The Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

Volume of a Solid

Date: 10/02/97 at 12:40:22
From: Brent M. Rasmussen
Subject: Calculus

Dear Dr. Math,

How would I go about solving the following problem:

The base of a solid is the region inside the circle x^2 + y^2 = 4.  
Find the volume of the solid if every cross section by a plane 
perpendicular to the x-axis is a square.

I would appreciate it if you could run me through the step by step 



Date: 10/10/97 at 00:00:06
From: Doctor Chita
Subject: Re: Calculus

Hi Brent:

There are several steps in this problem. The first thing to do is to 
make a sketch. 


Imagine that you are looking down on the x-y plane from above. 
The circle is the base of the solid, and squares perpendicular to the 
x-axis look like segments. One such square is shown in the sketch. 
Now for the solution:

(1) The center-radius equation of a circle is 

    (x - h)^2 + (y - k)^2 = r^2, 

    where h and k are the coordinates of the center of the circle, 
    and r is the radius. In the equation 

    x^2 + y^2 = 4, (h, k) = (0, 0), and r = 2. 

(2) The solid in the problem is composed of an infinite number of 
    squares perpendicular to the x-axis. The set of squares starts at 
    one end of the diameter at (-2, 0) and end at (2, 0). Each square 
    has a side s and an area of s^2, where s is the vertical distance 
    between two points on the circle. 

    From the equation of the circle,  y = |sqrt(4 - x^2)| and s = 2y. 
    The area of each square is 4y^2 or 4(4 - x^2).

(3) Now for the calculus: Each cross-section has thickness of delta-x. 
    If we let this thickness approach 0, then the volume of the solid 
    made up of this set of "infinitely thin" squares is equal to the 
    sum of the areas of the squares. This is equal to the sum of an 
    infinite number of areas, 4(4 - x^2) from -2 to 2. In other words, 
    the volume can be calculated by evaluating  the integral from -2 
    to 2:

    V = integral[4(4 - x^2)]dx

    V = 4 * integral(4 - x^2)dx

(4) Integrate the expression to get:

    V = 4 * [4x - x^3/3] from -2 to 2

(5) If V = F(x), evaluate the definite integral as F(b) - F(a), where 
    a and b are the limits of the integral and simplify:

    V = 4{[8 - (8)/3] - [-8 - (-8)/3]}

    V = 128/3

Hope this helps explain how to do this problem. A picture is worth a 
thousand words, however.

-Doctor Chita,  The Math Forum
 Check out our web site!   
Associated Topics:
High School Calculus
High School Geometry
High School Higher-Dimensional Geometry

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.