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Question Number 124651 by benjo_mathlover last updated on 05/Dec/20

	Answered by liberty last updated on 05/Dec/20
	$for −3≤x≤3 → x^2 +y^2 = 9 ; g(x)=(√(9−x^2 )) for 3≤x≤6→(x−(9/2))^2 +y^2 = (9/4) ; g(x)=−(√((9/4)−(x−(9/2))^2 )) therefore g(x)= { (((√(9−x^2 )) ; −3≤x≤3)),((−(√((9/4)−(x−(9/2))^2 )) ; 3≤x≤6)) :} thus f(6)=∫_(−3) ^( 3) g(x)dx + ∫_3 ^( 6) g(x) dx$
	$f o r - 3 ⩽ x ⩽ 3 \to x^{2} + y^{2} = 9; g (x) = \sqrt{9 - x^{2}}$ $f o r 3 ⩽ x ⩽ 6 \to {(x - \frac{9}{2})}^{2} + y^{2} = \frac{9}{4}$ $; g (x) = - \sqrt{\frac{9}{4} - {(x - \frac{9}{2})}^{2}}$ $t h e r e f o r e g (x) = {\begin{cases} \sqrt{9 - x^{2}}; - 3 ⩽ x ⩽ 3 \\ - \sqrt{\frac{9}{4} - {(x - \frac{9}{2})}^{2}}; 3 ⩽ x ⩽ 6 \end{cases}$ $t h u s f (6) = \int_{- 3}^{3} g (x) d x + \int_{3}^{6} g (x) d x$
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