Find the volume V of the described solid S.
The base of S is the region enclosed by the parabola
y = 3 − 2x2
and the x−axis. Cross-sections perpendicular to the y−axis are squares.
To find the volume of the described solid S, we need to integrate the cross-sectional area from the base to the top.
First, let's find the limits of integration. We need to determine the x-values where the parabola intersects the x-axis. Setting y = 0, we have:
0 = 3 - 2x^2
2x^2 = 3
x^2 = 3/2
x = ±√(3/2)
Since the parabola is symmetric about the y-axis, we only need to consider the positive x-values. Therefore, the limits of integration for x are from -√(3/2) to √(3/2).
Next, let's find an expression for the side length of the square cross-section. Since the cross-sections are perpendicular to the y-axis, we need to express the side length in terms of y. Since a square has all sides equal, the side length of the square cross-section will be equal to the height of the square.
The height of the square will be equal to the y-coordinate of the parabola at a specific x-value.
From the equation of the parabola, y = 3 - 2x^2, we can solve for x in terms of y:
2x^2 = 3 - y
x^2 = (3 - y)/2
x = ±√((3 - y)/2)
Again, we only need to consider the positive x-values, so x will be √((3 - y)/2).
Now, the side length of the square cross-section is equal to the height of the square, which is y. Therefore, the area of the square cross-section is y^2.
To find the volume, we need to integrate the area of the square cross-section over the interval [a, b], where a and b are the limits of integration.
Using the formula for volume, V = ∫[a, b] A(y) dy, where A(y) represents the area of the square cross-section, we have:
V = ∫[-√(3/2), √(3/2)] y^2 dy
Evaluating this integral will give us the volume of the solid S.