#double integral of an absolute value is killing me
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The absolute value gives 0 when x+y-2 = 0 so y=-x+2
When y>-x+2 the absolute value does nothing, since x+y-2 is already positive. When y<-x+2 the value becomes negative, and the absolute value corrects this by multiplying by -1
So try to split the target area into two regions, split up by the line y=-x+2. You get a D+ and a D-, integrate over these areas, for D+ you just integrate x+y-2 and for D- you integrate -(x+y-2)
TLDR, get rid of the absolute value by splitting your target area D into regions where the effect of the absolute value is known, and add up the integrals over these areas
yeah i figured the splitting part however either im doing something wrong or the integral itself is just yucky to calculate because y is restricted by (2x-x^2)^(1/2)
ok so i think i got it, you have to split this region in three: the part where it lies directly under the semi-circle, the one which lies under y=2-x and the one over y=2-x and under the semi circle. to find the limits of integration you gotta find the points where sqrt(2x-x^2)=2-x which are 1 and two, then you split the integral in three, first, integral from 0 to 1 of the integral from 0 to sqrt(2x-x^2) of 2-x-y dxdy (since in this part it x+y-2 takes negative values, so you have to compensate it because of the absolute value), then the integral from 1 to two of the integral from 0 to 2-x of 2-x-y dydx and finally the integral from 1 to two of the integral from 2-x to sqrt(2x-x^2) of x+y-2 dydx, you evaluate those, you add them up and you should get the answer. i got 1 as the answer but i don't know if it's correct. please let me know if it works and if you understood it.
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Yeah thank you so much I figured it out finally
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