Applications of the Definite Integral- Volume

In summary, the problem involves finding the volume of a pyramid with a 30 feet tall steeple as the height and square cross sections with side lengths of 3 feet at the base and 6 inches at the top. A general fact about pyramid shapes is used to solve the problem. The volume can be calculated by taking the difference of two shapes formed by shrinking the base and integrating.
  • #1
ashleyk
22
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Hi, I am having trouble visualizing this problem, if anyone can help me see it, I know how to do the integral part.

A church steeple is 30 feet tall with square cross sections. The square at the base has side 3 feet, the square at the top has side 6 inches, and the sides varies linearly in between. Compute the volume.

I assume the 3 feet and then 6 inches has to be in the same form so it would 36 inches and 6 inches. I also know that volume for a square would be length times width times height or one side raised to the 3rd power. I am just not sure to start.
 
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  • #2
A general fact about "pyramid" shapes, ie, shapes formed by starting with a base of a certain (2D) shape and sweeping out a volume by translating the shape and simultaneously linearly shrinking it to a point, is that their volume is 1/3 A*h, where A is the area of the base and h is the distance from the plane of the base to the tip. To prove this, note that as the side lengths linearly shrink, the area of the cross sections shrink quadratically, and integrating this out gives the result. In this problem, just take the difference of two such shapes.
 

What is the definite integral and how is it related to volume?

The definite integral is a mathematical concept used to find the area under a curve on a given interval. In the context of volume, the definite integral is used to find the volume of a three-dimensional object by calculating the area of cross-sections at various points along the object's length and adding them up.

What are some real-life applications of the definite integral in calculating volume?

The definite integral is commonly used in engineering, physics, and architecture to calculate the volume of objects such as tanks, pipes, and buildings. It is also used in fluid mechanics to determine the volume of liquids and gases in containers.

Can the definite integral be used to find volumes of irregularly shaped objects?

Yes, the definite integral can be used to find the volume of any three-dimensional object, regardless of its shape. This is because the integral allows for the calculation of the area of infinitely thin slices, making it suitable for calculating the volume of even the most irregular shapes.

What is the difference between using the definite integral and other methods, such as the disk or washer method, to calculate volume?

The disk and washer methods are specific techniques used to find the volume of objects with circular cross-sections, while the definite integral can be used for any shape. The disk and washer methods are often simpler and more straightforward to use, but the definite integral allows for more flexibility in calculating volumes of more complex objects.

Are there any limitations to using the definite integral to calculate volume?

One limitation of using the definite integral to calculate volume is that it requires a known function or equation to represent the object. If the object's shape is not defined by a function, other methods may need to be used. Additionally, the definite integral may be more time-consuming to calculate compared to other methods for certain shapes.

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