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## Apostol 2.13 - #15 Cavalieri Solids (Volume Integration)

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 $y=\sqrt(2^2-x)$ 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 $(2y)^2\sqrt(3)/4 = y^2\sqrt(3)$. Substituting for y, we have that $Area(x)=(4-x^2)\sqrt{3}$. Integrating, we find that
$\int_{-2}^2 A(x) dx=2\int_0^2 \sqrt{3}(4-x^2) dx = \frac{32\sqrt{3}}{3}$

The problem is that the book has $\frac{16\sqrt{3}}{3}$, and I want to make sure I didn't do it incorrectly.

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Recognitions:      Homework  Help  Check the area of your triangular cross-sections again.  If the base is  2y  , what is the height?

Quote by dynamicsolo 					 				 				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 $\sqrt{(2y)^2-y^2}=y\sqrt{3}$, so the area of this right triangle is $\frac{1}{2}y \times y\sqrt{3}$; however this is just one half of the area of the equilateral triangle. Therefore the area of the equilateral triangle is $y^2\sqrt{3}$.

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

Taking $s=2y$, we have that the area is equal to $\frac{(2y)^2\sqrt{3}}{4}=y^2\sqrt{3}$.

Recognitions:
Homework Help

## Apostol 2.13 - #15 Cavalieri Solids (Volume Integration)

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%...

Quote by dynamicsolo 					 				 				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.

Recognitions:      Homework  Help  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...

Quote by dynamicsolo 					 				 				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.

Recognitions:      Homework  Help  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.

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