The base of a solid is a circle of

radius = 4
Find the exact volume of this solid if the cross sections perpendicular to a given axis are equilateral right triangles.

The equation of the circle is:
x^2 + y^2 = 16

I have the area of the triangle (1/2bh) to be equal to 2sqrt(12)

(1/2 * 4 * sqrt12)

there are triangles that have vertical bases. They run parallel to the y-axis. The triangles are inside of the circle

Any help would be greatly appreciated! Thanks!

The volume of a cone is
V =(1/3)*(base area)*(height)

In your equilateral case,
height = sqrt3 * r.
Therefore
V = sqrt3*pi*r^3

323

To find the volume of the solid, we can use the formula for the volume of a cone since the cross sections are equilateral right triangles.

The formula for the volume of a cone is V = (1/3) * (base area) * (height).

In this case, the base of the solid is a circle of radius 4, so the base area is π * r^2 = π * 4^2 = 16π.

The height of each equilateral right triangle can be calculated as the length of one side of the triangle, which is the radius of the circle, multiplied by sqrt(3), as each side of an equilateral triangle is sqrt(3) times the length of the radius. So the height is sqrt(3) * 4 = 4sqrt(3).

Now, we can substitute the values into the formula to find the exact volume of the solid:

V = (1/3) * (16π) * (4sqrt(3))
= (16/3) * (4sqrt(3)) * π
= 64sqrt(3) * π / 3

Therefore, the exact volume of the solid is 64sqrt(3) * π / 3.