×

# 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

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

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

- 4 years, 3 months ago

Log in to reply

Thanks!!!.

- 4 years, 3 months 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!

- 4 years, 3 months 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.

- 4 years, 3 months ago

Log in to reply

Thanks Mark! :)

- 4 years, 3 months ago

Log in to reply

Great work!!

- 4 years, 3 months ago

Log in to reply

Nicely Done Mark !!

- 4 years, 3 months 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.

- 4 years, 3 months ago

Log in to reply

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

- 4 years, 3 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...