Calculate the following integral: double integral over D
Calculate the following integral: $$ \iint_D (x^2+y^2)\,dx\,dy, $$ where $D$ is the parallelogram with sides $y=x$, $y=x+a$, $y=a$, $y=3a$ ($a>0$).
Explanation of Method
The two slanted sides ($y=x$ and $y=x+a$) are parallel with slope $1$, and the two horizontal sides pin down $y\in[a,3a]$. Solving each slanted side for $x$ in terms of $y$ gives a clean strip $y-a\leqslant x\leqslant y$ for every $y$ in range, avoiding any splitting.
- Find the four vertices by intersecting the boundary lines.
- Set up the integral with $x$ inner (between the two slanted sides), $y$ outer.
- Integrate with respect to $x$, then $y$.
Step 1 · Find the vertices of the parallelogram
$y=x$ meets $y=a$ at $(a,a)$; $y=x$ meets $y=3a$ at $(3a,3a)$; $y=x+a$ meets $y=a$ at $(0,a)$; $y=x+a$ meets $y=3a$ at $(2a,3a)$. Vertices: $(0,a),(a,a),(3a,3a),(2a,3a)$.
Step 2 · Set up the integral
For each $y\in[a,3a]$, the left boundary is $y=x+a\Rightarrow x=y-a$, the right boundary is $y=x\Rightarrow x=y$:
$$ I = \int_a^{3a}\int_{y-a}^{y} (x^2+y^2)\,dx\,dy. $$
Step 3 · Integrate with respect to $x$
$$ \int_{y-a}^{y}(x^2+y^2)\,dx = \left[\frac{x^3}{3}\right]_{y-a}^{y} + y^2\left[x\right]_{y-a}^{y} = \frac{y^3-(y-a)^3}{3} + ay^2. $$
Expanding $(y-a)^3=y^3-3y^2a+3ya^2-a^3$:
$$ \frac{3y^2a-3ya^2+a^3}{3}+ay^2 = ay^2-ya^2+\frac{a^3}{3}+ay^2 = 2ay^2-a^2y+\frac{a^3}{3}. $$
Step 4 · Integrate with respect to $y$
$$ I = \int_a^{3a}\left(2ay^2-a^2y+\frac{a^3}{3}\right)dy = 2a\left[\frac{y^3}{3}\right]_a^{3a} - a^2\left[\frac{y^2}{2}\right]_a^{3a} + \frac{a^3}{3}[y]_a^{3a}. $$
$$ = 2a\cdot\frac{26a^3}{3} - a^2\cdot\frac{8a^2}{2} + \frac{a^3}{3}\cdot2a = \frac{52a^4}{3} - 4a^4 + \frac{2a^4}{3} = 18a^4-4a^4 = 14a^4. $$
Additional Notes
- Because the two slanted sides are parallel (both slope $1$), integrating with $x$ inner between them at every height $y$ requires no splitting — this is generally true for any parallelogram with one pair of horizontal sides.
Result
The solution to the integral is: