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.

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.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

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 \(\tfrac12 \times 10^2\pi = 50\pi\). – Mark Hennings · 3 years, 10 months ago

Log in to reply

– Krishna Jha · 3 years, 10 months ago

Thanks!!!.Log in to reply

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 · 3 years, 10 months ago

Log in to reply

– Mark Hennings · 3 years, 10 months 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.Log in to reply

– Pranav Arora · 3 years, 10 months ago

Thanks Mark! :)Log in to reply

– Piyal De · 3 years, 10 months ago

Great work!!Log in to reply

– Gabriel Merces · 3 years, 10 months ago

Nicely Done Mark !!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 · 3 years, 10 months ago

Log in to reply

– Krishna Jha · 3 years, 10 months ago

Yes, even I did think the same way...thanks for ur answerLog in to reply