Question

Find the area of the region bounded by \(x=2 \sin ^{2} \theta, y=2 \sin ^{2} \theta \tan \theta\), for \(0 \leq \theta \leq \frac{\pi}{2}\).

Short answer

The area is \( \frac{8}{3} \).

Step-by-step solution

Identify the Parametric Equations

The curve is given by the parametric equations:\[ x = 2 \sin^2 \theta, \quad y = 2 \sin^2 \theta \tan \theta \]for \( 0 \leq \theta \leq \frac{\pi}{2} \). Here \( \theta \) is the parameter.

Set Up the Integral for Area

The area \( A \) under a parametric curve given by \( x = f(t) \) and \( y = g(t) \) is: \[ A = \int_{\alpha}^{\beta} y \frac{dx}{d\theta} \, d\theta \]Substitute \( y = 2 \sin^2 \theta \tan \theta \) and find \( \frac{dx}{d\theta} \).

Differentiate x with Respect to θ

To find \( \frac{dx}{d\theta} \), differentiate \( x = 2 \sin^2 \theta \):\[ \frac{dx}{d\theta} = 4 \sin \theta \cos \theta = 2 \sin 2\theta \]

Substitute into the Area Integral

Substitute \( y = 2 \sin^2 \theta \tan \theta \) and \( \frac{dx}{d\theta} = 2 \sin 2\theta \) into the area formula:\[ A = \int_{0}^{\frac{\pi}{2}} \left( 2 \sin^2 \theta \tan \theta \right)(2 \sin 2\theta) \, d\theta \]

Simplify the Integral Expression

Simplify the integral:- \( 2 \sin^2 \theta \tan \theta = 2 \sin \theta \sec \theta \sin \theta \)- Substitute \( \sin 2\theta = 2 \sin \theta \cos \theta \)Now we have:\[ A = \int_{0}^{\frac{\pi}{2}} 8 \sin^3 \theta \cos \theta \sec \theta \, d\theta \]

Further Simplify and Perform Integration

Simplify the integrand: \( 8 \sin^3 \theta \cos \theta \sec \theta = 8 \sin^3 \theta \)Now perform the integration:\[ A = 8 \int_{0}^{\frac{\pi}{2}} \sin^3 \theta \, d\theta \]Use the identity \( \sin^3 \theta = \sin \theta (1 - \cos^2 \theta) \) to facilitate integration.

Solve the Integral

Given \( \int \sin^3 \theta \, d\theta = \int \sin \theta (1 - \cos^2 \theta) \, d\theta \), distribute:\[ \int \sin \theta \cos^2 \theta \, d\theta = \int \sin \theta \, d\theta - \int \sin \theta \cos^2 \theta \, d\theta \]Solve using substitution \( u = \cos \theta \), \( du = -\sin \theta \, d\theta \).

Evaluate the Integral

After substitution:\[ \int \sin \theta (1 - \cos^2 \theta) \, d\theta = -\cos \theta - \frac{1}{3}\cos^3\theta \]Evaluate from \( 0 \) to \( \frac{\pi}{2} \):\[ 8 \left[-\cos \theta - \frac{1}{3}\cos^3\theta \right]_{0}^{\frac{\pi}{2}} = 8 \left[0 + \frac{1}{3}(0)\right] - 8\left[-1 - 0\right] = \frac{8}{3} \]

Conclude the Solution

The final result is:The area bounded by the given parametric equations from \( \theta = 0 \) to \( \theta = \frac{\pi}{2} \) is \( \frac{8}{3} \).

Key concepts

Integral Calculus

Integral calculus is a fascinating area of mathematics dealing with the accumulation of quantities. Unlike differential calculus, which focuses on rates of change, integral calculus seeks to understand how quantities accumulate. For example, it helps us find areas under curves, volumes of shapes, and many other physical applications. An integral is essentially the sum of infinite tiny pieces that, when combined, give us the whole.
In this exercise, we use integral calculus to find the area of a region defined by parametric equations. When dealing with parametric curves, the formula for the area under a curve is slightly modified compared to the area under a typical function graph. Here, we need to integrate the product of the function describing the y-component and the derivative of the x-component, symbolically expressed as \[ A = \int_{\alpha}^{\beta} y \frac{dx}{d\theta} \, d\theta \].

This approach is necessary because the curve is traced in terms of a parameter, \( \theta \), which represents angles in this particular problem. Therefore, it helps us better analyze curves and problems that are not easily expressed with a single equation for y in terms of x.

Parametric Curves

Parametric curves are curves that are expressed through parameters—typically using a third variable, often \( t \) or \( \theta \), which separately defines the x and y coordinates. This provides flexibility, allowing curves that might be complex or impossible to express as single functions y = f(x).

• A parametric form allows the x and y components to be tied together by a parameter, offering a more comprehensive range of curves.
• It is particularly useful for modeling scenarios where one variable depends on another variable that doesn't fit traditional function forms.

In our exercise, the curves are represented by the parametric equations \( x = 2 \sin^2 \theta \) and \( y = 2 \sin^2 \theta \tan \theta \). Each value of \( \theta \) traces out a point, creating a continuous path or curve in the coordinate system as \( \theta \) varies from 0 to \( \frac{\pi}{2} \).

Using this method, unique and complex shapes can be explored without sticking to a single Cartesian equation. This makes parametric curves a powerful tool in both mathematics and related fields like physics and engineering.

Trigonometric Integration

Trigonometric integration involves integrating functions that contain trigonometric terms like sine, cosine, and tangent. Tackling these integrals requires familiarity with trigonometric identities and substitution methods to simplify the expressions.

The integral in our exercise, for example, simplifies to \( \int_{0}^{\frac{\pi}{2}} 8 \sin^3 \theta \, d\theta \). Here, the approach to solve it includes using identities like \( \sin^2 \theta = 1 - \cos^2 \theta \) to convert the integral into a simpler form.

• First, express powers of sine in terms of basic trigonometric identities.
• Then, use substitution where possible to change the variable of integration and simplify the integral's complexity.

In this process, substituting \( u = \cos \theta \) can help. This substitution simplifies the integrals into polynomials or functions that are easier to handle.

Understanding trigonometric integration is crucial as it appears in areas like physics, engineering, and geometry—not just pure mathematics. It’s about recognizing patterns and identities to make tough integrals look more like familiar ones.