Consider the regions \((x,y)|\hspace{2mm}x^{2}+y^{2}\leq100\) and \(sin(x+y)<0\).

Find the area of the region common to the above inequalities.

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TopNewestIf \(A_1 = \{(x,y)\,|\, x^2+y^2 \le 100 , \sin(x+y) > 0\}\) and \(A_2 = \{(x,y) \,|\, x^2+y^2 \le 100: \sin(x+y) < 0\}\), then a rotation through \(\pi\) about the origin maps \(A_1\) onto \(A_2\), and vice versa. Thus \(A_1\) and \(A_2\) have the same area. The sets are disjoint, and their union is the interior of the circle (with \(9\) lines removed), and so the area of each is \(\tfrac12 \times 10^2\pi = 50\pi\).

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Thanks!!!.

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Hello Mark!

What does this statement mean: "then a rotation through \(π\) about the origin maps \(A_1\) onto \(A_2\), and vice versa."? Can you please post a link which explains this?

Thanks!

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The point \((x,y)\) belongs to \(A_1\) if and only if the point \((-x,-y)\) belongs to \(A_2\). Changing the sign of both \(x\) and \(y\) is achieved by a rotation about the origin through \(\pi\) (or successive reflections in the \(x\)- and \(y\)-axes, or an enlargement scale factor \(-1\) through the origin). However you describe this transformation of the plane \(\mathbb{R}^2\), it preserves areas, so that \(A_1\) and \(A_2\) have the same area.

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Great work!!

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Nicely Done Mark !!

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area of x^2+y^2 is 100π.sin(x+y)>0 which implies 0>x+y>2π.....like that split the parts in which sin(x+y)>0if you see the graph it form a symetry and if you fold the circle you can see common region is 50π.

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Yes, even I did think the same way...thanks for ur answer

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