I assume the cross sections are taken perpendicular to the x-axis? This seems more likely than relative to the y-axis as far as easiness of calculation goes.
The base of each triangle is then determined by the distance between \(\sqrt{\cos x}\) and the x-axis, or simply \(\sqrt{\cos x}\). Because it's a right triangle, you know the legs' lengths occur in a 1:1 ratio. Since each triangular cross section has one of its legs as its base, the heights must be the same as their bases.
So, the area of any one cross-section is
\(A(x)=\dfrac12(\sqrt{\cos x})^2=\dfrac{\cos x}2\)
Then the volume of this region would be
\(\displaystyle\int_{-\pi/2}^{\pi/2}\frac{\cos x}2\,\mathrm dx=1\)
