# 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???)

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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 $R_2$.