Find the volume of the solid whose base is the semicircle y=sqrt(16−x^2) where −4 is less then or equal to x which is less then or equal to 4, and the cross sections perpendicular to the x-axis are squares.
Lengths of the sides of these squares are
equal to y.
The volume=Int(from -4 to 4)y^2dx=
2*Int(from 0 to 4)(16-x^2)dx=
2*(16x-x^3/3)(from 0 to 4)=
2(64-64/3)
To find the volume of the solid, we will integrate the area of the cross sections.
First, let's find the length of one side of the square cross section at each x-value. Since the cross sections are perpendicular to the x-axis, the side length will be equal to the height of the semicircle at that x-value.
The semicircle equation is y = sqrt(16 - x^2).
Since the cross sections are squares, the side length is:
s = 2 * sqrt(16 - x^2)
The limits of integration are -4 to 4, since the base of the solid is the interval where -4 ≤ x ≤ 4.
Now, let's integrate the area of each cross section to find the volume of the solid:
V = ∫[a, b] A(x) dx
V = ∫[-4, 4] (s)^2 dx
V = ∫[-4, 4] (2 * sqrt(16 - x^2))^2 dx
Let's simplify the integrand:
V = ∫[-4, 4] (4 * (16 - x^2)) dx
V = 4 ∫[-4, 4] (16 - x^2) dx
Now, let's integrate:
V = 4 [ ∫[4, -4] (16 * x - (x^3 / 3))] dx
V = 4 [(16 * (x^2)/2 - (x^4 / 12))] |[4,-4]
V = 4 [(8 * (16)/2 - (16 / 12)) - (-8 * (16)/2 + (16 / 12))]
V = 4 [(64 - (16 / 12)) + (64 + (16 / 12))]
V = 4 [(64 - (4/3)) + (64 + (4/3))]
V = 4 [128 + (4/3)]
V = 4 * [384/3 + 4/3]
V = 4 * (388 / 3)
Finally, we get the volume of the solid:
V = 1552 / 3
Therefore, the volume of the solid is 1552/3 cubic units.
To find the volume of the solid, we need to integrate the areas of the cross-sections.
First, let's visualize the solid. The base is a semicircle, given by the equation y = sqrt(16 - x^2), where -4 ≤ x ≤ 4. The cross-sections perpendicular to the x-axis are squares.
To find the area of each square cross-section, we need to determine the side length at each x-coordinate. Since the shape is symmetric about the y-axis, we can focus on the positive x-values (0 ≤ x ≤ 4) and then multiply the result by 2 to account for the entire shape.
Let's consider an infinitesimally small segment at some x-coordinate. The length of this segment is equal to the width of the square cross-section. We can obtain the width by subtracting the y-coordinate of the semicircle from the maximum y-value (which occurs at x = 0) of the semicircle.
The equation of the semicircle is y = sqrt(16 - x^2), so the y-coordinate at x is y = sqrt(16 - x^2). The maximum y-value is y = sqrt(16), which simplifies to y = 4.
Therefore, the width of the square cross-section at x is 4 - sqrt(16 - x^2).
To find the volume, we integrate the area of each square cross-section over the interval x = 0 to x = 4:
V = 2 * ∫[0 to 4] (width)^2 dx
= 2 * ∫[0 to 4] (4 - sqrt(16 - x^2))^2 dx
Now, we can expand and simplify the integrand:
V = 2 * ∫[0 to 4] (16 - 8√(16 - x^2) + (16 - x^2)) dx
= 2 * ∫[0 to 4] (32 - 8√(16 - x^2) - x^2) dx
Next, we integrate term by term:
V = 2 * (32x - 8∫√(16 - x^2) dx - ∫x^2 dx)
The first term is straightforward to integrate:
V = 2 * (32x - 8∫√(16 - x^2) dx - (x^3/3))
To integrate the second term, we can use the substitution u = 16 - x^2:
V = 2 * (32x - 8∫(u^0.5) * (-x) dx - (x^3/3))
= 2 * (32x + 8∫(u^0.5) * x dx - (x^3/3))
Now, we can integrate the second term:
V = 2 * (32x + 8 * (u^0.5) * (x^2/2) - (x^3/3))
Substituting back u = 16 - x^2:
V = 2 * (32x + 8 * (√(16 - x^2)) * (x^2/2) - (x^3/3))
Now, we can evaluate this expression from x = 0 to x = 4 to find the final volume of the solid:
V = 2 * ∫[0 to 4] (32x + 8 * (√(16 - x^2)) * (x^2/2) - (x^3/3)) dx
By plugging in the values for x, we can calculate the volume.