Question

Find the volume of the solid generated by revolving the region bounded by the graphs of y = 2x^2 + 1 and y = 2x + 5 about the x-axis. The volume of the solid generated by revolving the region bounded by the graphs of y = 2x^2 + 1 and y = 2x + 5 about the x-axis is cubic units.

Answer 1

To find the volume of the solid generated by revolving the region bounded by two curves about the x-axis, you can use the method of cylindrical shells.

Step-by-step explanation:

To find the volume of the solid generated by revolving the region bounded by the graphs of y = 2x^2 + 1 and y = 2x + 5 about the x-axis, we can use the method of cylindrical shells.

1. First, find the points of intersection between the two curves by setting them equal to each other: 2x^2 + 1 = 2x + 5. Solve for x to find the x-coordinates of the points of intersection.
2. Next, integrate the radius of the cylindrical shells with respect to x from one intersection point to another. The radius can be found by subtracting the y-coordinate of the parabola from the y-coordinate of the line: radius = (2x + 5) - (2x^2 + 1).
3. Then, integrate the formula for the circumference of a circle, 2πr, with respect to x from one intersection point to another.
4. Finally, evaluate the definite integral to find the volume of the solid.

The volume of the solid generated by revolving the region bounded by the given curves about the x-axis is cubic units.

Answer 2

The volume of the solid generated by revolving the region bounded by the graphs of y = 2x^2 + 1 and y = 2x + 5 about the x-axis is 9π cubic units.

Step-by-step explanation:

To find the volume of the solid generated by revolving the region bounded by the graphs of y = 2x^2 + 1 and y = 2x + 5 about the x-axis, we can use the method of cylindrical shells. First, we need to find the limits of integration by finding the x-coordinate where the two curves intersect. Setting the two equations equal to each other, we get:

2x^2 + 1 = 2x + 5

2x^2 - 2x - 4 = 0

Solving for x using the quadratic formula, we get x = (-(-2) ± sqrt((-2)^2 - 4(2)(-4))) / (2(2))

x = (2 ± sqrt(4 + 32)) / 4

x = (2 ± sqrt(36)) / 4

x = (2 ± 6) / 4

x = -1 or x = 2

So, the limits of integration are -1 and 2. Now, we can set up the integral to find the volume:

V = ∫[from -1 to 2] 2πx(y₂ - y₁)dx

V = ∫[from -1 to 2] 2πx((2x + 5) - (2x^2 + 1))dx

V = 2π∫[from -1 to 2] (2x^2 + 5x - 2x^2 - 1)dx

V = 2π∫[from -1 to 2] (5x - 1)dx

V = 2π[(5/2)x^2 - x]∣[from -1 to 2]

V = 2π[(5/2)(2)^2 - 2] - 2π[(5/2)(-1)^2 - (-1)]

V = 2π[10 - 2] - 2π[(5/2) - (-1)]

V = 2π[8] - 2π[(5/2) + 1]

V = 16π - 2π(7/2)

V = 16π - 7π

V = 9π

Therefore, the volume of the solid generated by revolving the region bounded by the given graphs about the x-axis is 9π cubic units.