Concrete sections for a new building have the dimensions (in...

Name: Problem 88, Chapter 7: Calculus
Uploaded: 2021-10-04T16:48:29-08:00
Duration: 4 min 8 s
Channel: Uma Kumari
Description: Concrete sections for a new building have the dimensions (in meters) and shape shown in the figure. (a) Find the area of the face of the section superimposed on the rectangular coordinate system. (b) Find the volume of concrete in one of the sections by multiplying the area in part (a) by 2 meters. (c) One cubic meter of concrete weighs 5000 pounds. Find the weight of the section.

Concrete sections for a new building have the dimensions (in meters) and shape shown in the figure. (a) Find the area of the face of the section superimposed on the rectangular coordinate system. (b) Find the volume of concrete in one of the sections by multiplying the area in part (a) by 2 meters. (c) One cubic meter of concrete weighs 5000 pounds. Find the weight of the section.

Concrete sections for a new building have the dimensions (in meters) and shape shown in the figure.
(a) Find the area of the face of the section superimposed on the rectangular coordinate system.
(b) Find the volume of concrete in one of the sections by multiplying the area in part (a) by 2 meters.
(c) One cubic meter of concrete weighs 5000 pounds. Find the weight of the section.

Key Concepts

Unit Conversion and Material Density

Understanding unit conversion is critical in engineering, especially when translating calculated volumes into weights. When the density or weight per unit volume of a material is known, this relationship is used to determine the overall weight of a section, which is vital for assessing loads and ensuring structural safety.

Area Under a Curve

Calculating the area under a curve is a key concept in many engineering applications, including structural design. By integrating the function that describes the boundary, designers can accurately determine the size of complex cross-sectional regions, which is essential for material estimation and structural integrity assessments.

Definite Integration

Definite integration involves calculating the accumulation of quantities, such as area, over a specified interval. This concept is fundamental in determining the exact area under a curve when the bounds are defined, which is a common requirement in engineering designs and analysis.

Volume Calculation via Cross-sectional Area

In many design problems, the volume of a structure or material is determined by calculating the area of a cross section and then multiplying it by the length or thickness of the material. This method is widely used in building design to ensure that the volume of materials used is sufficient and correctly estimated for construction purposes.

Transcript

00:01 In this question we have a figure that represents the concrete section for a new building.

00:10 We are required to find the area of the face of the section superimposed on the rectangular coordinate system.

00:17 After that, we are required to find the volume of the concrete.

00:20 And finally, we are required to find the weight of the section.

00:25 So let's see how to solve this question.

00:28 Since area of the face of the section is symmetric about y -axis, so the area of the face of the section is symmetric about y -axis.

00:34 Phase of the section can be written as area a is equal to 2 into integration 0 to 5 1 minus 1 under root 5 minus x d x plus 2 into 0 .5 into 1 .5 into 1.

01:01 0 .5 into 1 is the area of the 1 side and since we are considering the two sides so we are multiplying with 2.

01:10 And we are considering this expression because y is equal to 1 upon 3 under root 5 minus x is not defined for x greater than 5.

01:30 Now let's solve this expression.

01:35 So we can write 2 into x minus 1 upon 3 into minus 2 upon 3 5 5 5.

01:45 Minus x to the power 3 by 2 and the limits are 0 to 5 and the value of 2 into 0 .5 into 1 will be equals to 1.

01:57 Now let's substitute the limits.

01:59 So we can write 2 into 5 plus 2 upon 9 into 0 to the power 3 by 2 minus 0 minus 2 by 9 into 5 to the power 3 by 2 plus 1...

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