Question

Let the region R be the area enclosed by the function f(x) = x² , the horizontal line y = 3, and the y-axis. Find the volume of the solid

Answer 1

To find the volume of the solid bounded by the parabola y=x², y-axis, and the line y=3, set up the integral V = ∫_0^√3 πx²dx, evaluate, and simplify to find the volume to be 3π√3 cubic units.

Step-by-step explanation:

The question is asking to find the volume of a solid formed by revolving the region R around an axis. The region R is bounded by the parabola f(x) = x², the horizontal line y = 3, and the y-axis. To solve this problem, we use the disk method which involves taking an integral that represents the sum of the volumes of an infinite number of disks.

To find the limits for integration, we need to solve for x when f(x) = 3, which gives us x = √3. So we will integrate from 0 to √3. The volume (V) of the solid is given by the integral of the cross-sectional area (A) times the height (h). In this case, the cross-sectional area of a disk is πr² and the height is dx (the thickness of a disk). Since the radius of the disks (r) varies with x (because the boundary is the parabola), the radius is the distance from the y-axis to the point on the curve, which is the value of x.

The volume can be calculated as follows:

1. Set up the integral: V = ∫_0^√3 πx²dx
2. Integrate: V = π ∫_0^√3 x²dx
3. Evaluate the integral: V = π [x³/3]_0^√3
4. Calculate the volume: V = π [√3³/3 - 0/3]
5. Simplify: V = (π/3)(√3)³9 = 3π√3 cubic units

This provides the volume of the solid in question.