand and Area Element:
Substitute $x^2 + y^2 = r^2$ and $dy \, dx = r \, dr \, d\theta$ :
$I = \int_{0}^{\frac{\pi}{2}} \int_{0}^{1} \frac{1}{1 + r^2} \cdot r \, dr \, d\theta = \int_{0}^{\frac{\pi}{2}} \int_{0}^{1} \frac{r}{1 + r^2} \, dr \, d\theta$
Evaluate the Iterated Integral:
Factor the independent integrals:
$I = \left( \int_{0}^{\frac{\pi}{2}} d\theta \right) \left( \int_{0}^{1} \frac{r}{1 + r^2} \, dr \right)$ The angular integral is $[\theta]_0^{\frac{\pi}{2}} = \frac{\pi}{2}$ .
For the radial integral, let $u = 1 + r^2 \implies du = 2r \, dr \implies r \, dr = \frac{1}{2}du$ .
When $r = 0 \implies u = 1$ ; when $r = 1 \implies u = 2$ . $\int_{0}^{1} \frac{r}{1 + r^2} \, dr = \frac{1}{2} \int_{1}^{2} \frac{1}{u} \, du = \frac{1}{2}[\ln|u|]_{1}^{2} = \frac{1}{2}(\ln 2 - \ln 1) = \frac{1}{2}\ln 2$ Multiply the components: $I = \left(\frac{\pi}{2}\right) \left(\frac{1}{2}\ln 2\right) = \frac{\pi}{4}\ln 2$

Show your steps and calculations here.

Practice Problem 8

Set up the double integral for the area enclosed by the circle $x^2 + y^2 = a^2$ in both Cartesian and polar coordinates. Evaluate the polar setup and contrast the computational complexity of the two methods.

Solution Key:

Cartesian Coordinates Formulation:
Solving for $y$ yields $y = \pm\sqrt{a^2 - x^2}$ with $x \in [-a, a]$ . The Cartesian double integral is:
$\text{Area} = \int_{-a}^{a} \int_{-\sqrt{a^2 - x^2}}^{\sqrt{a^2 - x^2}} dy \, dx = \int_{-a}^{a} 2\sqrt{a^2 - x^2} \, dx$ Evaluating this integral requires the trigonometric substitution $x = a\sin\phi, dx = a\cos\phi \, d\phi$ , followed by the half-angle identity $\cos^2\phi = \frac{1 + \cos(2\phi)}{2}$ , requiring significant algebraic manipulation.
Polar Coordinates Formulation & Evaluation:
State the Polar Area Double Integral:
$\text{Area} = \int_{0}^{2\pi} \int_{0}^{a} r \, dr \, d\theta$ Evaluate using separability: $\text{Area} = \left( \int_{0}^{2\pi} d\theta \right) \left( \int_{0}^{a} r \, dr \right) = [ \theta ]_{0}^{2\pi} \left[ \frac{r^2}{2} \right]_{0}^{a} = (2\pi) \left( \frac{a^2}{2} \right) = \pi a^2$
Complexity Comparison:
Polar coordinates align with the rotational symmetry of the circle. The boundary curves become constant limits ( $r \in [0, a], \theta \in [0, 2\pi]$ ), completely eliminating radical expressions and reducing a trigonometric substitution problem to a basic power-rule calculation.

Show your steps and calculations here.

Regroup 2

Review why $dA = r \, dr \, d\theta$ and emphasize how the factor $r$ acts as a catalyst for algebraic $u$ -substitution.
Discuss how to quickly identify when an integral demands polar conversion (e.g., circular limits, $x^2 + y^2$ terms, or Gaussian exponents).

Skill Block 3

Worked Example A: Evaluating via Fubini's Separability Shortcut

Evaluate the double integral:

I = \iint_R r^2 \sin\theta \, dA

over the semicircular region

R = \{(r, \theta) \mid 0 \le r \le 2, \, 0 \le \theta \le \pi\}

Worked Solution:

Verify Separability Conditions:
1. The limits of integration are strictly constants: $r \in [0, 2]$ and $\theta \in [0, \pi]$ .
2. The combined integrand including the Jacobian $r$ is: $(r^2 \sin\theta) \cdot r = r^3 \sin\theta$ This factors completely into $g(r) = r^3$ and $h(\theta) = \sin\theta$ .
Apply Separability Formulation:
State Fubini's Separability Theorem in Polar Coordinates (defined in Separability in Polar Integrals):
$I = \left( \int_{0}^{\pi} \sin\theta \, d\theta \right) \left( \int_{0}^{2} r^3 \, dr \right)$
Evaluate the Independent Integrals:
Evaluate the angular integral:
$\int_{0}^{\pi} \sin\theta \, d\theta = [-\cos\theta]_{0}^{\pi} = (-\cos\pi) - (-\cos 0) = -(-1) - (-1) = 2$ Evaluate the radial integral: $\int_{0}^{2} r^3 \, dr = \left[ \frac{r^4}{4} \right]_{0}^{2} = \frac{16}{4} - 0 = 4$ Multiply the results: $I = 2 \times 4 = 8$

Worked Example B: Solid Volume Enclosed Under a Paraboloid

Find the volume of the solid enclosed below the paraboloid $z = 16 - x^2 - y^2$ and above the $xy$ -plane ( $z = 0$ ).

Worked Solution:

Determine the Base Domain in the $xy$ -Plane:
The solid terminates where the paraboloid meets the $xy$ -plane ( $z = 0$ ):
$16 - x^2 - y^2 = 0 \implies x^2 + y^2 = 16 \implies r^2 = 16 \implies r = 4$ The projection of the solid onto the $xy$ -plane is a circular disk of radius $4$ : $0 \le r \le 4, \quad 0 \le \theta \le 2\pi$
Formulate the Polar Volume Integral:
State the Double Integral Volume Formula in Polar Coordinates (defined in Volume and Surface Integration in Polar Coordinates):
$V = \iint_D z \, dA = \int_{0}^{2\pi} \int_{0}^{4} (16 - r^2) \, r \, dr \, d\theta$ Distribute the Jacobian factor $r$ : $V = \int_{0}^{2\pi} \int_{0}^{4} (16r - r^3) \, dr \, d\theta$
Evaluate via Separability:
Factor into independent single integrals:
$V = \left( \int_{0}^{2\pi} d\theta \right) \left( \int_{0}^{4} (16r - r^3) \, dr \right)$ Evaluate the radial integral: $\int_{0}^{4} (16r - r^3) \, dr = \left[ 8r^2 - \frac{r^4}{4} \right]_{0}^{4} = \left( 8(16) - \frac{256}{4} \right) - 0 = 128 - 64 = 64$ Evaluate the angular integral: $\int_{0}^{2\pi} d\theta = 2\pi$ Multiply: $V = 2\pi \times 64 = 128\pi$

Practice Problem 9: Concept & Reading Check

Open this Skill Block on the Memex to locate the information needed to answer the following:

(a) State the two specific mathematical conditions required before an iterated double integral in polar coordinates can be evaluated using Fubini's product separability shortcut.
(b) In Worked Example B, deconstruct how setting $z = 0$ determined both the radial limit $r = 4$ and the angular sweep $0 \le \theta \le 2\pi$ .
(c) In the linked engineering application Variable-Density Flywheel Rotational Inertia (Aerospace & Mechanical Engineering), explain why the moment of inertia integrand scales as $r^5 \, dr \, d\theta$ and how this benefits satellite attitude control.

Solution Key:

(a) As defined in Separability in Polar Integrals, the product separability shortcut requires:
1. Constant Integration Bounds: The region must be a polar rectangle ( $r \in [a, b]$ and $\theta \in [\alpha, \beta]$ ). If $r$ depends on $\theta$ , the inner evaluation produces functions of $\theta$ that cannot be factored out of the outer integral.
2. Multiplicative Integrand: The function must factor into the product of a purely radial function and a purely angular function ( $f(r, \theta) = g(r) \cdot h(\theta)$ ). Additive terms require term-by-term splitting or nested evaluation.
(b) The base of the solid in the $xy$ -plane occurs where the surface height reaches zero ( $z = 0$ ). Setting $16 - (x^2 + y^2) = 0 \implies x^2 + y^2 = 16 \implies r^2 = 16$ , which defines a circle of radius $R = 4$ . Because the solid is an upright circular paraboloid centered on the $z$ -axis with no angular restrictions, it spans the entire circumference $0 \le \theta \le 2\pi$ and extends from the central axis $r = 0$ to the outer boundary $r = 4$ .
(c) As detailed in Variable-Density Flywheel Rotational Inertia (Aerospace & Mechanical Engineering), the differential moment of inertia is the product of squared distance from the axis ( $r^2$ ), local area density ( $\rho(r) = 3r^2$ ), and the polar area element ( $dA = r \, dr \, d\theta$ ), giving $r^2 \cdot 3r^2 \cdot r = 3r^5$ . In satellite design, this steep $r^5$ radial dependence means material placed near the rim contributes dramatically more resistance to angular acceleration ( $\tau = I \alpha$ ), maximizing stored kinetic energy and stabilization torque while minimizing satellite launch mass.

Show your steps and calculations here.

Practice Problem 10

Find the area of the planar region enclosed by the cardioid curve $r = 1 + \cos\theta$ for $0 \le \theta \le 2\pi$ .

Solution Key:

Set Up the Nested Polar Integral:
State the General Polar Area Integral:
$\text{Area} = \int_{0}^{2\pi} \int_{0}^{1 + \cos\theta} r \, dr \, d\theta$ Because the upper limit depends on $\theta$ , this integral is non-separable and must be evaluated from the inside out.
Evaluate the Inner Radial Integral ( $dr$ ):
$\int_{0}^{1 + \cos\theta} r \, dr = \left[ \frac{r^2}{2} \right]_{0}^{1 + \cos\theta} = \frac{1}{2}(1 + \cos\theta)^2 = \frac{1}{2}\left(1 + 2\cos\theta + \cos^2\theta\right)$
Apply the Trigonometric Half-Angle Identity:
Substitute $\cos^2\theta = \frac{1 + \cos(2\theta)}{2}$ :
$\frac{1}{2}\left(1 + 2\cos\theta + \frac{1 + \cos(2\theta)}{2}\right) = \frac{1}{2}\left(\frac{3}{2} + 2\cos\theta + \frac{1}{2}\cos(2\theta)\right)$
Evaluate the Outer Angular Integral ( $d\theta$ ):
$\text{Area} = \frac{1}{2} \left[ \frac{3}{2}\theta + 2\sin\theta + \frac{1}{4}\sin(2\theta) \right]_{0}^{2\pi}$ Evaluate at the boundaries: $\text{Area} = \frac{1}{2} \left[ \left(\frac{3}{2}(2\pi) + 2\sin(2\pi) + \frac{1}{4}\sin(4\pi)\right) - 0 \right] = \frac{1}{2}(3\pi + 0 + 0) = \frac{3\pi}{2}$

Show your steps and calculations here.

Practice Problem 11

Find the volume of the solid that lies below the hemispherical dome $z = \sqrt{9 - x^2 - y^2}$ and above the $xy$ -plane ( $z = 0$ ).

Solution Key:

Identify the Base Domain in the $xy$ -Plane:
The hemisphere meets the $xy$ -plane where $z = 0$ :
$\sqrt{9 - x^2 - y^2} = 0 \implies x^2 + y^2 = 9 \implies r = 3$ The integration domain $D$ is the full circle of radius $3$ : $0 \le r \le 3, \quad 0 \le \theta \le 2\pi$
Formulate the Polar Volume Integral:
State the Double Integral Volume Formula in Polar Coordinates (defined in Volume and Surface Integration in Polar Coordinates):
$V = \iint_D z \, dA = \int_{0}^{2\pi} \int_{0}^{3} \sqrt{9 - r^2} \, r \, dr \, d\theta$
Evaluate via $u$ -Substitution and Separability:
Factor the independent angular integral:
$V = \left( \int_{0}^{2\pi} d\theta \right) \left( \int_{0}^{3} \sqrt{9 - r^2} \, r \, dr \right) = (2\pi) \int_{0}^{3} (9 - r^2)^{1/2} r \, dr$ For the radial integral, let $u = 9 - r^2 \implies du = -2r \, dr \implies r \, dr = -\frac{1}{2}du$ .
When $r = 0 \implies u = 9$ ; when $r = 3 \implies u = 0$ . $\int_{0}^{3} (9 - r^2)^{1/2} r \, dr = \int_{9}^{0} -\frac{1}{2} u^{1/2} \, du = \frac{1}{2} \int_{0}^{9} u^{1/2} \, du = \frac{1}{2} \left[ \frac{2}{3}u^{3/2} \right]_{0}^{9}$ $= \frac{1}{3}\left(9^{3/2} - 0\right) = \frac{1}{3}(27) = 9$ Multiply by the angular factor: $V = 2\pi \times 9 = 18\pi$ (Note: The geometric formula for a hemisphere of radius $R = 3$ confirms: $V = \frac{2}{3}\pi R^3 = \frac{2}{3}\pi (27) = 18\pi$ .)

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Regroup 3

Emphasize how the Jacobian factor $r$ naturally provides the exact substitution derivative for radical and exponential rotational surfaces.
Review why non-constant polar boundaries (like the cardioid $r = 1 + \cos\theta$ ) require strict inside-out nested integration and trigonometric half-angle reduction.

Synthesis Wrap-up

Polar double integration transforms cumbersome Cartesian radical boundaries into clean, calculable intervals by exploiting radial and rotational symmetry.
The Jacobian scaling factor $r$ ( $dA = r \, dr \, d\theta$ ) is mathematically essential to account for the diverging area of polar wedges.
When domains are polar rectangles and integrands are multiplicative, Fubini's product separability shortcut accelerates computation and avoids nested variable carrying.
Homework assignment: Complete the Knewton Alta assignment Double Integrals in Polar Coordinates, focusing on setting up radial boundaries for offset disks and computing solid volumes between curved quadric surfaces.