Extreme confusion with volume Calculus

[flyingpig]

Homework Statement

Find the volume of a solid with a circular base defined by $$x^2 + y^2 = 1$$ with parallel cross sections perpendicular to the base are equilateral triangles.


The Solution

$$A(x) = \frac{1}{2} \cdot 2\sqrt{1-x^2} * \sqrt{3}\sqrt{1-x^2} = \sqrt{3}(1-x^2)$$

$$\int_{-1}^{1} \sqrt{3}(1-x^2) dx = \frac{4\sqrt{3}}{3}$$

Question

Why is the height only $$\sqrt{3}\sqrt{1-x^2}$$ and not $$2 \sqrt{3}\sqrt{1-x^2}$$? I thought $$\sqrt{1-x^2}$$ is only the half the height of the circle

 

[LCKurtz]

y = sqrt(1-x²) is half the base of the triangle so

b = 2y, h = y sqrt(3)

A = (1/2)bh = (1/2)(2y)(ysqrt(3) = y²sqrt(3) = (1-x²)sqrt(3)

 

[flyingpig]

No I am asking about the height not the base.

 

[LCKurtz]

No I am asking about the height not the base.

Did you actually read what I wrote? Guess what h stands for.

 

[flyingpig]

h stands for height, but that's not the nature of the question. y = sqrt(1 + x^2) which is only half the height

 

[I like Serena]

Question

Why is the height only $$\sqrt{3}\sqrt{1-x^2}$$ and not $$2 \sqrt{3}\sqrt{1-x^2}$$? I thought $$\sqrt{1-x^2}$$ is only the half the height of the circle

The height of a unit equilateral triangle is $$\frac 1 2 \sqrt{3}$$
The side of the triangle would be, as you state, $$2 \sqrt{1-x^2}$$.
Note that in the product of the two the 1/2 and the 2 cancel against each other.