Finding Center of Mass for a Lamina in First Quadrant

kieranl
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Homework Statement



A lamina has the shape of the region in the first quadrant that is bounded by the graphs of y = sinx and y= cosx, between x = 0 and x = π/4. Find the centre of mass if the density is δ(x,y) = y.

Homework Equations



I know all the equations for moments and center of mass but I am confused about how to go about this problem. I don't know how to convert this to polar. The question is related to the lecture on polar integrals so I am assuming that's what has to be done?

The Attempt at a Solution

 
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Hi kieranl! :smile:
kieranl said:
A lamina has the shape of the region in the first quadrant that is bounded by the graphs of y = sinx and y= cosx, between x = 0 and x = π/4. Find the centre of mass if the density is δ(x,y) = y.

I know all the equations for moments and center of mass but I am confused about how to go about this problem. I don't know how to convert this to polar. The question is related to the lecture on polar integrals so I am assuming that's what has to be done?

Noooo …

you only convert coordinates if it makes the job easier …

for example, if the density function was δ(r,θ) …

in this case, δ = y (and also sin(rsinθ) is horrible ), so the easiest thing is to stay with the (x,y) coordinates. :wink:
 
Heya Kieranl

This is like your third curtin engineering question you've posted, so i'll assume you're probably doing mechanical engineering?

Alls I got to say is that this one took me like 5 pages of working out. I double checked it with maple so I know its right, but maybe i did it in some really long and complicated way. I didnt convert to polar format, although I wasnt sure if I was supposed to. I was just left with 4 integrals which I had to solve step by step.
 
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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