Correct Method for Finding Area Between 2 Curves Using Integration

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Homework Statement


The picture is found here (top picture):
http://www.iastate.edu/~statics/examples/centroid/centroida.html#ysubc



Homework Equations





The Attempt at a Solution


I get the wrong sign (ie: negative area) when I integrate to find the area between 2 curves when I integrate with respect to the y-axis (ie: when I use dy). So since my area is negative, I must be subtracting the wrong curve from the other curve (ie: instead of F-G, it should be G-F). Should it be [(top curve)-(bottom curve)] ? That's what I did, but it appears that [(bottom curve)-(top curve)] gives the correct answer. By the way, my work (attached pict) also shows me using dx just to prove to myself that it works, which it did. Thanks for any logic help.
 

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Think about finding the area by using a whole bunch of rectangles. Since you're integrating with respect to the y-axis, you'll have a lot of skinny rectangles oriented in the left-right direction. What's this about top/bottom then? Wouldn't you want to take the function on the right and subtract the function on the left to find the height of your rectangle?
 
makes total sense pizza (when you think about it physically). I didn't initially interpret it physically. The "top curve minus bottom curve" to find the area between 2 curves works only for dx; that's what the teacher told us. But for dy, the "top curve" is the right curve. I can see it now.
 
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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