I am doing multivariable calculus, and specifically double integrals. I am facing difficulties finding the domain of the integal, however i am given the following equations: $$1 ≤ 2x+y ≤ 2$$ $$0 ≤ x-2y ≤ 1$$
Through these two equations i am supposed to find the area of integrals for each of the variables i.e $x$ and $y$
I set them as simultaneous inequalities but it doesn't seem to help because i get the boundary for $x$ to be $2/5 ≤ x ≤ 1 $ and for $y$ to be $4/5 ≤ y ≤ 0 $ which is obviously WRONG because how can the lowest boundary for possibly have higher value than that in the higher.
Guys, I really appreciate you help, it means the world to me.
Because you said you are doing double integral, if you want to keep using $x$ $y$ in the integral, then you may need to following clumsy way.
We have $$1-2x \le y \le 2-2x$$ and $$\frac{x-1}{2} \le y \le \frac{x}{2}$$
Now for $x>1$, we have $$2-2x < \frac{x-1}{2}$$ and hence there is no solution.
For $x<2/5$, we have $$\frac{x}{2}<1-2x$$ and so there is also no solution.
For $4/5\le x\le 1$, we have $2-2x \le x/2$ and $1-2x \le (x-1)/2$, hence $$\frac{x-1}{2} \le y \le 2-2x$$
For $3/5 \le x \le 4/5$, we have $x/2 \le 2-2x$ and $1-2x \le (x-1)/2$, hence $$ \frac{x-1}{2} \le y \le \frac{x}{2}$$
For $2/5 \le x \le 3/5$, we have $x/2 \le 2-2x$ and $(x-1)/2 \le 1-2x$, hence $$1-2x \le y \le \frac{x}{2}$$
So your integral $$\iint_D f(x,y)dydx=\int_{4/5}^1 \int_{(x-1)/2}^{2-2x}f(x,y)dy dx$$ $$+\int_{3/5}^{4/5}\int_{(x-1)/2}^{x/2}f(x,y)dydx+\int_{2/5}^{3/5}\int_{1-2x}^{x/2}f(x,y)dydx$$