Apostol 2.13 - #15 Cavalieri Solids (Volume Integration)

In summary, Apostol's Solid Geometry textbook has an area equation that is off by a factor of 1/8, and the volume equation is off by a factor of 1/4.
  • #1
process91
106
0
A solid has a circular base of radius 2. Each cross section cut by a plane perpendicular to a fixed diameter is an equilateral triangle. Compute the volume of the solid.

First, we find a way to define a the distance of a chord of the circle perpendicular to the fixed diameter. The equation [itex]y=\sqrt(2^2-x)[/itex] from x=-2 to 2 gives half the chord, so 2y is equal to the chord's length. At any point x, the solid's area is an equilateral triangle, so all sides must have length equal to the chord of the circle, or 2y. Now the area of an equilateral triangle with side 2y is equal to [itex](2y)^2\sqrt(3)/4 = y^2\sqrt(3)[/itex]. Substituting for y, we have that [itex]Area(x)=(4-x^2)\sqrt{3}[/itex]. Integrating, we find that
[itex]\int_{-2}^2 A(x) dx=2\int_0^2 \sqrt{3}(4-x^2) dx = \frac{32\sqrt{3}}{3}[/itex]

The problem is that the book has [itex]\frac{16\sqrt{3}}{3}[/itex], and I want to make sure I didn't do it incorrectly.
 
Physics news on Phys.org
  • #2
Check the area of your triangular cross-sections again. If the base is 2y , what is the height?
 
  • #3
dynamicsolo said:
Check the area of your triangular cross-sections again. If the base is 2y , what is the height?

I did it by dropping an angle bisector from the top vertex of the equilateral triangle to create two right triangles. Then the base is y, and the hypotenuse is 2y. The pythagorean theorem yields the height equal to [itex]\sqrt{(2y)^2-y^2}=y\sqrt{3}[/itex], so the area of this right triangle is [itex]\frac{1}{2}y \times y\sqrt{3}[/itex]; however this is just one half of the area of the equilateral triangle. Therefore the area of the equilateral triangle is [itex]y^2\sqrt{3}[/itex].

This agrees with the formula for the area of an equilateral triangle given here:
http://www.mathwords.com/a/area_equilateral_triangle.htm

Taking [itex]s=2y[/itex], we have that the area is equal to [itex]\frac{(2y)^2\sqrt{3}}{4}=y^2\sqrt{3}[/itex].
 
  • #4
Sorry, yes: my fault for trying to deal with more than one matter at once. I am wondering if the solver for Apostol used symmetry and forgot to double the volume integration. I am getting the same answer you are.

Stewart does this as Example 7 in Section 6.2 with a radius of 1 and gets one-eighth our volume, which is consistent. Back-of-the-book answers aren't 100%...
 
  • #5
dynamicsolo said:
Sorry, yes: my fault for trying to deal with more than one matter at once. I am wondering if the solver for Apostol used symmetry and forgot to double the volume integration. I am getting the same answer you are.

Stewart does this as Example 7 in Section 6.2 with a radius of 1 and gets one-eighth our volume, which is consistent. Back-of-the-book answers aren't 100%...

Glad to see that you're getting the same answer as me. I felt pretty solid about this one, but Apostol's answers in the back are better percentage-wise than any other book I've seen. I've done every problem through the first 200 pages or so and only come up with a few legitimate discrepancies.
 
  • #6
What edition is Apostol up to now? Generally, Third and later Editions have the error rates in the answer sections down to about 0.25% or less...
 
  • #7
dynamicsolo said:
What edition is Apostol up to now? Generally, Third and later Editions have the error rates in the answer sections down to about 0.25% or less...

The most recent edition is the second, and it's from the 1960s. I don't think any new ones will be out any time soon, but it's a really solid text.
 
  • #8
Well, it's supposed to be a classic. But I suspect the percentage of errors in the answers could be somewhere in the 0.25% to 0.5% range (from my long experience with textbooks)...

I looked Apostol up and he's 88 this year. I doubt he's going to revise the book (though I've been surprised in the past); he's moved on to other projects.
 

1. What is the purpose of studying Cavalieri solids in volume integration?

Studying Cavalieri solids in volume integration allows us to understand and calculate the volume of irregular 3-dimensional shapes. This is useful in various fields such as engineering, architecture, and physics.

2. What is the definition of a Cavalieri solid?

A Cavalieri solid is a 3-dimensional shape whose cross-sections have equal areas at every height. This means that the shape remains unchanged when sliced horizontally or vertically.

3. How do you calculate the volume of a Cavalieri solid?

The volume of a Cavalieri solid can be calculated by integrating the cross-sectional area function over the height of the solid. This process is known as volume integration and is based on the Cavalieri's principle.

4. What is the difference between Cavalieri solids and regular solids?

Cavalieri solids have varying cross-sectional areas, whereas regular solids have constant cross-sectional areas. This means that regular solids have a uniform shape throughout, while Cavalieri solids may have a more irregular shape.

5. How can Cavalieri solids be applied in real-life situations?

Cavalieri solids can be applied in real-life situations where the shape of an object is irregular and cannot be easily measured or calculated using traditional methods. This includes calculating the volume of complex structures like buildings, bridges, and even human organs.

Similar threads

  • Calculus and Beyond Homework Help
Replies
10
Views
289
  • Calculus and Beyond Homework Help
Replies
34
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
416
  • Calculus and Beyond Homework Help
Replies
27
Views
2K
  • Calculus and Beyond Homework Help
Replies
12
Views
2K
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
904
  • Calculus and Beyond Homework Help
Replies
4
Views
725
  • Calculus and Beyond Homework Help
Replies
4
Views
913
  • Calculus and Beyond Homework Help
Replies
20
Views
386
Back
Top