Write the limits of integration for the double integral
Write the limits of integration for the double integral $\iint_D z(x,y)\,dx\,dy$ for both possible orders of integration. The region $D$ is a parallelogram with vertices $A(1,2)$, $B(2,4)$, $C(2,7)$, $D(1,5)$.
Explanation of Method
We verify the quadrilateral is a parallelogram, find the equations of its (slanted and vertical) sides, and describe it once as a vertically simple region ($dy\,dx$, which turns out simple) and once as a horizontally simple region ($dx\,dy$, which requires splitting into three pieces because the left/right boundary changes character with $y$).
- Verify opposite sides are parallel.
- Find the equations of all four sides.
- Describe $D$ for vertical strips (simple, one piece).
- Describe $D$ for horizontal strips (requires three pieces).
Step 1 · Verify the parallelogram
$$ \vec{AB}=(1,2),\quad \vec{DC}=(1,2),\quad \vec{AD}=(0,3),\quad \vec{BC}=(0,3). $$
Since $\vec{AB}=\vec{DC}$ and $\vec{AD}=\vec{BC}$, $ABCD$ is indeed a parallelogram. ✓
Step 2 · Find the equations of the sides
Side $AB$: slope $2$, so $y=2x$, for $1\leqslant x\leqslant2$.
Side $BC$: vertical line $x=2$, $4\leqslant y\leqslant7$.
Side $CD$: slope $2$, so $y=2x+3$, for $1\leqslant x\leqslant2$.
Side $DA$: vertical line $x=1$, $2\leqslant y\leqslant5$.
Step 3 · Describe the region
$D$ is bounded on the left by $x=1$, on the right by $x=2$, below by $y=2x$ (line $AB$), and above by $y=2x+3$ (line $CD$).
Step 4 · Order 1 — integrate $y$ first, then $x$
For each $x\in[1,2]$, $y$ runs from $y=2x$ up to $y=2x+3$:
$$ \iint_D z(x,y)\,dx\,dy = \int_1^2\int_{2x}^{2x+3} z(x,y)\,dy\,dx. $$
Step 5 · Order 2 — integrate $x$ first, then $y$
The vertical sides $x=1$ (for $2\leqslant y\leqslant5$) and $x=2$ (for $4\leqslant y\leqslant7$) overlap only in $4\leqslant y\leqslant5$, so the horizontal-strip description changes over three $y$-ranges:
For $2\leqslant y\leqslant4$: left boundary $x=1$, right boundary given by $y=2x\Rightarrow x=\frac y2$ (side $AB$).
For $4\leqslant y\leqslant5$: left boundary $x=1$, right boundary $x=2$ (both vertical sides active).
For $5\leqslant y\leqslant7$: left boundary given by $y=2x+3\Rightarrow x=\frac{y-3}{2}$ (side $CD$), right boundary $x=2$.
$$ \iint_D z(x,y)\,dx\,dy = \int_2^4\int_1^{y/2} z\,dx\,dy + \int_4^5\int_1^2 z\,dx\,dy + \int_5^7\int_{(y-3)/2}^2 z\,dx\,dy. $$
Summary Table
| Order | Limits |
|---|---|
| $dy\,dx$ | $1\leqslant x\leqslant2,\ 2x\leqslant y\leqslant2x+3$ |
| $dx\,dy$ (3 pieces) | $2\leqslant y\leqslant4$: $1\leqslant x\leqslant y/2$; $4\leqslant y\leqslant5$: $1\leqslant x\leqslant2$; $5\leqslant y\leqslant7$: $(y-3)/2\leqslant x\leqslant2$ |
Additional Notes
- The parallelogram is oriented so vertical strips (constant $x$) give a single simple integral, while horizontal strips (constant $y$) require splitting into three pieces because the slanted sides $y=2x$ and $y=2x+3$ only bound part of the $y$-range.
- This is a common feature of slanted parallelograms: one order of integration is much simpler than the other.
Result
The two equivalent forms are: