Question

The base of a solid in the xy-plane is a circle with a radius of 3. cross sections of the solid perpendicular to the x-axis are squares. set up the integral to arrive at the volume of the solid and solve.

Answer 1

`\large \boxed{144}`

Explanation:

1. Set up the integral.

The equation for the circle is

x² + y² = 9

The bottom corners of the square are at

`(x, \sqrt{9 - x^(2)})\text{ and } (x, -\sqrt{9 - x^(2)})`

The length (a) of a side is

`a = 2\sqrt{9 - x^(2)}`

and the area (A) of the square cross-section is

A = a² = 4(9 - x²)

The volume (V) of the solid is

`V = \displaystyle \int_(-3)^(3) {4(9 - x^(2))} dx`

2. Solve the integral

`\displaystyle \int_(-3)^(3) {4(9 - x^(2))} dx = 4\begin{bmatrix}9x - (1)/(3)x^(3)\end{bmatrix}_(-3)^(3)= 4[(27 - 9) - (-27 +9)] = 4[18 - (-18)]\\= 4[18 + 18] = 4 *36 = \mathbf{144}\\\\\text{The volume of the solid is $\large \boxed{\mathbf{144}}$}`