The figure shows a surface z = 5(x2 + y2) and a rectangle R in the xy-plane. z = 5(x2 + y?) R (a) Set up an iterated integral for the volume of the solid that lies under the surface and above R. Ap xp (b) Evaluate the iterated integral to find the volume of the solid.
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.
![**Finding the Volume Under a Surface Using Iterated Integrals**
The figure in this example shows a surface defined by the equation \( z = 5(x^2 + y^2) \) and a rectangle \( R \) in the xy-plane. The surface is a paraboloid opening upwards, and the rectangle \( R \) is situated in the plane defined by \( 1 \leq x \leq 2 \) and \( 1 \leq y \leq 3 \).
### Step-by-Step Process
#### (a) Setting Up the Iterated Integral
To find the volume of the solid that lies under the surface \( z = 5(x^2 + y^2) \) and above the rectangle \( R \), we need to set up an iterated integral. The volume \( V \) can be found by:
\[ V = \int_{1}^{2} \int_{1}^{3} 5(x^2 + y^2) \, dy \, dx \]
Here, the limits of integration for \( x \) are from 1 to 2, and for \( y \) are from 1 to 3. The function \( 5(x^2 + y^2) \) describes the height of the surface above the rectangle.
#### (b) Evaluating the Iterated Integral
Next, we need to evaluate the iterated integral to find the volume of the solid.
1. Integrate with respect to \( y \):
\[ \int_{1}^{3} 5(x^2 + y^2) \, dy = 5 \left[ x^2y + \frac{y^3}{3} \right]_{1}^{3} \]
\[ = 5 \left( x^2(3) + \frac{3^3}{3} - x^2(1) - \frac{1^3}{3} \right) \]
\[ = 5 \left( 3x^2 + 9 - x^2 - \frac{1}{3} \right) \]
\[ = 5 \left( 2x^2 + \frac{26}{3} \right) \]
\[ = \frac{10x^2}{3} + \frac{130}{3} \]
2. Integrate this result with respect to \( x \):
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