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### Volume of a Solid

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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
process.

Thanks,

Brent
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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!  http://mathforum.org/dr.math/
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Associated Topics:
High School Calculus
High School Geometry
High School Higher-Dimensional Geometry

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