Answer
$\approx 0.102$
Work Step by Step
The volume under the surface $z=f(x,y)$ and above the region $D$ in the xy-plane can be expressed as: $$\ Volume =\iint_{D} f(x,y) \space dA$$
The region $D$ can be defined as:
$$D=\left\{ (x, y) | x \leq y \leq \cos x, \ 0 \leq x \leq 0.739 \right\}
$$
Now, $V = \int_{0}^{0.739} \int_{x}^{\cos x}x dy \ dx \\ = \int_{0}^{0.739} [xy]_{x}^{\cos x} \ dx \\= \int_{0}^{0.739} x \cos x dx- \int_{0}^{0.739} x^2 dx \\=[ x \sin x -\int \sin x \ dx]_0^{0.739} -[\dfrac{x^3}{3}]_0^{0.739} \\=[ x \sin x+\cos x]_0^{0.739} -[\dfrac{(0.739)^3}{3}-0] \\ =(0.739) \sin (0.739)+\cos (0.739)-1 -0.1345 \\ \approx 0.102$
