What is the volume of the solid with given base and cross sections?
The base is the region enclosed by y=x^2 and y=3. The cross sections perpendicular to the y-axis are rectangles of height y^3.
To find the volume of the solid with the given base and cross sections, we can use the method of disks or washers.
First, we need to determine the limits of integration. Since the base is enclosed by y = x^2 and y = 3, we can find the x-coordinates of the points where these two curves intersect. Setting them equal to each other, we have:
x^2 = 3
Taking the square root of both sides, we get:
x = ± √3
Since the parabola y = x^2 is symmetric about the y-axis, we only need to consider the positive x-values.
Next, we need to determine the equation of the curve that gives the height of each rectangular cross section. In this case, the height is given by the function y^3.
Now, let's set up the integral to find the volume. Since the cross sections are perpendicular to the y-axis, we integrate with respect to y. The cross-sectional area of each rectangle is given by the product of its height (y^3) and the width (dy).
Therefore, the integral that represents the volume of the solid is:
V = ∫[from y=0 to y=3] (y^3) dy
Integrating this, we get:
V = ∫[from y=0 to y=3] y^3 dy = [y^4/4] [from y=0 to y=3] = (3^4/4) - (0^4/4) = 81/4
Hence, the volume of the solid is 81/4 cubic units.