Question

An iterated integral in polar coordinates is given. Sketch the region whose area is given by the iterated integral and evaluate the integral, thereby finding the area of the region. \(\int_{0}^{3 \pi / 2} \int_{0}^{\theta^{2}} r d r d \theta\)

Short answer

The area of the region is \(\frac{243\pi^5}{320}\).

Step-by-step solution

Understand the Iterated Integral in Polar Coordinates

The given iterated integral is \( \int_{0}^{3 \pi / 2} \int_{0}^{\theta^{2}} r \, dr \, d\theta \). This represents the area of a region in the polar coordinate plane. The limits for \( r \) are from 0 to \( \theta^2 \) and for \( \theta \) from 0 to \( \frac{3\pi}{2} \).

Sketch the Region of Integration

To sketch the region, plot the curve \( r = \theta^2 \) for \( \theta \) from 0 to \( \frac{3\pi}{2} \). This is a spiral that extends as \( \theta \) increases. The region lies between the pole and this spiral for the given \( \theta \) range.

Evaluate the Inner Integral

The inner integral is \( \int_{0}^{\theta^{2}} r \, dr \). This can be evaluated as follows: \[ \int_{0}^{\theta^{2}} r \, dr = \left[ \frac{r^2}{2} \right]_{0}^{\theta^{2}} = \frac{(\theta^{2})^2}{2} - 0 = \frac{\theta^4}{2} \].

Evaluate the Outer Integral

With the inner integral evaluated, substitute it into the outer integral: \( \int_{0}^{3\pi/2} \frac{\theta^4}{2} \, d\theta \). Evaluate this integral: \[ \int_{0}^{3 \pi / 2} \frac{\theta^4}{2} \, d\theta = \frac{1}{2} \int_{0}^{3\pi/2} \theta^4 \, d\theta \].

Integrate the Function

Calculate \( \int_0^{3\pi/2} \theta^4 \, d\theta \): \[ \int_0^{3\pi/2} \theta^4 \, d\theta = \left[ \frac{\theta^5}{5} \right]_0^{3\pi/2} = \frac{(3\pi/2)^5}{5} \].

Final Calculation

Simplify the expression: \( \frac{1}{2} \cdot \frac{(3\pi/2)^5}{5} = \frac{1}{2} \cdot \frac{243\pi^5}{32 \cdot 5} = \frac{243\pi^5}{320} \). This is the area of the region in polar coordinates.

Key concepts

Iterated Integrals

Iterated integrals are a powerful tool in calculus for evaluating the area of a region. They are especially useful when working with complex shapes or functions, such as in polar coordinates. In this case, an iterated integral involves a series of integral evaluations performed one after the other, moving from the innermost to the outermost integral.

To understand iterated integrals, consider the given problem: \( \int_{0}^{3 \pi / 2} \int_{0}^{\theta^{2}} r \, dr \, d\theta \).

• The inner integral defines the limits for \( r \), solving it first simplifies the integral inside.
• The outer integral is then evaluated over \( \theta \), using the result of the inner integral.

This step-by-step evaluation allows us to find the area of complex regions, even in polar coordinates.

Area of a Region

The concept of finding the area of a region using integration is crucial. It involves covering the region with infinitely small areas and adding them up, or integrating them, to find the total area.

In our problem, the area is bounded by the radial lines from the origin, defined by \( r = \theta^2 \) as \( \theta \) varies from \( 0 \) to \( \frac{3\pi}{2} \). This is a map of a spiral shape in the polar coordinate system.

• Understanding the spiral shape created by \( r = \theta^2 \) is key to sketching the region correctly.
• The iterated integral then computes the area within these polar boundaries.

Evaluating this integral involves considering both \( r \) and \( \theta \) to define the curve boundary and area enclosed.

Polar Coordinate Plane

The polar coordinate plane is an alternative to the Cartesian plane. Points are determined by a distance from the pole (origin) and an angle from the positive x-axis.

In polar coordinates, points are defined as \( (r, \theta) \), where:

• \( r \) is the radial distance from the pole.
• \( \theta \) is the angle in radians.

This coordinate system is particularly useful for regions and problems involving circular or rotational symmetry, such as the spiral described by \( r = \theta^2 \).
Translating Cartesian equations into polar coordinates can often make integration simpler, particularly for regions that are circular or spiral in nature.

Integration Techniques

The techniques of integration allow us to solve integrals by simplifying complex functions. In the context of polar coordinates, this often involves finding the area inside curves defined in polar terms.

Here’s how to handle a typical integration in polar coordinates:

• Setup the iterated integrals, identifying \( r \) and \( \theta \) limits.
• Evaluate the innermost integral first, storing results for further processing.
• Proceed to the outer integral, using results from the inner step.

In case of polar coordinates, often the integration involves special consideration of radial limits, like \( \int_{0}^{\theta^{2}} r \, dr \), reducing to familiar integral forms or geometric areas. Mastery of differential calculus and understanding of polar-symmetrical regions is critical in choosing the correct integration techniques to find areas or solve other problems.