Question

For the following exercises, find the area of the regions bounded by the parametric curves and the indicated values of the parameter.\(x=2 \cot \theta, y=2 \sin ^{2} \theta, 0 \leq \theta \leq \pi\)

Short answer

The area of the region is \(4\pi\).

Step-by-step solution

Understand the parametric equations

We are given the parametric equations \(x = 2 \cot \theta\) and \(y = 2 \sin^2 \theta\). The parameter \(\theta\) falls within the range \(0 \leq \theta \leq \pi\). These equations describe the curve over which we are to find the area.

Set up the integral for the area

The area \(A\) under a parametric curve given by \(x(t)\) and \(y(t)\) from \(\theta=a\) to \(\theta=b\) is computed using the formula \[A = \int_a^b y(\theta) \cdot \frac{dx}{d\theta} \, d\theta\]. Identify \(y(\theta) = 2 \sin^2 \theta\) and compute \(\frac{dx}{d\theta}\).

Calculate the derivative \(\frac{dx}{d\theta}\)

Differentiate \(x = 2 \cot \theta\) with respect to \(\theta\). The derivative is \(\frac{dx}{d\theta} = -2 \csc^2 \theta\), because the derivative of \(\cot \theta\) is \(-\csc^2 \theta\).

Substitute in the integral formula

Substitute \(y(\theta) = 2 \sin^2 \theta\) and \(\frac{dx}{d\theta} = -2 \csc^2 \theta\) into the integral:\[A = \int_0^\pi 2 \sin^2 \theta \cdot (-2 \csc^2 \theta) \, d\theta = -4 \int_0^\pi \sin^2 \theta \cdot \csc^2 \theta \, d\theta = -4 \int_0^\pi \sin^2 \theta \cdot \frac{1}{\sin^2 \theta} \, d\theta\].

Simplify the integral expression

The integral simplifies to \(-4 \int_0^\pi 1 \, d\theta\) because \(\sin^2 \theta \cdot \frac{1}{\sin^2 \theta} = 1\).

Integrate the simplified expression

The integral of 1 with respect to \(\theta\) over the interval \(0\) to \(\pi\) is simply the length of the interval: \(-4 \left[ \theta \right]_0^\pi = -4 (\pi - 0) = -4\pi\).

Interpret the result as area

Since area cannot be negative, the negative sign indicates the direction of integration was opposite to the normal positive direction. Therefore, the absolute value is taken: \(|-4\pi| = 4\pi\).

Key concepts

Area Under Curve

Finding the area under a curve is like trying to calculate the "space" that lies beneath it on a graph. When dealing with parametric curves, this requires an additional step because the curve is defined using equations dependent on a parameter, usually denoted by \( t \) or \( \theta \).

For a curve described by parametric equations \( x(t) \) and \( y(t) \), the area under the curve between two parameters, \( a \) and \( b \), is given by the integral

• \( A = \int_a^b y(t) \cdot \frac{dx}{dt} \, dt \)

This formula uses the product of \( y(t) \) with the derivative \( \frac{dx}{dt} \), which gives us the necessary dimensions to find the area.

Just like geometry involves shapes and the spaces inside them, finding the area under a parametric curve helps understand the "shape" created by the equations.

Integration

Integration is the process of summing or finding the total area under a curve. In calculus, it works like adding up many small pieces to find a whole.

When integrating parametric equations, it's important to match the parameters correctly. For example, using the correct limits of integration ensures accurate results.

• The integral \( \int y(t) \cdot \frac{dx}{dt} \, dt \) captures all small "strips" of area between the curve and the \( x \)-axis.

This method works for parametric curves using the setup from parametric equations.

The beauty of integration lies in its ability to connect small changes into a big picture. It's a powerful tool to give meaning to curves and lines drawn on a graph.

Parametric Equations

Parametric equations offer a different way to express curves by using a separate variable, like \( \theta \), to define both \( x \) and \( y \) coordinates.

In situations where equations aren't straightforward, parametric form simplifies the description. A classic example is the circle: rather than using the traditional \( x^2 + y^2 = r^2 \), you can use \( x = r \cos \theta \) and \( y = r \sin \theta \).

• They provide flexibility to represent complex motions and shapes.
• Introduce a third variable, making plotting viable even if \( x \) and \( y \) don’t easily relate in basic equation form.

In our example problems, we use \( \theta \) as the parameter, defining \( x \) and \( y \)'s behavior over a specified range.

These equations encapsulate the flow across a curve, making apparent the changes in both \( x \) and \( y \) as \( \theta \) varies.

Derivatives

Derivatives tell us how something changes. In the context of curves, a derivative provides the rate of change or the slope of the curve at any point.

For parametric curves, finding the derivative is crucial because it tells us how \( x \) changes with respect to the parameter. In our exercise, the derivative \( \frac{dx}{d\theta} \) is computed to help form the area integral.

• Knowing \( \frac{dx}{dt} \) allows us to account for changes in \( x \) concerning the entire curve described by a parameter.

The chain rule often helps here, as it generates derivatives for terms dependent on a third variable, such as \( \theta \).

The idea of derivatives extends beyond just slopes - in calculus, it signifies motion and predicts behavior, all through observing how one function responds as another changes.