Help ! Double integrals sign problem

[zetrox]

Hello guys i hope u re doing well ,im writing this post because im really confused in this case of integrals as u see in the picture the function (y= 3x-x^2 ) is above than the function (y=x),and the area is above the x-axis so why when im trying to calculate it , it gives a negative results , did i do something wrong, if no , please could u explain me why the result has a negative sign

 

[lex]

Above the line \(\displaystyle y=x, y>x\), and therefore with \(\displaystyle x>0, \hspace1ex xy>x^2 \text{ i.e. }x^2-xy<0\)

 

[Deleted member 4993]

How is the intergrand (x² - xy) shown is related to the problem?

 

[zetrox]

How is the intergrand (x² - xy) shown is related to the problem?

the full question to understand

 

[Deleted member 4993]

View attachment 28076
the full question to understand

You are NOT calculating area - you are calculating integral of a given function over a "domain" (area).

That integral does not HAVE to be positive.

 

[Eugenio]

View attachment 28076
the full question to understand

The integral does not have to be positive. Is not a volume but it's related. This relation can be expressed as

\(\displaystyle \iint_Sf(x,y)dxdy=(Volume\:upside\:z=0)-(Volume\:below\:z=0)\)

So in this case, the volume between the graph of f and the plane z = 0, below the plane, is greater than the volume upside the plane. And this is what this result is saying.