Answer:
A = 12√3
Step-by-step explanation:
first find the limits by finding the zeros
0 = 9 − 3x²
3x² = 9
x² = 3
x = ±[tex]\sqrt{3}[/tex]
[tex]A = \int\limits^a_b ({9 - 3x^2\\}) - 0\, dx[/tex]
where b = [tex]-\sqrt{3}[/tex] and a = [tex]\sqrt{3\\}[/tex]
A = 9x - x³ [tex]\left \{ {{\sqrt{3} } \atop {-\sqrt{3} }} \right.[/tex]
[tex]A = 9\sqrt{3} - \sqrt{3} ^3 - (9(-\sqrt{3)} - (-\sqrt{3} )^3)[/tex]
[tex]A = 9\sqrt{3} -3\sqrt{3} +9\sqrt{3} - 3\sqrt{3}[/tex]
[tex]A = 12\sqrt{3}[/tex]
