Answer:
[tex]\large \boxed{144}[/tex]
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
[tex](x, \sqrt{9 - x^{2}})\text{ and } (x, -\sqrt{9 - x^{2}})[/tex]
The length (a) of a side is
[tex]a = 2\sqrt{9 - x^{2}}[/tex]
and the area (A) of the square cross-section is
A = a² = 4(9 - x²)
The volume (V) of the solid is
[tex]V = \displaystyle \int_{-3}^{3} {4(9 - x^{2})} dx[/tex]
2. Solve the integral
[tex]\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 $\large \boxed{\mathbf{144}}$}[/tex]
