Volume problem: the base of a solid is a circular disk w/...

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ugh... this problem is really confusing me, im not sure how to start it. its..

The base of a solid is a circular disk with diameter AB of length 2a. Find the volume of the solid if each cross section perpendicular to AB is an equilater triangle.

Is the answer just 4/3*radical 3*a^3 ? or am i making it way too easy..?
 
Let's center the disk at the origin.

Since its diameter is 2a, its radius is a.

That gives us a circle equation of \(\displaystyle y=\sqrt{a^{2}-x^{2}\)

Let A(x) be the area of the equilateral triangle whose sides are length 2y



The area of the equilateral triangle is given by \(\displaystyle A(x)=\sqrt{3}{4}(2y)^{2}=\sqrt{3}y^{2}=\sqrt{3}(a^{2}-x^{2})\)

Now, we can integrate from -a to a.

\(\displaystyle \L\\\int_{-a}^{a}\sqrt{3}(a^{2}-x^{2})dx=2\sqrt{3}\int_{0}^{a}(a^{2}-x^{2})dx\\

=\frac{4a^{3}\sqrt{3}}{3}\)


HEY, that's what you got.
 
Re: Volume problem

Hello, aswimmer113!

The base of a solid is a circular disk with diameter AB of length 2a.
Find the volume of the solid if each cross section perpendicular to AB is an equilateral triangle.

Is the answer just \(\displaystyle \frac{4\sqrt{3}}{3}a^3\) . . . or am i making it way too easy?
Click to expand...
Your answer is correct!
Unless you got it by pure luck, you did a lot of work.
\(\displaystyle \;\;\)I doubt that you made it "way too easy".
Code: 
                 a|
                * * *   C
            *     |     *
          *       |     : *
         *        |     :  *
                  |     :y
        *         |     :   *
      A *---------+-----+---* B
      -a*         |     :   *a
                  |     :y
         *        |     :  *
          *       |     : *
            *     |     *
                * * *   D
                -a|
The equation of the circle is: \(\displaystyle \,x^2 + y^2\:=\:a^2\;\;\Rightarrow\;\;y\:=\:\sqrt{a^2\,-\,x^2}\)

The base of the equilateral triangle is: \(\displaystyle \,CD \:= \: 2y\:=\:2\sqrt{a^2\,-\,x^2}\)

The area of an equilateral triangle of side \(\displaystyle s\) is: \(\displaystyle \,A\:=\:\frac{\sqrt{3}}{4}s^2\)

Our triangle has area: \(\displaystyle \,A\;=\;\frac{\sqrt{3}}{4}(2\sqrt{a^2\,-\,x^2})^2\;=\;\frac{\sqrt{3}}{4}\cdot4(a^2\,-\,x^2)\;=\;\sqrt{3}(a^2\,-\,x^2)\)


The volume of the solid is: \(\displaystyle \L\,V\;=\;\int^{\;\;\;a}_{-a}\sqrt{3}(a^2\,-\,x^2)\,dx \;=\;2\,\times\,\sqrt{3}\int^{\;\;\;a}_0(a^2\,-\,x^2)\,dx\)

And we get: \(\displaystyle \L\,V \;= \;2\sqrt{3}\left[a^2x\,-\,\frac{1}{3}x^3\right]^a_0 \;= \;2\sqrt{3}\left[\left(a^3\,-\,\frac{1}{3}a^3\right)\,-\,(0\,-\,0)\right]\)

Therefore: \(\displaystyle \L\,V\;=\;\frac{4\sqrt{3}}{3}a^3\)


Edit: Too fast for me, Galactus!
 
yay! im beginning to depise this stuff a little bit less, thanks!
 
Re: Volume problem: the base of a solid is a circular disk w

aswimmer113 said:
ugh... this problem is really confusing me, im not sure how to start it. its..

The base of a solid is a circular disk with diameter AB of length 2a. Find the volume of the solid if each cross section perpendicular to AB is an equilater triangle.

Is the answer just 4/3*radical 3*a^3 ? or am i making it way too easy..?
Click to expand...
From your description of the solid you give, I envision a right circular cone with apex angle of 60º. If this is not the case, please let me know.

Dealing with one half the equilateral triangle vertical cross section, the base radius = "a", the upper vertex angle is 30º, the hypotenuse of the half triangle is 2a, and the height is sqrt(3).

The centroid of the triangle is a/3 away from the vertical line through the apex.

The area of the triangle is A = a^2sqrt(3)/2.

Rotating the triangle about the vertical height line with radius of a/3 sweeps out and defines the volume of the cone.

The distance the triangle's centroid travels is C = 2aPi/3.

The volume is therefore defined as V = a^2sqrt(3)/2[2aPi/3] = a^3Pi/sqrt(3).

The volume of a right circular cone is giiven as V = a^2Pi(h)/3 in any text book.

We have h = (a)sqrt(3).

Substituting, we again get V = a^2Pi(a)sqrt(3)/3 = a^3Pi/sqrt(3).

Using either the text book formula or the one derived by sweeping the area's centroid through 360º to derive the volume, we get the same result.

If I have made an incorrect assumption, please do correct me.
 
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