πŸ“š

Β >Β 

♾️ 

Β >Β 

🐢

8.7 Volumes with Cross Sections: Squares and Rectangles

5 min readβ€’june 18, 2024


AP Calculus AB/BC ♾️

279Β resources
See Units

Square and Rectangular Cross Sections

Volumes with cross sections can be a tricky concept to understand, but it is an important one for understanding calculus. In this guide, we will focus on how to find the volume of a solid with square and rectangular cross sections. First, let's define what is meant by a solid with cross sections. A solid with cross sections is a three-dimensional object that can be divided into many smaller two-dimensional shapes, such as squares or rectangles. The volume of the solid is found by adding up the volumes of all of these smaller shapes. To find the volume of a solid with square or rectangular cross sections, we will use the method of integration. This method involves breaking the solid into many small slices, each with a square or rectangle cross section, and then finding the volume of each slice. The total volume of the solid is found by adding up the volumes of all of these slices. Here are the steps to find the volume of a solid with square or rectangular cross sections: Determine the function that describes the cross section of the solid. This function will be in the form of y = f(x) for square cross sections and y = f(x) and x = g(y) for rectangular cross sections. Determine the limits of integration. These limits will be the range of x or y values for which the cross section is defined. Find the area of one cross section by plugging in the limits of integration into the function for the cross section. Use the definite integral to find the total volume of the solid by integrating the function for the cross section with respect to x or y and then multiplying by the width or height of the slice.

Here are a few examples:

Example 1:
Consider a solid with square cross sections, where the function describing the cross section is y = 2x. The limits of integration are from x = 0 to x = 2. To find the area of one cross section, we plug in x = 0 and x = 2 into the function and find that the area is 4. To find the total volume of the solid, we integrate the function squared (to account for both height and length!) with respect to x and multiply by the width of the slice, which is dx. This gives us the definite integral from 0 to 2 of (2x)^2 dx, or 4x^3/3 over 0 to 2, which equals 32/3.
Example 2:
Consider a solid with rectangular cross sections, where the function describing the cross section is y = 2x and x = 2y. The limits of integration are from y = 0 to y = 1. To find the area of one cross section, we plug in y = 0 and y = 1 into the function and find that the area is 2. To find the total volume of the solid, we integrate the function with respect to y and multiply by the height of the slice, which is dx. This gives us the definite integral from 0 to 1 of 2x dy, which equals 2.
Example 3:
Consider a solid with square cross sections, where the function describing the cross section is y = x^2. The limits of integration are from x = -1 to x = 1. To find the area of one cross section, we plug in x = -1 and x = 1 into the function and find that the area is 2. To find the total volume of the solid, we integrate the function with respect to x and multiply by the width of the slice, which is dx. This gives us the definite integral from -1 to 1 of x^2 dx, which equals [x^3/3] from -1 to 1 which is equal to 0.
Example 4:
Consider a solid with rectangular cross sections, where the function describing the cross section is y = x^2 and x = y. The limits of integration are from y = 0 to y = 1. To find the area of one cross section, we plug in y = 0 and y = 1 into the function and find that the area is 1/3. To find the total volume of the solid, we integrate the function with respect to y and multiply by the height of the slice, which is dx. This gives us the definite integral from 0 to 1 of x^2 dy, which equals [x^3/3] from 0 to 1 which is equal to 1/3.
Example 5:
Consider a solid with square cross sections, where the function describing the cross section is y = sqrt(x). The limits of integration are from x = 0 to x = 1. To find the area of one cross section, we plug in x = 0 and x = 1 into the function and find that the area is 2/3. To find the total volume of the solid, we integrate the function with respect to x and multiply by the width of the slice, which is dx. This gives us the definite integral from 0 to 1 of sqrt(x) dx, which equals [2/3*x^(3/2)] from 0 to 1 which is equal to 2/3.
It's important to note that the method of integration can be used to find the volume of solids with cross sections of other shapes as well, such as circles or triangles. The key is to find the area of one cross section and then use the definite integral to find the total volume of the solid.
In conclusion, finding the volume of a solid with square and rectangular cross sections involves using the method of integration. This method involves determining the function that describes the cross section, determining the limits of integration, finding the area of one cross section, and then using the definite integral to find the total volume of the solid. With practice, you will be able to find volumes of solids with cross sections of different shapes.

Browse Study Guides By Unit
πŸ‘‘Unit 1 – Limits & Continuity
πŸ€“Unit 2 – Fundamentals of Differentiation
πŸ€™πŸ½Unit 3 – Composite, Implicit, & Inverse Functions
πŸ‘€Unit 4 – Contextual Applications of Differentiation
✨Unit 5 – Analytical Applications of Differentiation
πŸ”₯Unit 6 – Integration & Accumulation of Change
πŸ’ŽUnit 7 – Differential Equations
🐢Unit 8 – Applications of Integration
πŸ¦–Unit 9 – Parametric Equations, Polar Coordinates, & Vector-Valued Functions (BC Only)
β™ΎUnit 10 – Infinite Sequences & Series (BC Only)
πŸ“šStudy Tools
πŸ€”Exam Skills

Fiveable
Fiveable
Home
Stay Connected

Β© 2024 Fiveable Inc. All rights reserved.


Β© 2024 Fiveable Inc. All rights reserved.