Find the volume of the solid formed by revolving the region bounded by the graphs of y=3-(x^2) and y=2 about the line y=2
Hint:
Why don't you translate the whole configuration down 2 units?
Then you would "find the volume of the solid formed by revolving the regions bounded by the graphs of y = 1 - x^2 and the y=2 (the x-axis) about the x-axis"
Hint #2
By the "Just-look-at-it Theorem" the x-intercepts are ±1 and there is symmetry, so you could just take it from 0 to 1 and double that.
To find the volume of the solid formed by revolving the region bounded by the graphs of y=3-(x^2) and y=2 about the line y=2, you can use the method of cylindrical shells.
Step 1: First, let's sketch the region bounded by the two graphs. We can see that the region lies between x = -√3 and x = √3 and is bounded above by y = 3 - x^2 and below by y = 2.
Step 2: Now, we need to find the height of each cylindrical shell at a given x-value. The height of each shell is the difference between the top curve (y = 3 - x^2) and the bottom curve (y = 2). So, the height is given by h(x) = (3 - x^2) - 2 = 1 - x^2.
Step 3: The radius of each cylindrical shell is the perpendicular distance from the axis of rotation (y = 2) to the x-axis. Since the axis of rotation is y = 2, the distance from the x-axis to y = 2 is 2 - 0 = 2.
Step 4: Now that we have the height and radius of each cylindrical shell, we can find the volume of each shell using the formula for the volume of a cylindrical shell: V = 2πrh, where r is the radius and h is the height.
Step 5: To get the total volume of the solid, we need to integrate the volume of each shell over the interval x = -√3 to x = √3:
V = ∫(from -√3 to √3) 2π(2)(1 - x^2) dx
Step 6: Simplify the integral and evaluate it:
V = 2π ∫(from -√3 to √3) (2 - 2x^2) dx
V = 2π [2x - (2/3)x^3] (from -√3 to √3)
V = 2π [2(√3) - (2/3)(√3)^3] - [2(-√3) - (2/3)(-√3)^3]
V = 2π [2√3 - 2(√3)^3/3]
V = 2π [2√3 - 2(3√3)/3]
V = 2π [2√3 - 2√3]
V = 2π (0)
V = 0
Therefore, the volume of the solid formed by revolving the region bounded by the graphs of y=3-(x^2) and y=2 about the line y=2 is 0.