The base of solid is right triangle whose base side has length a and whose perpendicular side has length (1/2)a. Find the volume of the solid if cross sections perpendicular to the base the triangle are semicircles.
The hypotenuse of the triangle is the line y = (a-x)/2
That means the diameter of the semicircle at (x,y) is y, so the area is
π/4 y^2 = π/4 (a-x)^2/4
The volume is thus
v = ∫[0,a] π/16 (a-x)^2 dx
To find the volume of the solid, we first need to determine how the radius of the semicircles varies depending on the position along the height of the solid.
Since the base is a right triangle with the base side of length "a" and the perpendicular side of length 1/2"a", we can use the Pythagorean Theorem to find the length of the hypotenuse:
c^2 = (a^2) + ((1/2 a)^2)
Simplifying the equation gives us:
c^2 = a^2 + (1/4 a^2)
c^2 = (5/4 a^2)
c = sqrt(5/4 a^2)
c = (1/2)sqrt(5) * a
Since the cross sections perpendicular to the base of the triangle are semicircles, the radius of each semicircle will be variable along the height of the solid. At any given height h, the radius r is given by:
r = (1/2)sqrt(5) * a - (1/2)sqrt(5) * h
To find the volume of the solid, we need to integrate the cross-sectional area over the height of the solid. The cross-sectional area of each semicircle is given by:
A = (1/2) * π * r^2
Integrating this over the height h from 0 to (1/2)a, we have:
V = ∫[0, (1/2)a] (1/2) * π * [(1/2)sqrt(5) * a - (1/2)sqrt(5) * h]^2 dh
Simplifying the expression and calculating the integral gives us the volume V of the solid.
To find the volume of the solid, we can integrate the areas of the cross sections along the height of the solid.
First, let's consider a cross section perpendicular to the base of the triangle. Since the cross sections are semicircles, the radius of each cross section will vary depending on the distance from the base.
Let's denote the height of the solid as h.
At a distance x from the base, the radius of the semicircle can be determined using the Pythagorean theorem. Let's call this radius R.
According to the Pythagorean theorem, we have:
R^2 = (1/2a)^2 + x^2
Simplifying this equation, we get:
R^2 = 1/4a^2 + x^2
Since the cross section is a semicircle, the area of each cross section is given by:
A(x) = (1/2)πR^2
Substituting R^2, we have:
A(x) = (1/2)π [1/4a^2 + x^2]
To find the volume, we need to integrate this expression with respect to x from 0 to h (the height of the solid).
V = ∫[0,h] A(x) dx
= ∫[0,h] (1/2)π [1/4a^2 + x^2] dx
Let's calculate this integral step-by-step.
V = (1/2)π * ∫[0,h] [1/4a^2 + x^2] dx
= (1/2)π * [ (1/4a^2) * x + (1/3) * x^3 ] |[0,h]
= (1/2)π * [ (1/4a^2) * h + (1/3) * h^3 ]
Therefore, the volume of the solid is given by:
V = (1/2)π * [ (1/4a^2) * h + (1/3) * h^3 ]