Double Integration with Polar Coordinates Pt. 2

Double Angle Formula for Cosine

• The double angle formula for cosine is given by: $\cos(2\theta) = \cos^2(\theta) - \sin^2(\theta)$.

• This formula is fundamental for simplifying trigonometric expressions or for integrating powers of sine and cosine by reducing the power.

Expression in Terms of Sine and Cosine

• To express $\cos(2\theta)$ solely in terms of sine, rewrite using the Pythagorean identity: $\cos^2(\theta) = 1 - \sin^2(\theta)$.

• Substituting this into the primary double angle formula yields: $\cos(2\theta) = (1 - \sin^2(\theta)) - \sin^2(\theta) = 1 - 2\sin^2(\theta)$.

• Conversely, to express $\cos(2\theta)$ solely in terms of cosine, rearrange the previous identity to solve for $\sin^2(\theta)$ ($\sin^2(\theta) = 1 - \cos^2(\theta)$):

• This gives: $\cos(2\theta) = \cos^2(\theta) - (1 - \cos^2(\theta)) = 2\cos^2(\theta) - 1$.

• These alternative forms are particularly useful when you need to manipulate expressions to contain only one type of trigonometric function.

Integrating Cosine Squared

• The integration formula for $\cos^2(\theta)$ is derived using the double angle identity $\cos(2\theta) = 2\cos^2(\theta) - 1$, which can be rearranged to $\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}$.

• Integrating this expression leads to: $\int \cos^2(x)dx = \int \frac{1 + \cos(2x)}{2} dx = \frac{1}{2}x + \frac{1}{4}\sin(2x) + C$.

Double Integrating in Polar Coordinates

• Double integrals in polar coordinates are particularly useful for evaluating integrals over regions that are circular, annular (ring-shaped), or involve polar curves, especially when the integrand contains $(x^2 + y^2)$.

• The transformation from Cartesian coordinates $(x, y)$ to polar coordinates $(r, \theta)$ is given by:

  • $x = r\cos(\theta)$

  • $y = r\sin(\theta)$

• Consequently, the term $(x^2 + y^2)$ simplifies to $r^2$. This often makes integrands much simpler.

• The differential area element $dA$ in Cartesian coordinates $(dx dy$ or $dy dx)$ transforms into $r dr d\theta$ in polar coordinates. The extra factor of $r$ (the Jacobian of the transformation) is crucial for correct area scaling.

• The general form of a double integral in polar coordinates is: $\iint<em>R f(x,y) dA = \iint</em>D f(r\cos\theta, r\sin\theta) r dr d\theta$.

Integral Problem Approach

• Problem given: Calculate $\int_{-1}^{1} \int_{0}^{\sqrt{1-x^2}} (x^2 + y^2) \ dy \ dx$.

• This integral is difficult to evaluate directly in Cartesian coordinates due to the square root in the upper limit of integration and the term $(x^2 + y^2)$.

• Convert to polar coordinates, which often simplifies integrals over circular or part-circular regions, and expressions involving $(x^2 + y^2)$. This integral is a prime candidate for polar conversion.

Visualization of Region

• The region described by the bounds $y=0$ to $y=\sqrt{1-x^2}$ and $x=-1$ to $x=1$ clearly represents the upper half of the unit circle.

  • Imagine the curve $y=\sqrt{1-x^2}$ as the upper semi-circle of $x^2+y^2=1$.

  • The limits for $x$ ($-1$ to $1$) define the full diameter along the x-axis, and $y=0$ as the lower bound means we are above this axis.

  • Visual Description: This region would appear as a semi-circle in the upper half of the xy-plane, centered at the origin with a radius of 1.

• In polar coordinates, this upper semi-circular region transforms simply:

  • The radius $r$ varies from 0 (the origin) to 1 (the unit circle's edge).

  • The angle $\theta$ varies from 0 (positive x-axis) to $\pi$ (negative x-axis, moving anti-clockwise through the upper half-plane).

Converting the Integral to Polar Coordinates

• Using the polar coordinate transformations: $x = r\cos(\theta)$ and $y = r\sin(\theta)$.

• The Cartesian term $(x^2 + y^2)$ transforms to $(r^2\cos^2(\theta) + r^2\sin^2(\theta)) = r^2(\cos^2(\theta) + \sin^2(\theta)) = r^2$.

• The differential area element $dy dx$ transforms into $r dr d\theta$.

• Expressing the integral in polar coordinates:
  $\int \int r^2 \cdot r \ dr d\theta$ which simplifies to $\int_{0}^{\pi} d\theta \int_{0}^{1} r^3 dr$.

• Applying integral properties allows separation of variables, as the limits of integration are constants and the integrand is a product of functions of $r$ and $\theta$ (in this case, just $r$ for the integrand).

• This results in two distinct and simpler integrals: $\int_{0}^{\pi} d\theta$ and $\int_{0}^{1} r^3 dr$.

Solving the Integrated Functions

• Solve the first integral with respect to $\theta$:
  $\int_{0}^{\pi} d\theta = [\theta]_{0}^{\pi} = \pi - 0 = \pi$.

• Solve the radial integral with respect to $r$:
  $\int_{0}^{1} r^3 dr = \left[\frac{r^4}{4}\right]_{0}^{1} = \frac{1^4}{4} - \frac{0^4}{4} = \frac{1}{4}$.

• Therefore, the result of the original integral is the product of these two results:
  $\frac{1}{4} \cdot \pi = \frac{\pi}{4}$. This represents the integral of $(x^2+y^2)$ over the upper unit semi-circle.

Volume of Solid Region

• Problem: Calculate the volume of the solid bounded by the paraboloid: $z = 1 - x^2 - y^2$ and the plane $z=0$.

• Visual Description: The paraboloid $z = 1 - x^2 - y^2$ is an inverted bowl shape, opening downwards, with its peak (vertex) at $(0,0,1)$. The plane $z=0$ is the floor, or the xy-plane.

• Intersection analysis with the plane $z=0$ (the xy-plane):

  • Setting $z=0$ in the paraboloid's equation yields: $0 = 1 - x^2 - y^2$.

  • Rearranging this gives $x^2 + y^2 = 1$, which is the equation of the unit circle in the xy-plane. This circle forms the base of the solid.

Setting Up the Integral

• Since the base of the solid is a unit circle, polar coordinates are ideal.

• In polar coordinates, the boundaries for the base are:

  • Radius $r$ from 0 (origin) to 1 (the edge of the unit circle).

  • Angle $\theta$ from 0 to $2\pi$ (a full revolution to cover the entire circle).

• The height of the solid at any point $(x,y)$ is given by $z = 1 - x^2 - y^2$.

• Converting this height function to polar coordinates: $1 - (r^2\cos^2(\theta) + r^2\sin^2(\theta)) = 1 - r^2$.

• The volume can be expressed as a double integral of the height function over the base area element $r dr d\theta$:
  $\int_0^{2\pi} \int_0^{1} (1 - r^2) r dr d\theta$.

• This simplifies to $\int_0^{2\pi} \int_0^{1} (r - r^3) dr d\theta$.

• Decomposing the integral allows for easier computation, separating it into two products of separate integrals because the integrand is a function of $r$ only and the limits are constant:
  $\left(\int_0^{2\pi} d\theta\right) \left(\int_0^{1} (r - r^3) dr\right)$.

Solving Volume Integral

• The $\theta$ integral evaluates to:
  $\int_0^{2\pi} 1 d\theta = [\theta]_0^{2\pi} = 2\pi - 0 = 2\pi$.

• The radial integral evaluates to:
  $\int_0^{1} (r - r^3) dr = \left[\frac{r^2}{2} - \frac{r^4}{4}\right]_0^{1} = \left(\frac{1^2}{2} - \frac{1^4}{4}\right) - \left(\frac{0^2}{2} - \frac{0^4}{4}\right) = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$.

• The final answer for the volume is the product of these two results:
  $V = 2\pi \cdot \frac{1}{4} = \frac{\pi}{2}$.

Understanding Polar Coordinates in Integration

• The function can be generally expressed as $z=f(x,y)$ and the area element in polar becomes: $dA = r dr d\theta$.

• Describing regions captured in polar coordinates where $r$ and $\theta$ depend on functional limits leads to complexities that simplify integration, especially for regions not easily described by rectangular bounds.

Steps for Converting and Evaluating Double Integrals in Polar Coordinates

1. Sketch the Region of Integration ($R$): Visualize the region described by the Cartesian limits to understand its shape. This is crucial for determining the correct polar bounds.

2. Transform the Integrand: Replace $x$ with $r\cos\theta$ and $y$ with $r\sin\theta$. Simplify any $(x^2+y^2)$ terms to $r^2$.

3. Determine Polar Limits for $r$ and $\theta$: Based on the sketch, find the range for $r$ (from an inner curve to an outer curve) and for $\theta$ (from a starting angle to an ending angle that covers the region).

   • For a full circle centered at the origin, $r$ goes from 0 to its radius, and $\theta$ goes from 0 to $2\pi$.

   • For a semi-circle, $\theta$ might go from 0 to $\pi$ or $\pi$ to $2\pi$, etc.

   • For an annular region, $r$ would go from an inner radius to an outer radius.

4. Replace $dA$ with $r dr d\theta$: Remember the crucial factor of $r$.

5. Set up the Polar Integral: Write the new integral with the transformed integrand, the new limits, and the polar differential $r dr d\theta$.

6. Evaluate the Integral: Integrate with respect to $r$ first (inner integral), then with respect to $\theta$ (outer integral), or vice versa if limits are functions.

Example Problems and Application

• Finding areas and volumes utilize understanding of:

  • Synergy between rectangular and polar coordinates, particularly when regions have circular symmetry or integrands simplify well in polar form.

  • Considerations of continuity and bounded regions in the plane.

Area of Polar Curves Example

• To find the area of one half of a polar rose curve defined by: $r = \cos(2\theta)$.

  • Visual Description: A polar rose curve with 4 petals. The expression $r = \cos(2\theta)$ completes its shape over $\theta$ from 0 to $\pi$.

  • Bounds for one petal need careful consideration. For example, for $\cos(2\theta)$, one petal forms between the angles where $r=0$. Since $\cos(2\theta)=0$ implies $2\theta = \pm \frac{\pi}{2}$, $\pm \frac{3\pi}{2}$ etc., we get $\theta = \pm \frac{\pi}{4}$, etc. So, one petal spans from $-\frac{\pi}{4}$ to $\frac{\pi}{4}$. The problem asks for "one half of a polar rose curve," implying half the area of one petal or half the area of the entire curve.

  • Let's assume "one half of a polar rose curve" refers to half of one petal for this context, so we integrate from $0$ to $\pi/4$. (If it referred to half the entire curve, the bounds or integral setup would differ).

• The formula for the area of a region bounded by a polar curve is $\frac{1}{2} \int_{\alpha}^{\beta} r^2 d\theta$.

• Solve by integrating: $\frac{1}{2} \int_{0}^{\pi/4} \left(\cos(2\theta)\right)^2 d\theta$ (Note: The previous integral was $\int_{0}^{\pi/2}$ which would be for one full petal from $\theta = 0$ to $\theta = \pi/2$ if $r=\cos(2\theta)$ was used for a sectorial part of a circle, but for a petal of a rose, $\theta$ goes from 0 to $\pi/4$ for half a petal, and 0 to $\pi/2$ for a quarter of the rose, and $-\pi/4$ to $\pi/4$ for one full petal).

  • Correction based on typical polar area formula: The area of one petal for $r = \cos(2\theta)$ is $\frac{1}{2} \int<em>{-\pi/4}^{\pi/4} (\cos(2\theta))^2 d\theta$ or $\int</em>{0}^{\pi/4} (\cos(2\theta))^2 d\theta$ (using symmetry).

  • The evaluation requires basic trigonometric identities, specifically $\cos^2(x) = \frac{1+\cos(2x)}{2}$.

  • $\frac{1}{2} \int (\cos(2\theta))^2 d\theta = \frac{1}{2} \int \frac{1 + \cos(4\theta)}{2} d\theta = \frac{1}{4} \left[\theta + \frac{1}{4}\sin(4\theta)\right]$. Evaluating from $0$ to $\pi/4$ gives:
    $\frac{1}{4} \left[\frac{\pi}{4} + \frac{1}{4}\sin(\pi) - (0 + \frac{1}{4}\sin(0))\right] = \frac{1}{4} \left[\frac{\pi}{4} + 0 - 0\right] = \frac{\pi}{16}$.

• Final area can then be repeated depending on symmetry of the function (e.g., 4 times for the full rose).