<s>[INST] Given a general region, find the double integral bounded between $y = x$ and $y=3x-x^2 $ $$ J  = \iint_R (x^2-xy)\,dx \,dy, $$
Suppose region R is bounded between $y = x$ and $y=3x-x^2 $
My attempt using vertical integration:
$$ \int^{x=2}_{x=0} \int^{y=3x-x^2}_{y=x} \left({x^2-xy}\right)dy\ dx$$
$$\int^2_0 \left[x^2y-x\frac{y^2}{2}\right]^{3x-x^2}_{x}\, dx$$
$$\int^2_0 \frac{-x^5+4x^4-4x^3}{2} \,dx $$
$$\boxed{J = -\frac{8}{15}}$$

My attempt using horizontal integration :
$$ \int^{y=2}_{y=0} \int^{x=y}_{x=3\,\pm \sqrt{9-y}} \left({x^2-xy}\right)dx\ dy$$
For $ x = 3+\sqrt{9-y}$
$$  \int^2_0 \left[\frac{x^3}{3}-\frac{x^2}{2}y\right]^y_{3+\sqrt{9-y} }\,dy$$
For $ x = 3-\sqrt{9-y}$
$$  \int^2_0 \left[\frac{x^3}{3}-\frac{x^2}{2}y\right]^y_{3-\sqrt{9-y} }\,dy$$

My doubts :
1.) How do I set my limit of integration for horizontal integration, if there is $\pm$ to be considered ?
2.) the answer as negative what does that imply in questions related to double integrals?
Could you guys please help
 [/INST] Note that in your second case, your region is divided into two (under and above the green line). When you find the inverse functions of $y(x)$'s you have $$x(y) = y$$
for $y = x$ and $$x(y) = \frac{3 \pm \sqrt{9-4y}}{2}$$ (separate them into two at $x = 3/2)$ for $y = 3x-x^2$
Now, move your pen from left to right to notice what should be the bounds for integration over each region.

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