Let R be the square region bounded by y = 2, x = 2, and the
x and y-axes. When R is rotated about the x-axis, what is the volume of the
resulting solid? What is the volume of the solid generated by rotating R about
the y-axis?
first, since it's a square, it does not matter which side is used as the rotational axis.
geometrically, we have a cylinder of height=2, radius=2
v = pi * 2^2 * 2 = 8pi
Analytically,
v = Integral(pi y^2 dx)[0,2]
= Integral(4pi dx)[0,2]
= 4pi x[0,2]
= 8pi
To find the volume of the solid generated by rotating the region R about the x-axis, we can use the method of cylindrical shells.
The region R is a square with sides of length 2, bounded by the x- and y-axes and the lines y = 2 and x = 2.
To find the volume, we need to integrate the area of the infinitesimally thin cylindrical shells, which can be expressed as height multiplied by circumference.
Step 1: Find the height of each cylindrical shell.
The height of each cylindrical shell is the difference between the y-coordinate of the region and the x-axis. In this case, the height is y = 2.
Step 2: Find the circumference of each cylindrical shell.
The circumference of each cylindrical shell is the same as the perimeter of the square region R, which is 4 times the side length. In this case, the circumference is 4.
Step 3: Set up the integral.
The integral to find the volume is given by V = ∫(2πrh)dx, where r is the distance from each cylindrical shell to the x-axis and dx is an infinitesimally small width.
Since we are rotating about the x-axis, r is equal to the x-coordinate of the cylindrical shell, and the infinitesimal width dx is equal to the infinitesimal height dy.
Step 4: Set the limits of integration.
The region R bounded by y = 2 and the x and y-axes is a square with sides of length 2. Thus, the limits of integration are from 0 to 2.
Step 5: Evaluate the integral.
The volume can then be calculated as V = ∫(0 to 2) (2πxy)dy = 2π∫(0 to 2) (2y)dy.
Evaluating the integral, we get V = 2π[y^2] from 0 to 2 = 2π(2^2 - 0^2) = 8π cubic units.
Therefore, the volume of the solid generated by rotating region R about the x-axis is 8π cubic units.
To find the volume of the solid generated by rotating R about the y-axis, we can follow a similar approach using the method of disks or washers.
The region R is a square with sides of length 2, bounded by the x- and y-axes and the lines y = 2 and x = 2.
Step 1: Find the radius of each disk or washer.
Since we are rotating about the y-axis, the radius is equal to the x-coordinate of each disk or washer. In this case, the radius is x = 2.
Step 2: Set up the integral.
The integral to find the volume is given by V = ∫(πr^2)dy, where r is the radius of each disk or washer and dy is an infinitesimally small height.
Since we are rotating about the y-axis, r is equal to 2 - y, and the infinitesimal height dy is equal to the infinitesimal width dx.
Step 3: Set the limits of integration.
The region R bounded by y = 2 and the x and y-axes is a square with sides of length 2. Thus, the limits of integration are from 0 to 2.
Step 4: Evaluate the integral.
The volume can then be calculated as V = ∫(0 to 2) (π(2-y)^2)dx = π∫(0 to 2) (4-4y+y^2)dx.
Evaluating the integral, we get V = π[x(4-4y+y^2)] from 0 to 2 = π(2)(4-4y+y^2) = 8π cubic units.
Therefore, the volume of the solid generated by rotating region R about the y-axis is 8π cubic units.