anonymous
  • anonymous
Find the center of mass of a thin plate covering the triangular region cut from the first quadrant by the line y = -x + 3, with density δ(x) = x.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Take rectangular strips of length dx at a distance x from origin perpendicular to y axis dm becomes density*V. Find V in terms of x and dx(You know the equation of the line). Then integrate from 0 to 3. y coordinate is same as that of a normal triangular plate.
anonymous
  • anonymous
|dw:1326959348751:dw|find side length and multiply by dx to find V
anonymous
  • anonymous
ok.thanks.

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anonymous
  • anonymous
i wanna ask one more question to u
anonymous
  • anonymous
\[\int\limits_{0}^{3}\]\[^{?}\sqrt{9-x^2}\]\[x^{2}\]dx
anonymous
  • anonymous
err i did not get that properly do you want to integrate sqrt(9-x^2) , what is that x^2 below it?
anonymous
  • anonymous
x^2.square root of 9-x^2 dx integrate this
anonymous
  • anonymous
from 0 to 3
anonymous
  • anonymous
put x=3cos(theta) i think just give me a minute.
anonymous
  • anonymous
ok
anonymous
  • anonymous
put x= 3cos t dt=-3sin t so the integral becomes -3(sin t) (3cos t)^2 sqrt(9 sin^2 t) = -81 sin^2 t cos ^2 t . sin ^2 t cos ^2 t=(sin ^2 (2t))/4.then sin ^2 (2t)=(1-cos 4t)/2 now you can integrate this easily.
anonymous
  • anonymous
in the first line it was dx=-3sint dt
anonymous
  • anonymous
Did you get it?
anonymous
  • anonymous
thnx
anonymous
  • anonymous
No problem.

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