Waste less time on Facebook — follow Brilliant.
×

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

No vote yet
1 vote

Comments

Sort by:

Top Newest

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 ago

Log in to reply

Thanks!!!.

Krishna Jha - 4 years ago

Log in to reply

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 ago

Log in to reply

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 ago

Log in to reply

@Mark Hennings Thanks Mark! :)

Pranav Arora - 4 years ago

Log in to reply

Great work!!

Piyal De - 4 years ago

Log in to reply

Nicely Done Mark !!

Gabriel Merces - 4 years ago

Log in to reply

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 ago

Log in to reply

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

Krishna Jha - 4 years ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...