Finding Centroid of Triangle in First Quadrant

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Homework Statement


Find the centroid of the triangular region in the first quadrant bounded by the line 4x+y=4, the x-axis, and the y-axis.


Homework Equations


\int\int\deltadA
xbar = My/M
ybar = Mx/M

The Attempt at a Solution


Is the boundary for the dy from 0 to 4, and dx from 0 to 4-4x?
 
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No, your limits for x can't contain x. They can only depend on y.
 
so its y-4/4?
also are my limits for dy correct?
 
No, that's not quite correct either. (I assume you mean (y-4)/4, not y-1.) You should always plug in a few values to do a sanity check. In this case, when y=0, you'd get a negative answer.

Your limits for y are fine. (But I'll note I'm saying this under the assumption that you set things up correctly.)

Did you draw a sketch of the region of integration?
 
the region would be the triangle from y=4 to y=0 and x=0 and x=1 right?
and is the limit 4-4x?
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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