The shaded region bounded by $y=3-x^2$, $y=x+x^2$ and $x=-1$ is rotated about the line $x=-1$ - HSC - SSCE Mathematics Extension 2 - Question 4 - 2002 - Paper 1

Using the cylindrical shells method, we express the volume $V$ as:

$V = 2\pi \int_{-1}^{1} (x + 1)(3 - x^2 - (x + x^2)) \, dx$

This integral computes the volume generated when rotating the shaded area about the line $x = -1$. The specific bounds are dictated by the intersection points and the given region.

To evaluate the integral:

$V = 2\pi \int_{-1}^{1} (x + 1)(3 - 2x - x^2) \, dx$

Expanding the integrand:

$= 2\pi \int_{-1}^{1} (3x + 3 - 2x^2 - x^3) \, dx$

Calculating the integral results in:

$= 2\pi \left[ \frac{3x^2}{2} + 3x - \frac{2x^3}{3} - \frac{x^4}{4} \right]_{-1}^{1}$

Evaluating from -1 to 1 gives the total volume as an exact value.

To demonstrate that $\angle DSR = \angle DAR$, we utilize the properties of cyclic quadrilaterals. The angles subtended by the same arc are equal, hence these angles are proven equal by referring to points $D$, $S$, and $R$.

This follows from the angles in a cyclic quadrilateral. Since $D$, $S$, and $T$ lie along chord $AC$, we can apply the inscribed angle theorem to establish this relationship.

Since $\angle DSR = \angle DAR$ and $\angle DST = \pi - \angle DCT$, we conclude that points $R$, $S$, and $T$ are indeed collinear as their angles correlate to lines intersecting at a common point.

To exceed 400, the first digit must be $4$, $5$, $6$, $7$, $8$, or $9$ (6 options). Considering the remaining two digits drawn from 1-9, we calculate the probabilities based on combinations exceeding specified limits.

The total arrangements for any three cards drawn is $\binom{9}{3} \cdot 3!$. The number of arrangements in descending order for any chosen three is just $1$. Therefore, the probability is given by:

$P = \frac{1}{\binom{9}{3}} = \frac{6}{84} = \frac{1}{14}.$