Problem 12 - APEX Calculus

Question

In Exercises \(11-14,\) an iterated integral in rectangular coordinates is given. Rewrite the integral using polar coordinates and evaluate the new double integral. $$ \int_{-4}^{4} \int_{-\sqrt{16-y^{2}}}^{0}(2 y-x) d x d y $$

Step-by-Step Solution

Answer

The integral evaluates to zero.

Step 1: Understand the Problem and Convert Coordinates

The given integral is \( \int_{-4}^{4} \int_{-\sqrt{16-y^{2}}}^{0}(2y - x)\, dx \, dy \). The region of integration is a semicircle above the \(x\)-axis with a radius of 4, centered at the origin (since \(x = -\sqrt{16-y^2}\) defines a semicircle). Switching to polar coordinates, where \(x = r\cos(\theta)\) and \(y = r\sin(\theta)\), the limits become \( r \) from 0 to 4 and \( \theta \) from \( \frac{\pi}{2} \) to \( \frac{3\pi}{2} \).

Step 2: Set Up the Integral in Polar Coordinates

In polar coordinates, the integral becomes \( \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \int_{0}^{4} (2r\sin(\theta) - r\cos(\theta)) r \, dr \, d\theta \). The term \((2y - x)\) was converted using polar substitutions: \(2y = 2r\sin(\theta)\) and \(x = r\cos(\theta)\). Multiply by \(r\) for the Jacobian of the polar coordinate transformation.

Step 3: Evaluate the Inner Integral

Evaluate \( \int_{0}^{4} (2r^2\sin(\theta) - r^2\cos(\theta)) \, dr \). This simplifies to \( (2\sin(\theta) - \cos(\theta)) \cdot \int_{0}^{4} r^2 \, dr \). The integral \( \int_{0}^{4} r^2 \, dr \) evaluates to \( \left. \frac{r^3}{3} \right|_0^4 = \frac{64}{3} \).

Step 4: Evaluate the Outer Integral

The outer integral is \( \frac{64}{3} \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} (2\sin(\theta) - \cos(\theta)) \, d\theta \). Evaluate each part separately: \( \int (2\sin(\theta)) \, d\theta = -2\cos(\theta) \) and \( \int (-\cos(\theta)) \, d\theta = -\sin(\theta) \). Combine these results and evaluate over the given limits.

Step 5: Calculate the Result

Substitute the limits into \(-2\cos(\theta) - \sin(\theta)\) to find the values at \(\theta = \frac{\pi}{2}\) and \(\theta = \frac{3\pi}{2}\), and then compute \((-2\cos(\theta) - \sin(\theta))\) resulting in zero due to the periodicity and symmetry of sine and cosine over this interval.

Key Concepts

Iterated Integral

Iterated integrals are a way to compute the value of a function over a specific area or volume. In simpler terms, an iterated integral allows you to sum up tiny bits of function values over a 2D or 3D region to find the total value.
Iterated integrals involve taking an integral within the scope of another integral, a method that is extremely useful in cases like this where you need to find areas under curves in two dimensions.

• For 2D problems as seen here, we handle two flat integrals: one for each direction (x and y).
• During the process, we usually integrate with respect to x first (holding y constant), and then integrate that result with respect to y.

Understanding iterated integrals lays the foundation for solving more complex problems. It involves layers of calculation, each adding a new dimension to the problem.

Coordinate Transformation

Coordinate transformation is a method of converting from one coordinate system to another. This is important especially when a problem is easier to solve in a different coordinate system.
In this exercise, converting from rectangular (or Cartesian) coordinates to polar coordinates simplifies the region of integration, which is a semicircle.

• Pole or center of the system is considered as origin with radius and angle being the two coordinates.
• The formulas used for this transformation are:
  • \( x = r \cos(\theta) \)
  • \( y = r \sin(\theta) \)
• In this problem, the transformation allows replacing troublesome square roots and simplifying limits considerably.

Using polar coordinates provides a straightforward path to evaluate the integral without dealing with complex roots.

Double Integral

A double integral is used to calculate the volume under a surface when the surface is defined over a two-dimensional region. Think of it as adding up an array of infinite 2D areas, one for each small element of the plane.
The original problem involves a double integral in rectangular coordinates that needs simplification.

• Each layer of integration computes a different dimension (first x, then y).
• In this exercise, you start by evaluating the inner integral with respect to x, then the outer one with respect to y.
• Double integrals provide the total accumulation over a 2D area, and when combined with polar coordinates, they become a powerful tool for finding areas in circular regions.

By converting to polar form, the double integral simplifies the calculation and evaluation processes.

Polar Integration

Polar integration is a specialized method to evaluate integrals over circular regions. It becomes particularly advantageous when working with regions symmetrical around a point (like a semicircle).
In polar integration, the term \( r \) in the integral represents the radius, which is multiplied by the function being integrated due to the Jacobian of the coordinate transformation.

• This Jacobian \( r \) naturally arises from converting area elements from rectangles to wedges.
• The limits for the integral are translated from Cartesian to angular bounds, which in this problem transformed from linear (like squares or rectangles) to radial bounds (circular or semicircles).
• Polar integration simplifies the semi-circular area into easily manageable regions, making calculation straightforward and clear.

Understanding polar integration provides a method for solving complex geometric problems with elegant solutions.