Problem 9 - University Calculus: Early Transcendentals

Question

Change the Cartesian integral into an equivalent polar integral. Then evaluate the polar integral. $$\int_{-1}^{1} \int_{0}^{\sqrt{1-x^{2}}} d y d x$$

Step-by-Step Solution

Answer

The polar integral evaluates to \(\frac{\pi}{2}\).

Step 1: Sketch the Region of Integration

First, we need to identify the region over which we are integrating in Cartesian coordinates. The outer integral \(\int_{-1}^{1} \) suggests that \(x\) ranges from \(-1\) to \(1\). For each fixed \(x\), the inner integral \(\int_{0}^{\sqrt{1-x^{2}}} \) shows that \(y\) ranges from \(0\) to \(\sqrt{1-x^{2}}\). This describes the upper half of a circle \(x^2 + y^2 = 1\) centered at the origin, but only the portion where \(y\) is non-negative.

Step 2: Convert the Limits to Polar Coordinates

In polar coordinates, the circle \(x^2 + y^2 = 1\) can be described by \(r = 1\). Given that \(y \geq 0\), the angle \(\theta\) ranges from \(0\) to \(\pi\). Thus, in polar coordinates, the limits become \(0 \leq r \leq 1\) and \(0 \leq \theta \leq \pi\).

Step 3: Set Up the Polar Integral

The differential area element \(dA\) in polar coordinates is \(r \, dr \, d\theta\), not just \(dr \, d\theta\). Consequently, the polar integral becomes:\[\int_{0}^{\pi} \int_{0}^{1} r \, dr \, d\theta.\]

Step 4: Evaluate the Inner Integral

Evaluate the inner integral with respect to \(r\):\[\int_{0}^{1} r \, dr = \left[ \frac{r^2}{2} \right]_{0}^{1} = \frac{1}{2}.\]

Step 5: Evaluate the Outer Integral

Substitute the result from Step 4 into the outer integral and compute:\[\int_{0}^{\pi} \frac{1}{2} \, d\theta = \frac{1}{2} \times \left[ \theta \right]_{0}^{\pi} = \frac{1}{2} \times \pi = \frac{\pi}{2}.\]

Step 6: Conclusion

The equivalent polar integral is \(\int_{0}^{\pi} \int_{0}^{1} r \, dr \, d\theta\), and its evaluated result is \(\frac{\pi}{2}\).

Key Concepts

Cartesian Integration

Cartesian integration involves evaluating an integral over a region defined in the Cartesian coordinate system, which uses coordinates \( x \) and \( y \). In the given exercise, the integral is represented in the form:\[\int_{-1}^{1} \int_{0}^{\sqrt{1-x^{2}}} d y \, d x\]This double integral signifies that we are finding the cumulative sum of values over a specified region. The outer integral\( \int_{-1}^{1} \) indicates that \( x \) varies from -1 to 1, while the inner integral \( \int_{0}^{\sqrt{1-x^{2}}} \) specifiesthat for each chosen \( x \), \( y \) will range between 0 and \( \sqrt{1-x^{2}} \).

• This integral describes a semi-circle in the plane above the x-axis, centered at the origin with a radius of 1.
• The challenge is that while Cartesian integrals are straightforward for rectangular regions, circular regions like this one can be cumbersome to handle directly.

Due to the circular nature of the region, it is beneficial to convert to polar coordinates, which can more intuitively represent this area.

Area Element

An essential part of transforming an integral to polar coordinates is redefining the area element. In Cartesian coordinates,the area element is \( dy \, dx \), combining small changes in \( x \) and \( y \). In polar coordinates, however, this changes dueto the nature of how the coordinates are defined.

Polar Coordinate System

In polar coordinates, each point is defined by \( (r, \theta) \) where \( r \) is the radius from the origin and \( \theta \) is the angle measured from the positive x-axis.

Defining the Polar Area Element

The polar area element is given by \( dA = r \, dr \, d\theta \). This accounts for:

• The change in circumference as you move further from the origin (hence the \( r \) factor).
• The small changes in the radial distance \( dr \) and the angular distance \( d\theta \).

This replacement of the area element is critical because it ensures that the transformation from Cartesian to polarintegrates the same values over the same region.

Integration Limits

Adjusting integration limits is required when switching from Cartesian to polar coordinates. Let's examine the conversion.The original Cartesian limits describe a semi-circle essentially, with \(x\) between -1 and 1 and \(y\) from 0 to\(\sqrt{1-x^2}\).

Conversion to Polar Coordinates

In polar coordinates:

• The radial limit \( r \) ranges from 0 to 1, as the furthest point from the origin within the circle is a radius of 1.
• The angular limit \( \theta \) goes from 0 to \( \pi \), capturing the top half of the circle (because \( y \geq 0\) corresponds to anglesbetween 0 and \( \pi \)).

These adjusted limits ensure that all points within the specified region are captured in the polar integral formulation.

Polar Integral Evaluation

The final step is solving the transformed polar integral, which often simplifies calculations for circular regions.Here's how it's done in our example.

Setting Up the Integral

With the polar area element and updated limits, the integral looks like:\[\int_{0}^{\pi} \int_{0}^{1} r \, dr \, d\theta\]This expression unifies the region of integration and accounts for each small wedge-shaped slice of the circular area.

Evaluating the Integral

• First, solve the inner integral over \( r \): \[\int_{0}^{1} r \, dr = \left[ \frac{r^2}{2} \right]_{0}^{1} = \frac{1}{2}\].This step calculates the cumulative size of the radial slices.
• Substitute the result into the outer integral, which evaluates how this accumulated value changes with \( \theta \): \[\int_{0}^{\pi} \frac{1}{2} \, d\theta = \frac{1}{2} \times \pi = \frac{\pi}{2}\].

This gives the final area of the semi-circle, confirming that polar integration can considerably simplify the computationfor circular regions.