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anonymous
 4 years ago
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.
anonymous
 4 years ago
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.

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Take 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
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1326959348751:dwfind side length and multiply by dx to find V

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i wanna ask one more question to u

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{0}^{3}\]\[^{?}\sqrt{9x^2}\]\[x^{2}\]dx

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0err i did not get that properly do you want to integrate sqrt(9x^2) , what is that x^2 below it?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0x^2.square root of 9x^2 dx integrate this

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0put x=3cos(theta) i think just give me a minute.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0put 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)=(1cos 4t)/2 now you can integrate this easily.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0in the first line it was dx=3sint dt
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