Can you find a shorter way to do this?

Let $f(x) = \sin x = y$ . Then

$\begin{aligned} V & = \int_0^1 \pi x^2 \ dy & \small \color{#3D99F6} \text{Since } y = \sin x \implies dy = \cos x \ dx \\ & = \pi \int_0^\frac \pi 2 x^2 \cos x\ dx & \small \color{#3D99F6} \text{By integration by parts} \\ & = \pi \left[x^2 \sin x - \int 2x \sin x \ dx \right]_0^\frac \pi 2 \\ & = \pi \left[\frac {\pi^2}4 + 2x \cos x - 2 \int \cos x \ dx \right]_0^\frac \pi 2 \\ & = \pi \left[\frac {\pi^2}4 - 2 \sin x \right]_0^\frac \pi 2 \\ & = \frac 14 \pi^3 - 2\pi \end{aligned}$

Therefore, $a+b+c = 1+4-2 = \boxed 3$ .

The volume V is the integral of π(x^2)dy within the limits 0 to 1, which is the same as the integral of π(x^2)cos(x)dx within the limits 0 to π/2. Integrating by parts, we get the indefinite integral as (x^2)sin(x)+2xcos(x)-2sin(x)+C. The value of the definite integral is (π/2)^2 -2. Hence the volume is ((π^3)/4)-2π, so that a=1, b=4, c=-2, making a+b+c=3