Calculate the following integral: double integral over D

Calculate the following integral: $$ \iint_D (x+y)\,dx\,dy, $$ where the region $D$ is bounded by the parabola $y^2=2x$ and the line $x+y=4$.

Step 1 · Find intersection points

From the line, $x=4-y$. Substituting into $y^2=2x$: $y^2=2(4-y)=8-2y \Rightarrow y^2+2y-8=0 \Rightarrow (y+4)(y-2)=0$, so $y=-4$ or $y=2$. At $y=2$: $x=2$; at $y=-4$: $x=8$. Intersection points: $(2,2)$ and $(8,-4)$.

Step 2 · Set up the integral

For each $y\in[-4,2]$, $x$ runs from the parabola $x=\frac{y^2}{2}$ to the line $x=4-y$:

$$ I = \int_{-4}^{2}\int_{y^2/2}^{4-y} (x+y)\,dx\,dy. $$

Step 3 · Integrate with respect to $x$

$$ \int_{y^2/2}^{4-y} (x+y)\,dx = \left[\frac{x^2}{2}+xy\right]_{y^2/2}^{4-y}. $$

At $x=4-y$: $\dfrac{(4-y)^2}{2}+(4-y)y = \dfrac{16-8y+y^2}{2}+4y-y^2 = 8-4y+\dfrac{y^2}{2}+4y-y^2 = 8-\dfrac{y^2}{2}$.

At $x=\dfrac{y^2}{2}$: $\dfrac{1}{2}\left(\dfrac{y^2}{2}\right)^2+\dfrac{y^2}{2}\cdot y = \dfrac{y^4}{8}+\dfrac{y^3}{2}$.

$$ \int_{y^2/2}^{4-y} (x+y)\,dx = 8-\frac{y^2}{2}-\frac{y^4}{8}-\frac{y^3}{2}. $$

Step 4 · Integrate with respect to $y$

$$ I = \int_{-4}^{2}\left(8-\frac{y^2}{2}-\frac{y^3}{2}-\frac{y^4}{8}\right)dy = \left[8y-\frac{y^3}{6}-\frac{y^4}{8}-\frac{y^5}{40}\right]_{-4}^{2}. $$

At $y=2$: $16-\dfrac{8}{6}-\dfrac{16}{8}-\dfrac{32}{40} = 16-\dfrac{4}{3}-2-\dfrac{4}{5} = 14-\dfrac{4}{3}-\dfrac{4}{5} = \dfrac{210-20-12}{15} = \dfrac{178}{15}$.

At $y=-4$: $-32-\dfrac{-64}{6}-\dfrac{256}{8}-\dfrac{-1024}{40} = -32+\dfrac{32}{3}-32+\dfrac{128}{5} = -64+\dfrac{32}{3}+\dfrac{128}{5} = \dfrac{-960+160+384}{15} = \dfrac{-416}{15}$.

Step 5 · Combine the results

$$ I = \frac{178}{15}-\left(-\frac{416}{15}\right) = \frac{594}{15} = \frac{198}{5}. $$

Additional Notes

• Treating $y$ as the outer variable avoids splitting the region into two pieces, since the parabola $y^2=2x$ is not a single-valued function of $x$ but is single-valued as $x=\frac{y^2}{2}$ in terms of $y$.