The base of a solid is the circle x2 + y2 = 9. Cross sections of the solid perpendicular to the x-axis are semi-circles. What is the volume, in cubic units, of the solid?

To find the volume of the solid, we need to integrate the area of the circular cross-sections perpendicular to the x-axis.

Given that the equation of the circle is x^2 + y^2 = 9, we can solve for y in terms of x by taking the square root of both sides:

y = sqrt(9 - x^2)

The radius of each semi-circle is given by y, so the area of each semi-circle is given by:

A = (1/2) * π * r^2
= (1/2) * π * y^2

Substituting the value of y, we get:

A = (1/2) * π * (sqrt(9 - x^2))^2
= (1/2) * π * (9 - x^2)

Now, to find the volume, we integrate the area function from -3 to 3 (since the range of x for the circle is -3 to 3):

V = ∫[from -3 to 3] (1/2) * π * (9 - x^2) dx

Evaluating the integral:

V = (1/2) * π * ∫[from -3 to 3] (9 - x^2) dx
= (1/2) * π * [9x - (x^3/3)] [from -3 to 3]
= (1/2) * π * [(9(3) - (3^3/3)) - (9(-3) - ((-3)^3)/3)]
= (1/2) * π * [27 - 9 - (-27 - 9)]
= (1/2) * π * [36]

Simplifying further, we get:

V = 18π

Therefore, the volume of the solid is 18π cubic units.

To find the volume of the solid, we need to integrate the areas of the semi-circles formed by the cross sections perpendicular to the x-axis.

First, let's consider a typical cross section of the solid. The circle x^2 + y^2 = 9 can be rewritten as y = sqrt(9 - x^2) or y = -sqrt(9 - x^2). Since we know the cross sections are semi-circles, we only need to consider the top half of the circle.

To find the area of a semi-circle, we can use the formula A = (1/2) * π * r^2, where r is the radius. In this case, the radius is the y-coordinate of the circle, which is sqrt(9 - x^2).

To find the limits of integration, we need to determine the x-values at which the semi-circles start and end. From the equation of the circle, we know that -3 ≤ x ≤ 3. Therefore, the limits of integration for the volume are -3 to 3.

The volume V of the solid can be found by integrating the area of the semi-circle over the range of x:

V = ∫[-3, 3] (1/2) * π * (sqrt(9 - x^2))^2 dx.

Simplifying,

V = (1/2) * π * ∫[-3, 3] (9 - x^2) dx.

Now we can integrate:

V = (1/2) * π * [9x - (1/3)x^3] |[-3, 3].

Evaluating the definite integral,

V = (1/2) * π * [9(3) - (1/3)(3)^3] - [(1/2) * π * (9(-3) - (1/3)(-3)^3)]

V = (1/2) * π * [27 - 9] - [(1/2) * π * (-27 - 9)]

V = (1/2) * π * [18] - [(1/2) * π * (-36)]

V = 9π + 18π

V = 27π.

Therefore, the volume of the solid is 27π cubic units.

the description is of a hemisphere

of radius 3

v = 2/3 π r³