Area common to two regions

Consider the regions (x,y)x2+y2100(x,y)|\hspace{2mm}x^{2}+y^{2}\leq100 and sin(x+y)<0sin(x+y)<0.

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

Note by Krishna Jha
6 years, 1 month ago

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If A1={(x,y)x2+y2100,sin(x+y)>0}A_1 = \{(x,y)\,|\, x^2+y^2 \le 100 , \sin(x+y) > 0\} and A2={(x,y)x2+y2100:sin(x+y)<0}A_2 = \{(x,y) \,|\, x^2+y^2 \le 100: \sin(x+y) < 0\}, then a rotation through π\pi about the origin maps A1A_1 onto A2A_2, and vice versa. Thus A1A_1 and A2A_2 have the same area. The sets are disjoint, and their union is the interior of the circle (with 99 lines removed), and so the area of each is 12×102π=50π\tfrac12 \times 10^2\pi = 50\pi.

Mark Hennings - 6 years, 1 month ago

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Nicely Done Mark !!

Gabriel Merces - 6 years, 1 month ago

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Great work!!

Piyal De - 6 years, 1 month ago

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Hello Mark!

What does this statement mean: "then a rotation through ππ about the origin maps A1A_1 onto A2A_2, and vice versa."? Can you please post a link which explains this?

Thanks!

Pranav Arora - 6 years, 1 month ago

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The point (x,y)(x,y) belongs to A1A_1 if and only if the point (x,y)(-x,-y) belongs to A2A_2. Changing the sign of both xx and yy is achieved by a rotation about the origin through π\pi (or successive reflections in the xx- and yy-axes, or an enlargement scale factor 1-1 through the origin). However you describe this transformation of the plane R2\mathbb{R}^2, it preserves areas, so that A1A_1 and A2A_2 have the same area.

Mark Hennings - 6 years, 1 month ago

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@Mark Hennings Thanks Mark! :)

Pranav Arora - 6 years, 1 month ago

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Thanks!!!.

Krishna Jha - 6 years, 1 month ago

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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.

Vignesh Subramanian - 6 years, 1 month ago

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

Krishna Jha - 6 years, 1 month ago

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