The base of a solid in the xy-plane is the first-quadrant region bounded y = x and y = x^2. Cross sections of the solid perpendicular to the x-axis are equilateral triangles. What is the volume, in cubic units, of the solid? May I have all the explanation with the answer, please. Thanks in advantage
Draw the graphs. Each triangle has a base equal to the distance between the graphs: x-x^2
The area of an equilateral triangle of side s is √3/4 s^2
So, add up the volumes of all those thin triangles and you get
v = ∫[0,1] √3/4 (x-x^2)^2 dx = 1/(40√3)
To find the volume of the solid, we need to integrate the area of each cross section over the interval of the base.
First, let's find the limits of integration. The base of the solid is the region bounded by y = x and y = x^2 in the first quadrant. To find the points of intersection, we set the two equations equal to each other:
x = x^2
Rearranging, we get:
0 = x^2 - x
Factoring, we have:
0 = x(x - 1)
So the points of intersection are x = 0 and x = 1.
Now, let's find the area of a cross section at any given x-value. Since the cross sections are equilateral triangles, we can find the height of each triangle as the difference between the two functions:
height = x^2 - x
The base of each triangle is given by the length of the interval: Δx = dx
The area of each triangle is given by:
A = (1/2) * base * height
A = (1/2) * dx * (x^2 - x)
So, the volume of the solid can be expressed as:
V = ∫[0 to 1] A dx
V = ∫[0 to 1] (1/2) * (x^2 - x) dx
To find the integral, we can evaluate it step by step:
V = (1/2) * ∫[0 to 1] (x^2 - x) dx
V = (1/2) * [ (x^3/3) - (x^2/2) ] evaluated from 0 to 1
V = (1/2) * [ (1/3) - (1/2) - (0) ]
V = (1/2) * [ (1/3) - (1/2) ]
V = (1/2) * [ (2/6) - (3/6)]
V = (1/2) * [ -1/6 ]
V = -1/12
Since we are working with volume, the negative sign doesn't make physical sense. Therefore, the absolute value of the volume is:
|V| = 1/12
So, the volume of the solid is 1/12 cubic units.
To find the volume of the solid, we need to break it down into infinitesimally thin slices and calculate the volume of each slice.
Let's consider an infinitesimally thin slice at a specific value of x. This slice will be perpendicular to the x-axis and have a width of dx.
The height of this slice, h(x), will be the difference between the y-values of the two bounding curves. In this case, the two bounding curves are y = x and y = x^2. Therefore, the height of the slice can be expressed as h(x) = x^2 - x.
Since the cross section of the solid is an equilateral triangle, we know that the height of the triangle is equal to the length of its base. In this case, the height of the triangle is h(x), and thus, the length of the base of the triangle is also h(x).
The formula for the volume of a prism is V = A * h, where A is the area of the base and h is the height. In this case, the area of the equilateral triangle base can be calculated using the formula A = (√3/4) * s^2, where s is the length of one side of the triangle.
Since the base is an equilateral triangle, the length of the base (s) is equal to the height of the slice (h(x)). Therefore, the area of the base of each slice can be calculated as A(x) = (√3/4) * [h(x)]^2.
To find the volume of the entire solid, we need to integrate the areas of all the slices from x = 0 to x = 1 and sum them up. The integral becomes:
V = ∫[0,1] A(x) * dx
= ∫[0,1] (√3/4) * [x^2 - x]^2 * dx
Simplifying and evaluating the integral:
V = (√3/4) * ∫[0,1] [x^4 - 2x^3 + x^2] * dx
= (√3/4) * [x^5/5 - x^4/2 + x^3/3] |[0,1]
= (√3/4) * [(1/5) - (1/2) + (1/3)]
= (√3/4) * [(3/10) - (5/10) + (10/30)]
= (√3/4) * [(3/10) - (5/10) + (1/3)]
= (√3/4) * [3/10 - 5/10 + 10/30]
= (√3/4) * [3/10 - 1/2]
= (√3/4) * [-1/5]
= -√3/20
Therefore, the volume of the solid is -√3/20 cubic units. Note that the negative sign does not affect the magnitude of the volume in this context.