The volume of the region bounded by the y-axis, y=4 and y=x²?
I'm only seeing two dimensions here, so I believe you mean area instead of volume.
You can start by visualizing all of the equations on a 2-D plane.
And, if you are interested in volume, rotated about which axis?
To find the volume of the region bounded by the y-axis, y=4, and y=x^2, we can use the method of cylindrical shells.
First, let's visualize the region in question. The y-axis, y=4, and y=x^2 form a bounded region in the xy-plane. This region is symmetric about the y-axis and resembles an elongated bowl opening up towards positive y-values.
To calculate the volume, we will integrate the area of infinitesimally thin cylindrical shells that are vertical to the x-axis. Each shell will have a height of Δy and a thickness of Δx.
Now, let's break down the steps to calculate the volume:
1. Determine the limits of integration:
Since the region is bounded by y=4 and y=x^2, we need to find the x-values that correspond to these y-values.
Setting y=4, we get:
4 = x^2
Taking the square root of both sides:
x = +/- 2
Therefore, our limits of integration for x will be from -2 to 2.
2. Set up the integral for the volume using cylindrical shells:
The volume of each cylindrical shell is given by the product of its circumference (2πr) and height (Δy).
The radius, r, of each cylindrical shell is simply the x-value at a given y, which is x=y^(1/2).
The height, Δy, of each cylindrical shell is the infinitesimal change in y.
Therefore, the volume of the region can be calculated by integrating the product of 2πr and Δy with respect to y, from y=0 (y-axis) to y=4:
V = ∫[0, 4] (2πr * Δy) dy
= ∫[0, 4] (2πy^(1/2) * Δy) dy
3. Evaluate the integral:
Integrate the expression (2πy^(1/2) * Δy) with respect to y, from y=0 to y=4:
V = ∫[0, 4] (2πy^(1/2) * dy)
= [2π * (2/3)y^(3/2)] [0, 4]
= (4π/3) * [(2/3)^(3/2) * 4^(3/2)]
≈ 16.755 cubic units (rounded to three decimal places)
Therefore, the volume of the region bounded by the y-axis, y=4, and y=x^2 is approximately 16.755 cubic units.