Problem 4 - Calculus Early Transcendentals

Question

1-4 Find the area of the region that is bounded by the given curve and lies in the specified sector. $$r=\sqrt{\sin \theta}, \quad 0 \leqslant \theta \leqslant \pi$$

Step-by-Step Solution

Answer

The bounded area is 1 square unit.

Step 1: Understand the Polar Equation

The equation given is \(r = \sqrt{\sin \theta}\), which is in polar form. It describes how the radius \(r\) changes with the angle \(\theta\). We are interested in calculating the area of the region bounded by this curve from \(\theta = 0\) to \(\theta = \pi\).

Step 2: Formula for the Area in Polar Coordinates

The formula to find the area of a region in polar coordinates is \( A = \frac{1}{2} \int_{\alpha}^{\beta} r^2 \, d\theta \). Here \(r = \sqrt{\sin \theta}\), \(\alpha = 0\), and \(\beta = \pi\). So, the expression becomes: \[ A = \frac{1}{2} \int_{0}^{\pi} (\sqrt{\sin \theta})^2 \, d\theta. \]

Step 3: Simplify the Expression

We simplify \((\sqrt{\sin \theta})^2\) to \(\sin \theta\). Thus, the integral changes to: \[ A = \frac{1}{2} \int_{0}^{\pi} \sin \theta \, d\theta. \]

Step 4: Solve the Integral

Integrate \(\sin \theta\) with respect to \(\theta\). The antiderivative of \(\sin \theta\) is \(-\cos \theta\). Therefore, the integral becomes: \[ A = \frac{1}{2} \left[ -\cos \theta \right]_{0}^{\pi}. \]

Step 5: Evaluate the Integral at the Limits

Substitute the limits into the antiderivative: \[ A = \frac{1}{2} \left(-\cos(\pi) - (-\cos(0))\right). \] Plug in the values: \(\cos(\pi) = -1\) and \(\cos(0) = 1\). Therefore, \[ A = \frac{1}{2} \left( -(-1) - (1) \right) = \frac{1}{2} (1 + 1) = \frac{1}{2} \times 2 = 1. \]

Step 6: Conclusion

The area of the region that is bounded by the curve \(r = \sqrt{\sin \theta}\) in the sector \(0 \leq \theta \leq \pi\) is 1 square unit.

Key Concepts

Area in Polar Coordinates

When dealing with polar coordinates, one of the main challenges is finding the area bounded by a curve. Unlike Cartesian coordinates where you may use simple geometric formulas, here the process involves integrating over a range of angles. The formula used is: \[ A = \frac{1}{2} \int_{\alpha}^{\beta} r^2 \, d\theta \]This formula helps us find the area of a sector defined by a curve in polar form.

• \(\alpha\) and \(\beta\) are the angles that define the limits of integration.
• \(r\) is the radius described by the curve as a function of \(\theta\).

Using this formula requires computing the integral, which can be more involved compared to Cartesian problems. It's important to ensure that the limits are appropriate for the region of interest.

Integration

Integration is a fundamental concept used to calculate areas, among other things, in polar coordinates. In our type of problem, it helps quantify the area enclosed by a polar curve.To integrate in polar coordinates:

• First, determine the function in terms of \(r\) that needs to be squared and integrated.
• Next, evaluate the integral over the specified range of \(\theta\).
• Finally, apply the limits to find the net area.

For the given example, once we transform \((\sqrt{\sin \theta})^2\) into \(\sin \theta\), the integration process starts. Evaluate the integral \(\frac{1}{2} \int_{0}^{\pi} \sin \theta \, d\theta\), using basic integration techniques - knowing that the antiderivative of \(\sin \theta\) is \(-\cos \theta\). This allows us to calculate it efficiently and find the exact area under the curve.

Trigonometric Functions

Trigonometric functions play a crucial role in polar coordinates because the radius \(r\) is often expressed as a function of \(\theta\). These functions, like \(\sin \theta\) and \(\cos \theta\), have properties that are especially useful when working with polar equations, because:

• They help describe periodic behaviors and rotations.
• They have well-defined integrals, which simplifies calculation processes.

In our example, \(r = \sqrt{\sin \theta}\) involves \(\sin \theta\). Squaring it simplifies directly to \(\sin \theta\), facilitating the integration step. Understanding the behavior of these trigonometric functions significantly aids in both visualizing the curve and solving for areas. These functions align with angles in a unit circle, providing a natural framework for polar coordinate systems.

Polar Equations

Polar equations define curves using a radius, \(r\), and an angle, \(\theta\), instead of the classic \(x\) and \(y\) coordinates. They enable descriptive efficiency for circular and spiral patterns often found in nature and applications.In polar form:

• The radius \(r\) is a function of \(\theta\).
• Equations may resemble \(r = f(\theta)\). This is how they describe the relationship between the radius and the angle.

For our specific case, the equation \(r = \sqrt{\sin \theta}\) indicates how the radius changes as \(\theta\) progresses from 0 to \(\pi\). Such equations often require conversion to Cartesian form for graph representation, but in computations for area, staying in polar form simplifies accessibility to the formula for polar area calculation.