Determine the center of mass of a triangular lamina shown whose density at any point is equal to its distance from the x-axis. The triangle has vertices at (0,0)(3,9)(7,0).
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Where's the "triangular lamina shown"? – Mar 04 '14 at 01:42
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Sorry I have no idea how to post the picture. The lamina is the line from point (0,0) to (3,9) – ayv2 Mar 04 '14 at 01:45
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What's the link? Tell me and I'll edit the question. – Mar 04 '14 at 01:46
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There are standard formulas, calculate the mass and the moments about the axes. – André Nicolas Mar 04 '14 at 01:49
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It can be done by one-variable methods, but because of the tag we use two variables. Let $T$ be our triangle. Then the mass $m$ is given by $$m=\iint_T y \,dx\,dy.$$ The moment $A$ about the $y$-axis is given by $$\iint_T xy \,dx\,dy.$$ The moment $B$ about the $x$-axis is given by $$\iint_T y^2 \,dx\,dy.$$ The centre of mass is $\left(\frac{A}{m},\frac{B}{m}\right)$.
There are $3$ double integrals to evaluate.
André Nicolas
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