the region bounded by the quarter circle (x^2) + (y^2) =1. Find the volume of the following solid.
The solid whose base is the region and whose cross-sections perpendicular to the x-axis are squares.
"The solid whose base is the region and whose cross-sections perpendicular to the x-axis are squares."
means that z=2y
but since y=2sqrt(1-x^2) (on the circle), so z=2sqrt(1-x^2)
For example, at x=0, z=1,
at x=1, z=0.
The volume of the solid is then
∫∫∫dx dy dz
where the limits of integration are
for z: 0 to 2sqrt(1-x^2)
for y: -sqrt(1-x^2) to sqrt(1-x^2)
for x: -1 to 1
To find the volume of the solid, we need to integrate the area of each square cross-section along the x-axis.
First, let's visualize the given region. The equation of the quarter circle is (x^2) + (y^2) = 1. This represents the upper-right quarter of a unit circle centered at the origin (0,0) in the xy-plane.
To find the volume, we'll consider an infinitesimally small section of the region. Suppose we have a small interval dx along the x-axis. Each corresponding square cross-section will have a side length of 2y, where y is the y-coordinate at that particular x-value.
To find the y-coordinate, we can solve the equation of the quarter circle for y:
y = sqrt(1 - x^2)
The area of each square cross-section is (2y)^2 = 4y^2. Therefore, the volume element is the area of the square cross-section times the small interval dx:
dV = 4y^2 * dx
To find the total volume, we need to integrate the volume element over the range of x-values that define the region. Since the quarter circle is bounded by x = 0 and x = 1, the integral will be:
V = ∫[0,1] 4y^2 dx
Substituting y = sqrt(1 - x^2), we can rewrite the equation:
V = ∫[0,1] 4(1 - x^2) dx
Simplifying further,
V = 4∫[0,1] (1 - x^2) dx
V = 4[ x - (1/3)x^3 ] from 0 to 1
Evaluating the integral,
V = 4[1 - (1/3)] - 4[0 - 0]
V = 4/3
Therefore, the volume of the solid whose base is the region bounded by the quarter circle and whose cross-sections perpendicular to the x-axis are squares is 4/3 cubic units.