Answer
$$\dfrac{\pi}{4}$$
Work Step by Step
Consider $I= \int_0^{\pi} \int_0^{\sin x} y dy dx= \int_0^{\pi} [\dfrac{y^2}{2}]_0^{\sin x} dx $
Apply the formula: $ \sin ^2 x= \dfrac{1- \cos 2 x}{2}$
Now, $I=(\dfrac{1}{4}) \int_0^{\pi} (1-\cos 2x) \space dx \\= (\dfrac{1}{4}) [\pi -\dfrac{\sin 2 \pi}{2} ] \\=\dfrac{\pi}{4}$$
