Double Integration in Polar Coordinates Pt. 1

Introduction to Polar Integration

• Objective: Estimate the volume through polar coordinates by calculating the area of a base in polar form. This method is particularly useful for regions with circular symmetry or integrands involving $(x^2 + y^2)$.

Area of the Base

• Definition: The infinitesimal area element in polar coordinates is denoted as $\Delta A$, which represents the small area segment in the polar plane that we are working with.
• Calculation: To derive this area, we consider an annular sector (a portion of a ring). This involves subtracting the area of one circular sector from another.

Sector Area Formulation

• Formula Reference: The area of a sector of a circle is given by: $\text{Area} = \frac{1}{2} r^2 \theta$
  • Where $r$ is the radius and $\theta$ is the central angle in radians.
• In terms of our application: The incremental central angle for calculating a tiny sector area is defined as $\Delta \theta$. Thus, the area of a small sector at radius $r$ is $\frac{1}{2} r^2 \Delta \theta$.

Difference of Sectors

• Calculation Process for $\Delta A$: We calculate the area of an annular sector by taking the difference between two sector areas formed by two consecutive radii.
  • If we denote the outer radius as $r<em>i$ and the inner radius as $r</em>{i-1}$, the area of the annular sector is:
    ${Area} = \frac{1}{2} \Delta \theta (r<em>i^2 - r</em>{i-1}^2)$
• Apply the Difference of Squares: The term $(r<em>i^2 - r</em>{i-1}^2)$ can be factorized as: $r<em>i^2 - r</em>{i-1}^2 = (r<em>i - r</em>{i-1})(r<em>i + r</em>{i-1})$
  • This factorization is crucial for simplifying our calculations into a more intuitive form.

Simplifying the Area Calculation

• Consequently: Substituting the factorization into the area formula:
  ${Area} = \frac{1}{2} \Delta \theta (r<em>i - r</em>{i-1})(r<em>i + r</em>{i-1})$
• Recognizing the terms:
  • Let $\Delta r = r<em>i - r</em>{i-1}$ be the radial width of the annular sector.
  • Let $r<em>i^* = \frac{r</em>i + r_{i-1}}{2}$ be the average radius of the annular sector.
• Therefore, the expression for the differential area element simplifies to: $\Delta A = r_i^* \Delta r \Delta \theta$
  • This form highlights that the area element is approximately a rectangle with dimensions $\Delta r$ (radial length) and $r_i^* \Delta \theta$ (arc length).

Riemann Sum for Double Integral Estimation

• Transitioning to the Riemann Sum: To approximate the volume under a surface $f(x,y)$ over a region in polar coordinates, we use a Riemann sum. The function $f(x,y)$ is converted into polar coordinates, $(r, \theta)$, by substituting:
  • $x = r<em>i^* \cos(\theta</em>j^*)$
  • $y = r<em>i^* \sin(\theta</em>j^*)$
  • Here, $r<em>i^<em>$ and $\theta</em>j^</em>$ are sample points within the small annular sector, and the indices $i$ and $j$ represent subdivisions in the $r$ and $\theta$ directions, respectively.
• The Riemann sum for a function $f(x,y)$ over a polar region is:
  $\sum<em>{i=1}^{m} \sum</em>{j=1}^{n} f(r<em>i^* \cos(\theta</em>j^<em>), r<em>i^</em> \sin(\theta</em>j^<em>)) \cdot r_i^</em> \Delta r \Delta \theta$

Limit Process for Double Integrals

• As the number of subdivisions $m$ and $n$ approaches infinity ($m \to \infty, n \to \infty$), the Riemann sum approximation leads to a double integral. The sum becomes an exact integral: $\iint<em>R f(x,y) \, dA = \iint</em>D f(r \cos \theta, r \sin \theta) r \, dr \, d\theta$
  • Where $R$ is the region in Cartesian coordinates, and $D$ is the corresponding region in polar coordinates.
  • In this integral, $r$ is typically bounded by $[a, b]$ and $\theta$ by $[\alpha, \beta]$.

Transformation in Polar Coordinates

• Theorem for Double Integrals: Given that a function $f(x,y)$ is continuous over a region, the differential area element $dA$ in Cartesian coordinates transforms into polar coordinates as: $dA = r \, dr \, d\theta$
  • The extra factor of $r$ is the Jacobian determinant of the transformation from Cartesian to polar coordinates, and it is physically intuitive as it accounts for the stretching of area elements as $r$ increases.

Example of Polar Integration for Volume

• Problem: Calculate the volume under the surface $z = 3x + 4y^2$ over the region $D$ in the first quadrant, bounded by the circles $x^2 + y^2 = 1$ and $x^2 + y^2 = 4$.
• Region Description and Conversion:
  • The region is an annular sector in the first quadrant.
  • In Cartesian coordinates, the bounds are $1 \le x^2 + y^2 \le 4$ and $x \ge 0, y \ge 0$.
  • Converting to polar coordinates:
  • $x^2 + y^2 = r^2$, so $1 \le r^2 \le 4 \implies 1 \le r \le 2$.
  • The first quadrant means $0 \le \theta \le \frac{\pi}{2}$.
  • Convert the function $z = f(x,y)$ to polar coordinates:
  • Substitute $x = r \cos \theta$ and $y = r \sin \theta$:
    $f(r, \theta) = 3(r \cos \theta) + 4(r \sin \theta)^2 = 3r \cos \theta + 4r^2 \sin^2 \theta$

Integration Setup

• Volume Integration Steps: Formulate the double integral in polar coordinates: $V = \int<em>{\alpha}^{\beta} \int</em>{a}^{b} f(r, \theta) r \, dr \, d\theta$
  • For our example, the integral becomes:
    $V = \int<em>{0}^{\pi/2} \int</em>{1}^{2} (3r \cos \theta + 4r^2 \sin^2 \theta) r \, dr \, d\theta$
    $V = \int<em>{0}^{\pi/2} \int</em>{1}^{2} (3r^2 \cos \theta + 4r^3 \sin^2 \theta) \, dr \, d\theta$

Example Integration Simplification

• Evaluate the Inner Integral (with respect to $r$):
  • Perform distribution and integration:
    $\int_{1}^{2} (3r^2 \cos \theta + 4r^3 \sin^2 \theta) \, dr$
  • Integrate term by term with respect to $r$, treating $\theta$ as a constant:
    $= \left[ r^3 \cos \theta + r^4 \sin^2 \theta \right]_{r=1}^{r=2}$
  • Apply the Fundamental Theorem of Calculus with bounds from 1 to 2:
    $= (2^3 \cos \theta + 2^4 \sin^2 \theta) - (1^3 \cos \theta + 1^4 \sin^2 \theta)$
    $= (8 \cos \theta + 16 \sin^2 \theta) - (\cos \theta + \sin^2 \theta)$
    $= 7 \cos \theta + 15 \sin^2 \theta$
• Evaluate the Outer Integral (with respect to $\theta$):
  • Now integrate the result with respect to $\theta$ from $0$ to $\pi/2$:
    $V = \int_{0}^{\pi/2} (7 \cos \theta + 15 \sin^2 \theta) \, d\theta$
  • Addressing common integrals: For $\sin^2 \theta$, use the power-reducing identity:
    $\sin^2 \theta = \frac{1 - \cos(2\theta)}{2}$
  • Substitute the identity:
    $V = \int<em>{0}^{\pi/2} \left( 7 \cos \theta + 15 \left( \frac{1 - \cos(2\theta)}{2} \right) \right) \, d\theta$$V = \int</em>{0}^{\pi/2} \left( 7 \cos \theta + \frac{15}{2} - \frac{15}{2} \cos(2\theta) \right) \, d\theta$
  • Integrate term by term:
    $= \left[ 7 \sin \theta + \frac{15}{2} \theta - \frac{15}{2} \cdot \frac{1}{2} \sin(2\theta) \right]<em>{0}^{\pi/2}$$= \left[ 7 \sin \theta + \frac{15}{2} \theta - \frac{15}{4} \sin(2\theta) \right]</em>{0}^{\pi/2}$
  • Apply the bounds:
    $= \left( 7 \sin(\pi/2) + \frac{15}{2} (\pi/2) - \frac{15}{4} \sin(\pi) \right) - \left( 7 \sin(0) + \frac{15}{2} (0) - \frac{15}{4} \sin(0) \right)$
    $= \left( 7(1) + \frac{15\pi}{4} - \frac{15}{4} (0) \right) - \left( 7(0) + 0 - 0 \right)$
    $V = 7 + \frac{15\pi}{4}$
• Final Result: The volume under the given surface over the specified region is $$7 + \frac{15