anonymous
  • anonymous
Show geometrically why ∫sqrt(2-x^2) = (pi/4)+(1/2)
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
\[\int\limits_{0}^{1}\sqrt{2-x ^{2}}dx = \Pi /4+1/2\]
amistre64
  • amistre64
geometrically? wouldnt that just be a graph of the sqrt part?
anonymous
  • anonymous
You have to like break it into two peices

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anonymous
  • anonymous
a sector of a circle and a triangle
amistre64
  • amistre64
oh, then the graph would be useful at anyrate
amistre64
  • amistre64
http://www.wolframalpha.com/input/?i=sqrt%282-x%5E2%29%2C+real%2C+x%3D-2to2%2C+y%3D0to2 the wolf hates when I try to define stuff ....
JamesJ
  • JamesJ
right exactly. The segment of the circle has angle pi/4, or 1/8 of a complete circle of 2pi and radius sqrt(2). Hence the segment of circle has area \[ \frac{1}{8}\pi r^2 = \frac{2}{8}\pi = \frac{\pi}{4} \] Now, what about the triangle?
anonymous
  • anonymous
well the triangle wld be a special triangle
JamesJ
  • JamesJ
No, it would be a right angled triangle.
anonymous
  • anonymous
1.1 and sqrt(2)
anonymous
  • anonymous
ya a right triangle
JamesJ
  • JamesJ
what's the length of the base, b? And the height, h?
anonymous
  • anonymous
|dw:1327361596715:dw|
JamesJ
  • JamesJ
yes, exactly. And its area?
anonymous
  • anonymous
(b*H)/2
anonymous
  • anonymous
so it is a 1/2
anonymous
  • anonymous
ohhhhh I see
JamesJ
  • JamesJ
Nice question.
anonymous
  • anonymous
Thanks :D

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