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 ri and the inner radius as r{i-1}, the area of the annular sector is:
      {Area} = \frac{1}{2} \Delta \theta (ri^2 - r{i-1}^2)
  • Apply the Difference of Squares: The term (ri^2 - r{i-1}^2) can be factorized as: ri^2 - r{i-1}^2 = (ri - r{i-1})(ri + r{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 (ri - r{i-1})(ri + r{i-1})
  • Recognizing the terms:
    • Let \Delta r = ri - r{i-1} be the radial width of the annular sector.
    • Let ri^* = \frac{ri + 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 = ri^* \cos(\thetaj^*)
    • y = ri^* \sin(\thetaj^*)
    • Here, ri^ and \thetaj^ 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{i=1}^{m} \sum{j=1}^{n} f(ri^* \cos(\thetaj^), ri^ \sin(\thetaj^)) \cdot r_i^ \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: \iintR f(x,y) \, dA = \iintD 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{\alpha}^{\beta} \int{a}^{b} f(r, \theta) r \, dr \, d\theta
    • For our example, the integral becomes:
      V = \int{0}^{\pi/2} \int{1}^{2} (3r \cos \theta + 4r^2 \sin^2 \theta) r \, dr \, d\theta
      V = \int{0}^{\pi/2} \int{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{0}^{\pi/2} \left( 7 \cos \theta + 15 \left( \frac{1 - \cos(2\theta)}{2} \right) \right) \, d\theta V = \int{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]{0}^{\pi/2} = \left[ 7 \sin \theta + \frac{15}{2} \theta - \frac{15}{4} \sin(2\theta) \right]{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