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Curves

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Joined: 05 Oct 2008
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24 Dec 2008, 04:31
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Curves $$x^2 + y^2 = 4$$ and $$y = |x|$$ enclose a sector on the top part of XY-plane. What is the area of this sector?

* $$\frac{\pi}{4}$$
* $$\frac{\pi}{2}$$
* $$\pi$$
* $$2\pi$$
* $$3\pi$$
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Joined: 01 Apr 2008
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Name: Ronak Amin
Schools: IIM Lucknow (IPMX) - Class of 2014
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24 Dec 2008, 05:08
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x^2 + y^2 = 4 is a circle with center (0,0) and radius 2. Plot it on a paper.
y=x is a line passing through (0,0) at 45degrees in the second quadrant.
y=-x is a line passing through (0,0) at 45 degrees in the first quadrant.

so the angle between two lines is 2(pi/4) = pi/2 = 90 degrees
area of sector = (pi * r^2 ) / 4 = pi
ans. C
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Joined: 31 Oct 2009
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02 Dec 2009, 13:22
I plotted both curves on paper and it was a little easier for me to see that way.

The first curve is a circle about the origin with a radius of 2
The second curve is a "V" shape--two $$45^{\circ}$$ lines going through the top 2 quadrants meeting at the origin

The area of the first curve = $$\pi2^2=4\pi$$
The area of the first curve in the top 2 quadrants is $$\frac{4\pi}{2}=2\pi$$

Each "leg" of the "V" of the second curve clearly bisects one quarter of the circle created by the first curve, so the area is $$\frac{2\pi}{2}=\pi$$

Where does this question come from?
Re: Curves   [#permalink] 02 Dec 2009, 13:22
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