# Double Integral question

## Homework Statement

Evaluate ##\int\int_{R} (x+2)(y+1) \; dx \; dy## where ##R## is the pentagon with vertices ##(\pm 1,0)##, ##(\pm 2,1)## and ##(0,2)##.

## The Attempt at a Solution

After drawing ##R## I split ##R## into two sections ##R_1## (left half) and ##R_2## (right half) so we now calculate $$\int\int_{R_1} (x+2)(y+1) \; dx \; dy+\int\int_{R_2} (x+2)(y+1) \; dx \; dy.$$
From ##R_1## we have ##y = 2+\frac{x}{2} \Longleftrightarrow x = 2y-4##.and ##y = -1+x \Longleftrightarrow x = y+1##. From ##R_2## we have ##y = 2-\frac{x}{2} \Longleftrightarrow x = 4-2y## and ##y = -1-x \Longleftrightarrow x = -1-y##.
Does this inner integral become
$$\int_{y+1}^{2y-4} (x+2)(y+1) \; dx+\int_{-1-y}^{4-2y} (x+2)(y+1) \; dx.$$
Then what does the outer integral become (clearly the bounds of integration are scalars but what???)

Mark44
Mentor

## Homework Statement

Evaluate ##\int\int_{R} (x+2)(y+1) \; dx \; dy## where ##R## is the pentagon with vertices ##(\pm 1,0)##, ##(\pm 2,1)## and ##(0,2)##.

## The Attempt at a Solution

After drawing ##R## I split ##R## into two sections ##R_1## (left half) and ##R_2## (right half) so we now calculate $$\int\int_{R_1} (x+2)(y+1) \; dx \; dy+\int\int_{R_2} (x+2)(y+1) \; dx \; dy.$$
Put the limits of integration in both integrals, instead of just ##R_1## and ##R_2##. In both integrals, since you're integrating with respect to x first, the resulting outer integrand won't have any terms involving x.
squenshl said:
From ##R_1## we have ##y = 2+\frac{x}{2} \Longleftrightarrow x = 2y-4##.and ##y = -1+x \Longleftrightarrow x = y+1##. From ##R_2## we have ##y = 2-\frac{x}{2} \Longleftrightarrow x = 4-2y## and ##y = -1-x \Longleftrightarrow x = -1-y##.
Does this inner integral become
$$\int_{y+1}^{2y-4} (x+2)(y+1) \; dx+\int_{-1-y}^{4-2y} (x+2)(y+1) \; dx.$$
Then what does the outer integral become (clearly the bounds of integration are scalars but what???)

My solution is ##-\frac{106}{3}## which is clearly wrong as it has to be positive.
My limits of integration on ##y## is ##0## and ##-2## for ##R_1## and ##0## and ##2## for ##R_2##.