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My objective:

Find the center of mass of a thin triangular plate bounded by the y-axis and the lines $y= 7x+3$ and $y= 36-4x$. Assume that the density is given by $\delta(x,y) = 7x+2y+2$.

In class we were given the formula $\bar x = \frac{\iiint_V x\delta dV}{\iiint_V \delta dV}$ to find the x center, so I set it up like $\bar x = \frac{\int_{0}^{3}\int_{0}^{36} x(7x+2y+2)\,dydx}{\int_{0}^{3}\int_{0}^{36} (7x+2y+2)\,dydx} = \frac{156}{97}$, but that isn't correct.

Is this approach wrong or did I make a mistake somewhere?

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    $\begingroup$ I think you integrated over a rectangle rather than a triangle. I think you have to adjust the boundaries of the inner integral to the given boundaries. $\endgroup$ – flawr Oct 27 '14 at 21:06
  • $\begingroup$ @flawr That's it. Thanks a lot! $\endgroup$ – Zach Saucier Oct 27 '14 at 21:11
  • $\begingroup$ In your last revision, the ratio you computed was invalid only because of the incorrect integrand on top. The bounds of integration were correct. So you really do have everything you need in order to compute $\overline{x}$. If you succeed, I'd encourage you to answer your own question so that we can give feedback. $\endgroup$ – Semiclassical Oct 27 '14 at 23:36
  • $\begingroup$ @flawr Feel free to post an answer. If you don't within a day or two I will $\endgroup$ – Zach Saucier Oct 28 '14 at 1:21
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I was taking the integral of a rectangle when I should have been taking it about the triangle.

As a result, the inner bounds should be changed and the the correct integral is $\bar x = \frac{\int_{0}^{3}\int_{7x+3}^{36-4x} x(7x+2y+2)\,dydx}{\int_{0}^{3}\int_{7x+3}^{36-4x} (7x+2y+2)\,dydx}$

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