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

Answer:

$\large \boxed{144}$

Step-by-step 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

$\displaystyle \int_{-3}^{3} {4(9 - x^{2})} dx = 4\begin{bmatrix}9x - \frac{1}{3}x^{3}\end{bmatrix}_{-3}^{3}= 4[(27 - 9) - (-27 +9)] = 4[18 - (-18)]\\= 4[18 + 18] = 4 \times36 = \mathbf{144}\\\\\text{The volume of the solid is \large \boxed{\mathbf{144}} }$