Sketch the solid region in space whose volume is given by the double integral (do not evaluate the integral): 2 sin e (4 – r sin 0) r dr d0 .
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.
![## Educational Exercise: Visualizing the Solid Region Defined by a Double Integral
### Task:
Sketch the solid region in space whose volume is defined by the following double integral. *Important note: Do not evaluate the integral.*
### Given Integral:
\[
\int_{0}^{\pi} \int_{0}^{2 \sin \theta} (4 - r \sin \theta) \, r \, dr \, d\theta
\]
### Interpretation:
1. **Integral Bounds Analysis:**
- The outer integral, \(\int_{0}^{\pi} d\theta\), suggests that \(\theta\) varies from \(0\) to \(\pi\), covering the top half of a polar coordinate system, sweeping from right to left.
- The inner integral, \(\int_{0}^{2 \sin \theta} dr\), indicates that the radius \(r\) varies from \(0\) to \(2 \sin \theta\).
2. **Function within the Integral:**
- The expression \((4 - r \sin \theta)\) represents a surface in polar coordinates. The term modifies how the height of the region changes with respect to \(r\) and \(\theta\).
3. **Shape and Geometry:**
- The polar coordinate form implies a rotationally symmetric shape about the vertical (\(z\)-axis) within the angle interval \([0, \pi]\).
- The limits and expression indicate constructing a shape that evolves as \(\theta\) sweeps around the specified angular extent. The bounding function \(r = 2 \sin \theta\) creates a semi-circular boundary in the \(xy\)-plane when projected.
### Sketching Approach:
- Sketch the \(xy\)-plane region by plotting \(r = 2 \sin \theta\) for \(\theta \in [0, \pi]\), which results in a semi-circle of radius 2, centered at the origin.
- The function \((4 - r \sin \theta) \, r\) determines the height at each point within this semi-disk, which provides a varying height across the semi-circular region, decreasing outward.
This integral provides an exercise in understanding how regions and volumes can be defined and visualized using polar coordinates and double integrals.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fffc06da2-7bd8-42a3-b8a1-3f8b831a8034%2F0f60af9f-0043-4e7e-ad6f-00dad5751c99%2Focuosw0i_processed.png&w=3840&q=75)
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