A figure is formed by revolving the region bounded by $f(x) = \cos{(x)}$ and $g(x) = \sin{(x)}$ from $0$ to $\dfrac{\pi}{4}$ about the line $y=-1$. This figure is formed by integration of two concentric circles with radii $f$ and $g$ where $f > g$. Since the axis of revolution is shifted down $1$, the functions are now $1$ greater than the axis $\implies V=\int_{0}^{\pi/4}((\cos{(x)}+1)^2-(\sin{(x)}+1)^2)dx$.
Is this true?
Yep,
You're almost right, just forgot the $\pi$.
$V=\pi \int\limits_0^{\frac{\pi }{4}} (\cos (\theta )+1)^2-(\sin (\theta )+1)^2 \, d\theta =\left(2 \sqrt{2}-\frac{3}{2}\right) \pi$