What's the Volume?

Question

Let B be the solid whose base is the circle x^2 + y^2 = 196 and whose vertical cross sections perpendicular to the x-axis are equilateral triangles. Compute the volume of B.

V=

Answer 1

What's the Volume?

Let B be the solid whose base is the circle x^2 + y^2 = 196 and whose vertical cross sections perpendicular to the x-axis are equilateral triangles. Compute the volume of B.

V=

HELLO again Maisie!!!!,

The key to solving this type of problem, is to graph, or visualize in my case, a circle with a radius 14, centered at the origin. The basis of your cross sections which you will sum up in the integral, are equilateral triangles. The length of a side on these equilateral triangles will vary as a function of x .

The area of an equilateral triangle can be determined as follows...

_{ohhh silly geometry... such a pointless class,}

From a point at its' center, divide an equilateral triangle into 3 equal triangular parts. The angle at the vertex of each part will be 120°. Now draw a vertical line (or Apothem) from the vertex, to the base. That cuts the 120° degree triangle in half, leaving two, 60° right triangles. Use a 30, 60, 90 triangle to get the area:

Ι\

Ι..\

Ι..θ\

You should have a 30, 60, 90, with base x/2, which is opposite the 60° angle.

Set: x/2 = s√3 and solve for s

x/(2√3) = s

The height of the 30, 60, 90 triangle is opposite the 30° angle. which is just s

So, the area of the 30, 60, 90 is calculated using 1/2b*h

A = 1/2 * (x/2) * x/(2√3)

A = x²/(8√3)

Multiply this by 6, since there are 6 congruent trianlge slices that make up the equilateral triangle wedge we are finding the area for.

A_Δ = 3x²/(4√3)

**Alternatively, we could have used the fact that Area = 1/2(Apothem)*(Perimeter)

Recall that the apothem of the equilateral triangle, will be the side opposite the 30° angle, in the 30, 60, 90 triangle which is x/(2√3)

Area = 1/2(Apothem)*(Perimeter)

Area = 1/2(x/(2√3))*(3x) = 3x²/(4√3)

Confirmed.

Now, before we can finally integrate, notice that the length of one side of the equilateral triangle will be that of 2y. Check your graph. You should have a circle, centered at the origin, now make a slice that runs perpendicular to the x-axis, this represents the base of one of your equilateral triangles, Notice it has a length of 2y.... So let us solve for y, as a function of x.

x² + y² = 196

y = √(196 - x²)

Since the area of the equilateral triangles, depends on the length of the base = 2y, substitute 2y for x:

A_Δ = 3x²/(4√3)

A_Δ = 3(2y)²/(4√3)

Now, let's integrate:

V = ∫_-14¹⁴ [ 3(2y)²/(4√3) ]dx

Simplify, using symmetry, and subbing y = √(196 - x²)

V = 2 ∫₀¹⁴ [ 3(2√(196 - x²))²/(4√3) ]dx

V = 2 ∫₀¹⁴ [ 3(4(196 - x²)/(4√3) ]dx

V = 2 ∫₀¹⁴ [ 3(196 - x²)/(√3) ]dx

V = (6/√3) ∫₀¹⁴ [ (196 - x²) ]dx

V = (6/√3) [ 196x - 1/3*x³ ]₀¹⁴

V = (6/√3) [ 196(14) - 1/3*(14)³ ] - [ 0 ]

V = (6/√3) [ 196(14) - 1/3*(14)³ ]

V = (6/√3) [ 14³ (1 - 1/3) ]

V = ... ?

That may or may not be correct. You decide!