Calculate the following integral: double integral over D
Calculate the following integral: $$ \iint_D \frac{x}{2}\,dx\,dy, $$ where the region $D$ is bounded by the lines $x=0$, $x=2+\sin y$, $y=0$, $y=2\pi$.
Explanation of Method
With $y$ as the outer variable, the inner integral in $x$ is elementary; the resulting integrand in $y$ expands into $1+\sin y+\frac{\sin^2y}{4}$, each term integrable over a full period using standard identities.
- Integrate with respect to $x$ from $0$ to $2+\sin y$.
- Expand $(2+\sin y)^2$.
- Integrate term by term over $[0,2\pi]$, using $\sin^2y=\frac{1-\cos2y}{2}$.
Step 1 · Set up the integral
$$ I = \int_0^{2\pi}\int_0^{2+\sin y} \frac x2\,dx\,dy. $$
Step 2 · Integrate with respect to $x$
$$ \int_0^{2+\sin y} \frac x2\,dx = \frac12\left[\frac{x^2}{2}\right]_0^{2+\sin y} = \frac{(2+\sin y)^2}{4}. $$
Step 3 · Expand the expression
$(2+\sin y)^2=4+4\sin y+\sin^2y$, so
$$ \frac{(2+\sin y)^2}{4} = 1+\sin y+\frac{\sin^2y}{4}. $$
Step 4 · Integrate with respect to $y$
$$ I = \int_0^{2\pi} 1\,dy + \int_0^{2\pi} \sin y\,dy + \frac14\int_0^{2\pi} \sin^2y\,dy. $$
First: $\int_0^{2\pi}1\,dy=2\pi$. Second: $[-\cos y]_0^{2\pi}=-1+1=0$. Third, using $\sin^2y=\frac{1-\cos2y}{2}$:
$$ \int_0^{2\pi}\sin^2y\,dy = \frac12\left[y-\frac{\sin2y}{2}\right]_0^{2\pi} = \frac12(2\pi)=\pi. $$
Step 5 · Combine the results
$$ I = 2\pi + 0 + \frac14\cdot\pi = \frac{8\pi+\pi}{4} = \frac{9\pi}{4}. $$
Additional Notes
- The right boundary $x=2+\sin y$ never becomes negative (since $\sin y\geqslant-1$), so it remains a valid single-valued right boundary for all $y\in[0,2\pi]$.
- Integrating $\sin y$ and $\sin^2y$ over a full period $[0,2\pi]$ is a standard shortcut: the odd-harmonic term vanishes, while the squared term averages to $\frac12$ of the interval length.
Result
The solution to the integral is: