A curve P is defined by $y = -\sin^2 x$ where $-\pi \le x \le 0$. The finite region R is bounded by the curve P, the y-axis and the lines $x = -\frac{\pi}{2}$, $y = -2$.

1. Find the first derivative of the equation of the curve and set it to zero to locate stationary points.

$y = -\sin^2 x$
$\frac{dy}{dx} = -2\sin x \cos x$
For stationary points, $\frac{dy}{dx} = 0$:
$-2\sin x \cos x = 0$
$\sin x = 0 \quad \text{or} \quad \cos x = 0$

2. Solve for $x$ in the given domain and evaluate $y$ to find the exact coordinates of the minimum point.

Given the domain $-\pi \le x \le 0$:
If $\sin x = 0$, $x = -\pi \text{ or } 0$.
If $\cos x = 0$, $x = -\frac{\pi}{2}$.

Evaluating the corresponding $y$-values:
At $x = -\pi$, $y = -\sin^2(-\pi) = 0$
At $x = 0$, $y = -\sin^2(0) = 0$
At $x = -\frac{\pi}{2}$, $y = -\left(\sin\left(-\frac{\pi}{2}\right)\right)^2 = -(-1)^2 = -1$

Comparing the $y$-values, the minimum occurs at $x = -\frac{\pi}{2}$.
The coordinates of the minimum point are $(-\frac{\pi}{2}, -1)$.

1. Set up the definite integral for the area of the region bounded by the given boundaries.

The area of the region $R$ is given by the integral of the upper boundary minus the lower boundary:
$\text{Area} = \int_{-\frac{\pi}{2}}^{0} (-\sin^2 x - (-2)) \,dx$
$\text{Area} = \int_{-\frac{\pi}{2}}^{0} (2 - \sin^2 x) \,dx$

2. Use the double angle identity to express $\sin^2 x$ in an integrable form.

Using the double angle identity $\sin^2 x = \frac{1 - \cos 2x}{2}$:
$\text{Area} = \int_{-\frac{\pi}{2}}^{0} \left(2 - \left(\frac{1 - \cos 2x}{2}\right)\right) \,dx$
$= \int_{-\frac{\pi}{2}}^{0} \left(\frac{3}{2} + \frac{1}{2}\cos 2x\right) \,dx$

3. Integrate and evaluate using the limits to find the exact area.

Integrating the expression:
$\text{Area} = \left[ \frac{3}{2}x + \frac{1}{4}\sin 2x \right]_{-\frac{\pi}{2}}^{0}$
Evaluating at the limits:
$= \left( \frac{3}{2}(0) + \frac{1}{4}\sin(0) \right) - \left( \frac{3}{2}\left(-\frac{\pi}{2}\right) + \frac{1}{4}\sin(-\pi) \right)$
$= 0 - \left( -\frac{3\pi}{4} + 0 \right)$
$= \frac{3\pi}{4}$

1. Set up the integral for the volume of revolution about the $y$-axis by splitting the region horizontally.

Let the total volume be $V$. The region $R$ must be split into two horizontal parts: a lower rectangle for $-2 \le y \le -1$, and the upper region between the curve and the $y$-axis for $-1 \le y \le 0$.

For $-2 \le y \le -1$, rotating the rectangle (from $x=-\frac{\pi}{2}$ to $x=0$) around the $y$-axis forms a solid cylinder of radius $\frac{\pi}{2}$.
$V_1 = \int_{-2}^{-1} \pi \left(-\frac{\pi}{2}\right)^2 dy = \left[ \frac{\pi^3}{4} y \right]_{-2}^{-1} = \frac{\pi^3}{4}$

For $-1 \le y \le 0$, the region is bounded by the curve $x(y)$ on the left and $x=0$ on the right.
$V_2 = \int_{-1}^{0} \pi x^2 dy$

The total volume is $V = \frac{\pi^3}{4} + \int_{-1}^{0} \pi x^2 dy$.

2. Use substitution to change the variable of integration from $y$ to $x$ for the second integral.

Given the curve $y = -\sin^2 x$, we find the differential $dy$:
$\frac{dy}{dx} = -2\sin x \cos x = -\sin 2x \implies dy = -\sin 2x \, dx$

Determine the new limits:
When $y = -1$, $-\sin^2 x = -1 \implies \sin x = -1 \implies x = -\frac{\pi}{2}$.
When $y = 0$, $-\sin^2 x = 0 \implies \sin x = 0 \implies x = 0$.

Substitute these into the integral for $V_2$:
$V_2 = \int_{-1}^{0} \pi x^2 dy = \int_{-\pi/2}^{0} \pi x^2 (-\sin 2x) dx = -\pi \int_{-\pi/2}^{0} x^2 \sin 2x \, dx$

3. Evaluate the integral $-\pi \int x^2 \sin 2x \, dx$ using integration by parts.

Let $u = x^2$ and $\frac{dv}{dx} = \sin 2x$. Then $\frac{du}{dx} = 2x$ and $v = -\frac{1}{2}\cos 2x$.

$-\pi \int_{-\pi/2}^{0} x^2 \sin 2x \, dx = -\pi \left( \left[ -\frac{1}{2}x^2 \cos 2x \right]_{-\pi/2}^{0} - \int_{-\pi/2}^{0} -x \cos 2x \, dx \right)$

Evaluate the boundary terms:
$\left[ -\frac{1}{2}x^2 \cos 2x \right]_{-\pi/2}^{0} = 0 - \left( -\frac{1}{2}\left(-\frac{\pi}{2}\right)^2 \cos(-\pi) \right) = 0 - \left( -\frac{1}{2} \cdot \frac{\pi^2}{4} \cdot (-1) \right) = -\frac{\pi^2}{8}$

Substituting this back, the expression for $V_2$ becomes:
$V_2 = -\pi \left( -\frac{\pi^2}{8} + \int_{-\pi/2}^{0} x \cos 2x \, dx \right) = \frac{\pi^3}{8} - \pi \int_{-\pi/2}^{0} x \cos 2x \, dx$

4. Apply integration by parts a second time to evaluate $\int x \cos 2x \, dx$.

Let $u = x$ and $\frac{dv}{dx} = \cos 2x$. Then $\frac{du}{dx} = 1$ and $v = \frac{1}{2}\sin 2x$.

$\int_{-\pi/2}^{0} x \cos 2x \, dx = \left[ \frac{1}{2}x \sin 2x \right]_{-\pi/2}^{0} - \int_{-\pi/2}^{0} \frac{1}{2}\sin 2x \, dx$

Evaluate the terms:
$\left[ \frac{1}{2}x \sin 2x \right]_{-\pi/2}^{0} = 0 - 0 = 0$
$- \int_{-\pi/2}^{0} \frac{1}{2}\sin 2x \, dx = \left[ \frac{1}{4}\cos 2x \right]_{-\pi/2}^{0} = \frac{1}{4}(\cos 0 - \cos(-\pi)) = \frac{1}{4}(1 - (-1)) = \frac{1}{2}$

5. Combine the results to state the final exact volume.

Substitute the evaluated integral back into $V_2$:
$V_2 = \frac{\pi^3}{8} - \pi \left(\frac{1}{2}\right) = \frac{\pi^3}{8} - \frac{\pi}{2}$

Add this to the volume of the lower cylindrical section ($V_1$) to find the total volume:
$V = \frac{\pi^3}{4} + \left( \frac{\pi^3}{8} - \frac{\pi}{2} \right) = \frac{3\pi^3}{8} - \frac{\pi}{2}$