Optimization
Optimization comes from the same root as "optimal". "Optimal" means the highest. When you do the optimization process, that is when you are "making it best" to maximize everything and to achieve optimal results, a set of parameters is the base for the selection of the best element for a given system.
Integration
Integration means to sum the things. In mathematics, it is the branch of Calculus which is used to find the area under the curve. The operation subtraction is the inverse of addition, division is the inverse of multiplication. In the same way, integration and differentiation are inverse operators. Differential equations give a relation between a function and its derivative.
Application of Integration
In mathematics, the process of integration is used to compute complex area related problems. With the application of integration, solving area related problems, whether they are a curve, or a curve between lines, can be done easily.
Volume
In mathematics, we describe the term volume as a quantity that can express the total space that an object occupies at any point in time. Usually, volumes can only be calculated for 3-dimensional objects. By 3-dimensional or 3D objects, we mean objects that have length, breadth, and height (or depth).
Area
Area refers to the amount of space a figure encloses and the number of square units that cover a shape. It is two-dimensional and is measured in square units.
![### Problem Statement:
**Question:**
Find the area of the region bounded below by the x-axis and above by the curve
\[ x = 2 \sin^2(\theta), \; y = 3 \sin^2(\theta) \tan(\theta) \; \text{with} \; 0 \leq \theta \leq \frac{\pi}{2}. \]
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**Explanation:**
To find the area of the region described by the given parametric equations and interval, we need to employ techniques involving integrals and parametric forms. Here is a structured approach:
1. **Set up the integral:**
- Identify the parametric equations for \( x \) and \( y \):
\[ x = 2 \sin^2(\theta) \]
\[ y = 3 \sin^2(\theta) \tan(\theta) \]
2. **Use the given interval:**
- The bounds for \(\theta\) are:
\[ 0 \leq \theta \leq \frac{\pi}{2} \]
3. **Determine the form of the area integral:**
- The area \(A\) under the curve can be expressed in terms of \(\theta\) using the following integral form for parametric equations:
\[ A = \int_{\alpha}^{\beta} y \frac{dx}{d\theta} \, d\theta \]
4. **Calculate the derivative \(\frac{dx}{d\theta}\):**
- Compute \(\frac{dx}{d\theta}\):
\[ x = 2 \sin^2(\theta) \]
Using the chain rule,
\[ \frac{dx}{d\theta} = 4 \sin(\theta) \cos(\theta) \]
Simplify using the double angle identity \(\sin(2\theta) = 2 \sin(\theta) \cos(\theta)\):
\[ \frac{dx}{d\theta} = 2 \sin(2\theta) \]
5. **Set up the integral for the area:**
- Substitute \( y \) and \(\frac{dx}{d\theta} \) into the integral:
\[ A = \int_{0}^{\frac{\pi}{2}} 3 \sin^2(\](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fee6a40b4-eeaf-409d-8d6e-00937c376b0e%2F1bb0cdcd-7dbb-4b2f-9c16-a32ecc71d031%2Fokxq6cj_processed.jpeg&w=3840&q=75)
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