Problem 14 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 13-18, 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. $$ \int_{0}^{2 \pi} \int_{1}^{3} r d r d \theta $$

Step-by-Step Solution

Answer

The area of the region is \(8\pi\).

Step 1: Understand the Limits of Integration

The iterated integral given in the problem is \( \int_{0}^{2\pi} \int_{1}^{3} r \, dr \, d\theta \). This integral is in polar coordinates. The inner integral, \( \int_{1}^{3} r \, dr \), corresponds to the radial part, indicating the radial distance \( r \) ranges from 1 to 3. The outer integral, \( \int_{0}^{2\pi} \, d\theta \), corresponds to the angular part, indicating the angle \( \theta \) sweeps from 0 to \( 2\pi \).

Step 2: Sketch the Region in Polar Coordinates

In polar coordinates, the region is defined by the radial line from \( r=1 \) to \( r=3 \) and the full circular sweep from \( \theta=0 \) to \( \theta=2\pi \). This forms an annular (ring-shaped) region in the plane, bounded by circles with radii of 1 and 3 centered at the origin.

Step 3: Evaluate the Inner Integral

Evaluate the inner integral \( \int_{1}^{3} r \, dr \). Using the power rule for integration, \( \frac{r^2}{2} \bigg|_{1}^{3} \), the calculation becomes \( \frac{3^2}{2} - \frac{1^2}{2} = \frac{9}{2} - \frac{1}{2} = 4 \).

Step 4: Evaluate the Outer Integral

Substitute the result from the inner integral into the outer integral: \( \int_{0}^{2\pi} 4 \, d\theta \). Since 4 is a constant, the integral becomes \( 4 \times \theta \bigg|_{0}^{2\pi} = 4(2\pi) - 4(0) = 8\pi \).

Step 5: Conclude the Total Area

After evaluating the iterated integral, the total area of the region defined by the integrals is \( 8\pi \). This makes sense because the integral calculates the area of the annular region in polar coordinates.

Key Concepts

Iterated Integral

An iterated integral is used to calculate areas or volumes by integrating in stages, one variable at a time. In the problem above, the iterated integral is given as \( \int_{0}^{2\pi} \int_{1}^{3} r \, dr \, d\theta \). This expression means that we are integrating firstly with respect to \( r \) and then with respect to \( \theta \), also known as a double integral in polar coordinates.

• The inside integral, \( \int_{1}^{3} r \, dr \), is performed first and involves the radial direction.
• The outside integral, \( \int_{0}^{2\pi} \, d\theta \), is performed second, capturing the angular rotation around the origin.

Breaking down the integration in this manner helps in solving complex areas by handling one dimension at a time.

Annular Region

An annular region, as illustrated in the problem, is a ring-shaped area between two concentric circles. This region is bound by one circle of radius 1 and another of radius 3, with both circles centered at the origin.

• In polar coordinates, this is described with each point having a certain radial distance \( r \) from the origin and extending over all angles \( \theta \) from \( 0 \) to \( 2\pi \).
• This defines a complete circular sweep, making sure the area enclosed is a ring.

Understanding the geometric shape helps to visualize what area the integration calculates, ensuring clarity on the concept of iterated integrals in polar coordinates.

Integration in Polar Coordinates

Integrating in polar coordinates involves changing the traditional rectangular coordinates (\( x, y \)) to a system based on \( r \) and \( \theta \). This approach is particularly useful when dealing with regions or objects that naturally fit into circular shapes.

• The variable \( r \) represents how far away a point is from the origin.
• The angle \( \theta \) shows the direction of the point from the positive x-axis.
• The integration then takes into the account of radius and rotation aspect through \( dr \) and \( d\theta \) respectively.

This transformation simplifies solving integrals over circular domains and leads to easier computation in the context of circular symmetry.

Area Calculation

When calculating the area of a region defined in polar coordinates, iterated integrals allow each differential element to contribute to the total area.

• In the current exercise, solving \( \int_{1}^{3} r \, dr \) gives the partial area moving radially outward from 1 to 3.
• Continuing with \( \int_{0}^{2\pi} 4 \, d\theta \) represents sweeping an entire circle, completing the annular region calculation.
• The process results in the total area \( 8\pi \), confirming now how the double integration method evaluates the region accurately.

This systematic approach gives a comprehensive way of area calculation for rings, circles, and sectors effectively in mathematics.

Radial Distance

Radial distance is an essential concept within polar coordinates, indicating how far each point is from the origin.

• Represented by \( r \), it essentially measures the radius of a circle at any point \( \theta \).
• In our iteration, \( r \) ranges from 1 to 3, meaning each point lies between these distances, forming the area we integrate over.
• This element defines the inner boundary (\( r=1 \)) and outer boundary (\( r=3 \)) of our annular region.

Comprehending radial distance is crucial since it directly impacts the integration bounds and the subsequent area calculation.