The base of a solid is the circle x^2 + y^2 = 9. Cross sections of the solid perpendicular to the x-axis are equilateral triangles. What is the volume, in cubic units, of the solid?
36√3
36
18√3
18
idk
So the cross sections (parallel to the base) are also circles, of formula
x^2+y^2=r^2, where r is radius.
Now given the vertical cross sections are equilaterial, that means the sides at x or y at zero, is equal to twice the radius of the base, or 6.
so the height of the solid is 6/sqrt3 (sketch a equilateral triangle of side 6), or h=2.44
at this point, you dont need calculus and integration (you can do that), but knowing the volume of a cone is 1/3 base area * height, or
volume= 1/2 (PI*3^2)(2.44)
To find the volume of the solid, we need to integrate the area of each circular cross section along the x-axis. Since the cross sections are equilateral triangles, we need to determine the side length of each triangle first.
The equation of the circle is x^2 + y^2 = 9. Solving for y, we get y = √(9 - x^2).
The base of the equilateral triangle is the diameter of the circular cross-section, which is 2y. Since the diameter of the circle is 6 (2 * 3), the side length of the equilateral triangle is equal to the diameter divided by √3. Therefore, the side length of each triangle is 6/√3 = 2√3.
To find the volume, we need to integrate the area of each triangular cross-section with respect to x.
The area of each equilateral triangle is (sqrt(3) / 4) * (side length)^2. Substituting in the side length, we get the area as (sqrt(3) / 4) * (2√3)^2 = 3√3.
Now, we integrate the area of each cross-section from x = -3 to x = 3 (the limits of the circle).
The volume V of the solid is given by:
V = ∫[from -3 to 3] 3√3 dx
Integrating, we get:
V = 3√3 * ∫[from -3 to 3] dx
V = 3√3 * [x] [from -3 to 3]
V = 3√3 * (3 - (-3))
V = 3√3 * 6
V = 18√3 cubic units
Therefore, the volume of the solid is 18√3 cubic units.
To find the volume of the solid, we need to determine the area of a typical cross-section and then integrate it over the given range.
The equation of the circle is x^2 + y^2 = 9, which represents a circle with a radius of 3 units.
We know that the cross-sections perpendicular to the x-axis are equilateral triangles. The base of each equilateral triangle is the line segment joining the ends of the circle, which has a length of 2 * 3 = 6 units.
We need to find the height of the equilateral triangle. To do this, we draw a line segment from the center of the circle (origin) to one vertex of the equilateral triangle. This line segment divides the equilateral triangle into two congruent right-angled triangles. The length of the line segment is the radius of the circle, which is 3 units.
Using the Pythagorean theorem, we can find the height of the equilateral triangle:
height^2 = 6^2 - 3^2 = 36 - 9 = 27
height = sqrt(27) = 3 sqrt(3) units
Now, the area of an equilateral triangle can be calculated using the formula:
area = (sqrt(3) / 4) * side^2
In this case, the side of the equilateral triangle is 6 units, so:
area = (sqrt(3) / 4) * 6^2 = (sqrt(3) / 4) * 36 = 9 sqrt(3) square units
To find the volume of the solid, we integrate the area over the range of x-values that the circle covers. Since the equation of the circle is x^2 + y^2 = 9, we can solve for y:
y = sqrt(9 - x^2)
Now, we can set up the integral:
V = ∫[a,b] A(x) dx = ∫[a,b] (9 sqrt(3)) dx
The limits of integration, a and b, represent the range of x-values over which the circle covers. In this case, since the base of the solid is a circle, the range is -3 to 3.
V = ∫[-3,3] (9 sqrt(3)) dx = 9 sqrt(3) ∫[-3,3] dx = 9 sqrt(3) [x]_(-3 to 3)
Evaluating the integral:
V = 9 sqrt(3) [3 - (-3)] = 9 sqrt(3) * 6 = 54 sqrt(3) cubic units
Therefore, the volume of the solid is 54 sqrt(3) cubic units.