The base of a solid is the region enclosed by y=x^3 and the x-axis on the interval [0,4]. Cross sections perpendicular to the x-axis are semicircles with diameter in the plain of the base. Write an integral that represents the volume of the solid.

I drew a picture, but I learned how to do this so long ago (in calc AB) that I don't remember how to start.

So, add up all the little wafers with diameter x^3, perpendicular to the x-y plane.

They would have their centers at (x,y/2) with radius y/2 = x^3/2. The volume of a small semi-circular wafer of radius r and thickness dx is 1/2 πr^2 dx

v = ∫[0,4] 1/2 π(x^3/2)^2 dx
= π/8 ∫[0,4] x^6 dx
...

No problem! I can help you remember how to approach this problem.

To find the volume of a solid, we need to integrate the areas of its cross sections. In this case, the cross sections are semicircles with diameters lying in the plane of the base.

Here's how you can start:

1. First, let's consider a generic vertical cross section. The radius of each semicircle will be half of the distance between the curve y = x^3 and the x-axis at a given x-value.

2. To determine the radius, we need to express the curve y = x^3 in terms of x. Since the curve lies on the interval [0,4], we can simply use the equation y = x^3.

3. The distance between the curve y = x^3 and the x-axis is given by the expression x^3 - 0 = x^3.

4. The radius of each semicircle is half of this distance, so the radius is (1/2)(x^3) = x^3/2.

5. The area of each semicircle is given by A = (π/2)(r^2), where r is the radius. In this case, the area becomes A = (π/2)(x^3/2)^2 = (π/2)(x^6/4) = (π/8)(x^6).

6. To find the volume of the solid, we need to integrate the areas of the cross sections over the interval [0,4]. The integral representation can be written as:
V = ∫[0,4] (π/8)(x^6) dx.

Therefore, the integral that represents the volume of the solid is ∫[0,4] (π/8)(x^6) dx.