# Composing an iterated double integral given the equation of three lines which form a triangle.

I am being asked to compose two iterated double integrals given a specific region. I have to compose a type I and a type II iterated integral for the region (I do not need to evaluate the integral, just compose it) The region given is; the region R to the right of the line $x = −2$, below the line $y = 3$, and above the line $y = 1/2x$.

The answer I came up with for the integrals was $\int_{-2}^6(\int_{-1}^{(1/2)x} f(x,y) dy) dx$ and $\int_{-1}^3(\int_{-2}^{2y} f(x,y) dx) dy$

Do these answers make sense? Having a bit of difficulty with this concept when given equations of lines.

No, the domain of integration is above $y=\frac12x$.

As shown in the figure below. If the outer integral is with respect to $x$ then the limits for $y$ are $y=\frac12x$ and $y=3$. If the outer integral is $y$ then the limits with respect to $x$ are $-2$ and $x=2y.$ Accordingly the two integrals are

$$(a) \,\, \int_{-2}^6\ \int_{\frac x2}^3\ f(x,y) \ dy \ dx$$

and

$$(b) \,\, \int_{-1}^3\ \int_{-2}^{2y}\ f(x,y) \ dx \ dy.$$