Question

Draw the solid region whose volume is given by the following double integrals. Then find the volume of the solid. $$\int_{0}^{6} \int_{1}^{2} 10 d y d x$$

Short answer

Based on the given double integral, the solid region is a rectangular prism with dimensions 6 in the x-direction, 1 in the y-direction, and a constant height of 10 in the z-direction. The volume of this solid region is computed by evaluating the double integral, resulting in a volume of 60 cubic units.

Step-by-step solution

Identify the solid's region

A double integral, such as the one given in the exercise, can be interpreted as a volume if the integrand represents the "height" of the solid. In this case, the integrand is 10, which represents a constant height. The limits of integration, \(x\) from 0 to 6 and \(y\) from 1 to 2, represent the bounds of the solid in the \(xy\)-plane. So we can deduce the following points as vertices: - (0,1) - (0,2) - (6,1) - (6,2) Since our function is a constant, the solid region is a rectangular prism with the following dimensions: - \(x\)-length: 6 - \(y\)-width: 2-1 = 1 - \(z\)-height: 10 Now let's find the volume of this solid region.

Compute the volume of the solid

Given the equation from the exercise, we can compute the volume of the solid by evaluating the double integral: $$\int_{0}^{6} \int_{1}^{2} 10 dy dx$$ First, integrate with respect to \(y\): $$\int_{0}^{6} \left[10y\bigg|_{1}^{2}\right] dx$$ Evaluate the inner integral: $$\int_{0}^{6} (10\cdot2 - 10\cdot1) dx$$ Simplify: $$\int_{0}^{6} 10 dx$$ Next, integrate with respect to \(x\): $$10\left[x\bigg|_{0}^{6}\right]$$ $$10(6-0)$$ Finally, compute the volume: $$\text{Volume} = 10\cdot6 = 60$$ Thus, the volume of the given solid region is 60 cubic units.

Key concepts

Volume of a Solid

Understanding the volume of a solid is crucial in many mathematical and real-world applications. A solid's volume refers to the amount of three-dimensional space it occupies. When dealing with geometry, volume can be calculated using various methods depending on the shape of the solid. In the case of a double integral, we often interpret the integral as the volume beneath a surface described by the integrand and above a certain region in the plane.

In the exercise provided, the volume of the solid is evaluated using a double integral with a constant function as the integrand. This implies that the solid has a uniform height across the entire region. Recognizing this allows us to simplify the problem significantly by linking it to familiar geometric shapes, like the rectangular prism, making the volume calculation straightforward.

Rectangular Prism

A rectangular prism is a three-dimensional shape with six faces, all of which are rectangles. It is an important geometric shape as it represents the simplest form of a box, a structure we often encounter in the real world. In the exercise context, the double integral simplifies to finding the volume of a rectangular prism.

Once you've identified the region and the constant height from the integral, the solid can be visualized as a prism with:

• An "x" length of 6 units (from the limits of integration on the x-axis).
• A "y" width of 1 unit (from the limits of integration on the y-axis).
• A "z" height determined by the integrand, here being 10 units.

Understanding that the solid is a rectangular prism allows you to use a straightforward approach to calculate its volume as "length × width × height".

Integral Evaluation

Integral evaluation is a process used for calculating the area, volume, or another specific measure by integrating a function over a certain interval or region. For double integrals, it's about integrating a function over a two-dimensional region. In this example, the double integral evaluates to the volume of a solid.

The step-by-step process involves integrating the given function first with respect to one variable (here "y"), fixing the other variable as a constant. Upon completion, you then proceed to integrate the resulting function with respect to the second variable ("x" in this instance). This systematic approach allows us to accurately determine the total measure (like volume) over the specified range.

Bounds of Integration

The bounds of integration play a crucial role in defining the region over which the integration takes place. They essentially frame the limits for the variable being integrated and define the extent of the region under consideration.

For the given problem, the bounds for "x" are from 0 to 6 and for "y", from 1 to 2. These bounds define a rectangular area in the "xy" plane. When paired with the constant height given by the integrand, these bounds allow us to visualize and calculate the volume of the rectangular prism by integrating over this defined region.

Understanding how to set and interpret these bounds is essential for correctly solving double integrals, particularly when they are used for calculating volumes or other spatial properties.