Integral Volume of a Solid

Let's take circle as $x^{2}$ + $y^{2}$ =1 centered at origin, since all cross sections are equilateral triangles , $y=\displaystyle\sqrt{1-x^{2}}$ will be half of the base length of any triangle,

$\sqrt{4y^{2} -y^{2} }$ =height of triangle= $\sqrt{3 } y$ ,

Area(x) = $\dfrac{1}{2}\sqrt{3 } y$ $2y$ = $\sqrt{3 } y^{2}$

$\displaystyle\int_{-1}^{1} Area(x) dx$ =Integral volume of triangles=Volume of the solid

$\displaystyle\int_{-1}^{1} \sqrt{3 }(1-x^{2}) dx$ = $-\sqrt {3} (\dfrac{x^{3}}{3} - x )$ $\Biggr|_{-1}^{1}$ = 4 $\dfrac{\sqrt{3}}{3}$

The function for a circle with radius one centered at the origin is,

$x^2 + y^2 = 1$

Note that this is half the height of each equilateral triangle.

The area of an equalterial triangle,

$A = \frac{\sqrt{3}}{4} s^2$

Each base of the triangle can be represented by the unit circle function above. As the height of the circle centered at 1 only gives us half of the base of the triangle, we have to double it. Thus, the area of each cross section at point x on the circle is,

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

We now want the sum of all of these cross-sections times an infitesimal $\Delta x$ ,

$\lim_{\Delta x \to 0} \sum_{i=1}^{n}{\frac{\sqrt{3}}{4} (2*\sqrt{1-i^2})^2} \Delta x = \int_{-1}^{1} \frac{\sqrt{3}}{4} (2*\sqrt{1-x^2})^2 dx$

The function is even,

$2\int_{0}^{1} \sqrt{3} (1-x^2) dx = \frac{4\sqrt{3}}{3}$

$2*(3) + 4 = \boxed{10}$

Using $x^2 + y^2 = 1$ as the equation of a unit circle, and $\frac{\sqrt{3}}{4}s^2$ as the area of an equilateral triangle with side $s$ , the volume of the solid is $\displaystyle \int_{-1}^1 \frac{\sqrt{3}}{4}(2\sqrt{1 - x^2})^2 = \frac{4\sqrt{3}}{3}$ , so $A = 4$ , $B = 3$ , and $A + 2B = \boxed{10}$ .