Answer
$$ \iint \limits_{D} xd A=\pi$$
Work Step by Step
Given $$ \iint \limits_{D} xd A$$
$$D=\{(x, y) |0\leq x \leq \pi,0\leq y \leq sinx\}$$
So, we have
\begin{aligned} A&=\iint \limits_{D} x \ d A\\
&=\int_{0}^{\pi} \int_{0}^{sinx} x \ d y \ d x\\
&=\int_{0}^{\pi} x\left[ y\right]_{0}^{sinx} \ d x\\&
= \int_{0}^{\pi}x sinx \ d x\\
\end{aligned}
So, by the partition technique, let $$u=x \Rightarrow du= dx$$ $$dv=\sin x \ dx \Rightarrow v= - \cos x $$ So, we get
\begin{aligned} I &=\left[uv \right]_{0}^{\pi}- \int_{0}^{pi} vdu\\ &
=\left[-x \cos x \right]_{0}^{\pi}+ \int_{0}^{\pi} \cos x \ d x\\ &=\left[-x \cos x \right]_{0}^{\pi}+\left[ \sin x \right]_{0}^{\pi} \\
&=\left[-\pi \cos\pi -0\right]+\left[ \sin \pi -\sin 0 \right] \\
&= \pi \end{aligned}
