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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 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 $$\tfrac12 \times 10^2\pi = 50\pi$$.

- 4 years ago

Thanks!!!.

- 4 years ago

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!

- 4 years ago

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.

- 4 years ago

Thanks Mark! :)

- 4 years ago

Great work!!

- 4 years ago

Nicely Done Mark !!

- 4 years ago

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.

- 4 years ago

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

- 4 years ago