How Do You Calculate the Mass of a Triangle with Variable Density?

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hi i am having great difficulties doing this i was wondering if anyone could start me of thanks,
a triangle is formed by the x, y - axes and the line x+2y=4. the density is given by f(x,y)= 2y. I know that i need to convert the line for y but from there on i have no idea, much abliged
 
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take the double integer of 2y where y ranges from 0 to (x+4)/2 and x ranges from 0 to 4
 
thanks for that i have now got to the part of 10x/2 dx I am not to sure how to integrate this, should i do it in parts or substitution?
 
i think i got it now the answer is 40kg? is this correct
 
mp252 said:
i think i got it now the answer is 40kg? is this correct

No, the area of the triangle is 4 and the maximum density is 4. That can't possibly be right. You'd better show us what you did. BTW why was MiniST wrong in saying the limit of integration is (4+x)/2?
 
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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