5. Find the volume of the solid generated by revolving the region bounded by y = x2, y = 0 and x = 1 about
(a) the x-axis
(b) the y-axis
(c) theline x=2
(d) the line y = −2
I'll do (a) and you can apply the logic to the others.
Using discs, the small discs have thickness dx
v = ∫[0,1] πr^2 dx
where r=y=x^2
v = π∫[0,1] x^4 dx
= π * 1/5 x^5 [0,1]
= π/5
Using shells, the thickness is dy, so we have
v = ∫[0,1] 2πrh dy
where r=y and h=(1-x)=(1-√y)
v = 2π∫[0,1] y(1-√y) dy
= 2π (1/2 y^2 - 2/5 y^5/2) [0,1]
= 2π (1/2 - 2/5)
= π/5
For rotation about the line y=-2,
discs have a hole in them
shells have a radius y+2
Give it a shot; try both ways to validate your answers
To find the volume of the solid generated by revolving the given region, we can use the method of cylindrical shells.
(a) Revolving about the x-axis:
Step 1: First, let's sketch the region that we are revolving.
The region is bounded by y = x^2, y = 0, and x = 1. It is a parabolic shape that lies between the x-axis and x = 1.
Step 2: Now, imagine a vertical strip with width dx at a distance x from the y-axis. When this strip is revolved around the x-axis, it forms a cylindrical shell.
Step 3: The height of the cylindrical shell is given by the difference between the upper and lower y-coordinates of the strip, which is y = x^2 - 0 = x^2.
Step 4: The width of the cylindrical shell is dx.
Step 5: The circumference (circ) of the cylindrical shell is equal to 2π times the distance from the y-axis, which is x.
Step 6: Using the formula for the volume of a cylindrical shell, the volume of the solid generated is given by:
dV = 2πx(x^2)dx
Step 7: To find the total volume, we integrate dV from x = 0 to x = 1:
V = ∫[0, 1] 2πx(x^2)dx
V = 2π ∫[0, 1] x^3 dx
V = 2π [x^4 / 4] evaluated from 0 to 1
V = 2π [(1^4 / 4) - (0^4 / 4)]
V = 2π (1 / 4)
V = π / 2
Therefore, the volume of the solid generated by revolving the region about the x-axis is π / 2.
(b) Revolving about the y-axis:
Step 1: Similar to step 1 in part (a), let's sketch the region that we are revolving.
The region is still bounded by y = x^2, y = 0, and x = 1. It is still a parabolic shape that lies between the x-axis and x = 1.
Step 2: This time, imagine a horizontal strip with width dy at a distance y from the x-axis. When this strip is revolved around the y-axis, it forms a cylindrical shell.
Step 3: The height of the cylindrical shell is given by the difference between the right and left x-coordinates of the strip, which is x = √y.
Step 4: The width of the cylindrical shell is dy.
Step 5: The circumference (circ) of the cylindrical shell is equal to 2π times the distance from the y-axis, which is y.
Step 6: Using the formula for the volume of a cylindrical shell, the volume of the solid generated is given by:
dV = 2πy(√y)dy
Step 7: To find the total volume, we integrate dV from y = 0 to y = 1:
V = ∫[0, 1] 2πy(√y)dy
V = 2π ∫[0, 1] y^(3/2) dy
V = 2π [2/5 y^(5/2)] evaluated from 0 to 1
V = 2π [2/5 (1)^(5/2) - 2/5 (0)^(5/2)]
V = 2π [2/5 - 0]
V = 4π / 5
Therefore, the volume of the solid generated by revolving the region about the y-axis is 4π / 5.
To find the volume of the solid generated by revolving a region about an axis, we can use the method of disks or the method of washers, depending on the shape of the region.
(a) Revolving the region bounded by y = x^2, y = 0, and x = 1 about the x-axis:
To find the volume using the method of disks, we divide the region into infinitesimally thin disks perpendicular to the x-axis. The volume of each disk is given by the formula V = π * (radius)^2 * height.
In this case, the radius of each disk is the value of y (since we are revolving around the x-axis), and the height is the differential dx. Therefore, the volume of each disk is given by dV = π * (y)^2 * dx.
To find the total volume, we integrate the volume of each disk over the range of x values that defines the region. In this case, the range is from x = 0 to x = 1. Therefore, the integral for the volume is:
V = ∫[0,1]π * (y)^2 * dx
To express the integral in terms of y, we need to rewrite the equation of the curve y = x^2 in terms of y. Solving for x, we get x = √y.
Substituting these values into the integral, we have:
V = ∫[0,1]π * (√y)^2 * dx
V = ∫[0,1]π * y * dx
V = π * ∫[0,1]y * dx
Integrating with respect to x, the bounds change to y = 0 to y = 1, so the integral becomes:
V = π * ∫[0,1]y * dx
V = π * [x] [0,1]
V = π * (1 - 0)
V = π
Therefore, the volume of the solid generated by revolving the region about the x-axis is π.
(b) Revolving the region bounded by y = x^2, y = 0, and x = 1 about the y-axis:
To find the volume using the method of washers, we divide the region into infinitesimally thin washers with a hollow center, created by revolving the region about the y-axis. The volume of each washer is given by the formula V = π * (outer radius)^2 * height - π * (inner radius)^2 * height.
In this case, the outer radius is the value of x (since we are revolving around the y-axis), and the inner radius is 0. Therefore, the volume of each washer is given by dV = π * (x)^2 * dy.
To find the total volume, we integrate the volume of each washer over the range of y values that define the region. In this case, the range is from y = 0 to y = 1. Therefore, the integral for the volume is:
V = ∫[0,1]π * (x)^2 * dy
To express the integral in terms of x, we need to rewrite the equation of the curve y = x^2 in terms of x.
Substituting these values into the integral, we have:
V = ∫[0,1]π * (x)^2 * dy
V = π * ∫[0,1](x)^2 * dy
Integrating with respect to y, the bounds change to x = 0 to x = √y, so the integral becomes:
V = π * ∫[0,1](x)^2 * dy
V = π * ∫[0,1]x^2 * dy
V = π * [y^3/3] [0,1]
V = π * (1/3 - 0)
V = π/3
Therefore, the volume of the solid generated by revolving the region about the y-axis is π/3.