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October 1st, 2012, 02:00 AM   #1
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find the area of the region

i got trouble of this question, anyone can help?

The figure shows a shaded region consisting of all
points inside a square (with sides of 2 cm) that are closer to the centre than to the sides of the square. Find the area of the region. You must show all appropriate working.

thanks.
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 1.png (103.5 KB, 1166 views)

October 1st, 2012, 02:45 AM   #2
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From: St. Augustine, FL., U.S.A.'s oldest city

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Math Focus: Calculus/ODEs
Re: find the area of the region

The locus of the set of points equidistant to a point and a line is by definition a parabola. I would use symmetry to find 1/8 of the area as in the diagram:

[attachment=0:2lsnpr1j]area001.jpg[/attachment:2lsnpr1j]

We have placed the center of the square at the origin. The square in the diagram is the upper right quadrant of the entire square. The parabola with its focus at the origin and its directrix as the line $y=1$ is:

$y=\frac{2-x^2}{4}$

This intersects the line $y=x$ in the first quadrant for $x=\sqrt{6}-2$

Hence, the requested area is:

$A=8\int_{0}\,^{\sqrt{6}-2} \frac{2-x^2}{4}-x\,dx=2\int_{0}\,^{\sqrt{6}-2} 2-4x-x^2\,dx=8$$\sqrt{6}-\frac{7}{3}$$$
Attached Images
 area001.jpg (10.0 KB, 1165 views)

 October 1st, 2012, 04:01 AM #3 Member   Joined: Oct 2012 Posts: 30 Thanks: 0 Re: find the area of the region i'm sorry but can i ask how you get the x=sqrt{6}-2 thanks.
 October 1st, 2012, 08:19 AM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,209 Thanks: 517 Math Focus: Calculus/ODEs Re: find the area of the region $\frac{2-x^2}{4}=x$ $2-x^2=4x$ $x^2+4x-2=0$ Take the positive root, by the quadratic formula: $x=\frac{-4+\sqrt{4^2-4(1)(-2)}}{2(1)}=\frac{-4+\sqrt{24}}{2}=\frac{-4+2\sqrt{6}}{2}=\sqrt{6}-2$
 October 1st, 2012, 03:46 PM #5 Member   Joined: Oct 2012 Posts: 30 Thanks: 0 Re: find the area of the region thanks so much!!
 October 2nd, 2012, 08:56 AM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,209 Thanks: 517 Math Focus: Calculus/ODEs Re: find the area of the region I made an error in my computation above, specifically in finding the equation of the parabola. It should be: $y=\frac{1-x^2}{2}$ Equating this to $x$, we find the intersection is at: $x=\sqrt{2}-1$ And so the area A in question is: $A=4\int_0\,^{\sqrt{2}-1}1-2x-x^2\,dx=\frac{4}{3}$$4\sqrt{2}-5$$$
 October 3rd, 2012, 12:05 AM #7 Member   Joined: Oct 2012 Posts: 30 Thanks: 0 Re: find the area of the region so did you divide that by 4 in stand of 8?
 October 3rd, 2012, 12:12 AM #8 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,209 Thanks: 517 Math Focus: Calculus/ODEs Re: find the area of the region I still divided the region into 8 equal portions, but the denominator for the parabolic curve was 2 instead of 4, so the outside constant is 4 instead of 2.
 October 3rd, 2012, 04:32 AM #9 Newbie   Joined: Oct 2012 Posts: 4 Thanks: 0 Re: find the area of the region Sorry, but how did you get the equation y = (1-x^2)/2? I tried using the distance formula and equated it with y and I got y = (x^2 + 1)/2?
 October 3rd, 2012, 04:48 AM #10 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,209 Thanks: 517 Math Focus: Calculus/ODEs Re: find the area of the region We want the point(s) (x,y) equidistant from the point (0,0) and the line $y=1$, hence: $x^2+y^2=(y-1)^2$ $x^2+y^2=y^2-2y+1$ $x^2=-2y+1$ $2y=1-x^2$ $y=\frac{1-x^2}{2}$

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# the shaded region in the figure consists of all points whose distances from the center of the square are less than their distances from

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