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Area common to two regions

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.

Note by Krishna Jha
4 years, 3 months ago

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If \(A_1 = \{(x,y)\,|\, x^2+y^2 \le 100 , \sin(x+y) > 0\}\) 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 \(\tfrac12 \times 10^2\pi = 50\pi\).

Mark Hennings - 4 years, 3 months ago

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Thanks!!!.

Krishna Jha - 4 years, 3 months ago

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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!

Pranav Arora - 4 years, 3 months ago

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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.

Mark Hennings - 4 years, 3 months ago

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@Mark Hennings Thanks Mark! :)

Pranav Arora - 4 years, 3 months ago

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Great work!!

Piyal De - 4 years, 3 months ago

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Nicely Done Mark !!

Gabriel Merces - 4 years, 3 months 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 - 4 years, 3 months ago

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

Krishna Jha - 4 years, 3 months ago

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