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

The volume $V$ of the solid of revolution can be calculated using the method of cylindrical shells:

1. The radius of a typical shell is given by the distance from the axis of rotation to the function, which is $x + 1$ (since we are rotating around $x = -1$).
2. The height of the shell is the difference between the outer function ($y=3-x^2$) and the inner function ($y=x+x^2$).
3. The formula for the volume of cylindrical shells is: $V = \int_{a}^{b} 2\pi (radius)(height) \,dx$.
4. Here, $a$ and $b$ are the points of intersection. Thus, $V = \int_{-1}^{\frac{-1 + \sqrt{13}}{2}} 2\pi (x + 1) ((3 - x^2) - (x + x^2)) \,dx$.

To evaluate the integral, we need to simplify the height:

1. The height is: $3 - x^2 - x - x^2 = 3 - 2x^2 - x$.
2. Therefore, the volume integral becomes: $V = 2\pi \int_{-1}^{\frac{-1 + \sqrt{13}}{2}} (x + 1)(3 - 2x^2 - x) \,dx$.
3. Expanding this and simplifying the expression will lead to: $V = 2\pi \int_{-1}^{\frac{-1 + \sqrt{13}}{2}} (3x + 3 - 2x^3 - x^2 - 2x) \,dx$.
4. Integrate term by term, compute the definite integral, and simplify the result to find the volume.

To show that $\angle DSR = \angle DAR$, we leverage properties of cyclic quadrilaterals. By the Inscribed Angle Theorem:

1. $\angle DSR$ subtends arc $DR$ and $\angle DAR$ subtends the same arc $DR$.
2. Therefore, by the theorem, $\angle DSR = \angle DAR$.

To show this, we can use similar properties of angles in cyclic quadrilaterals:

1. Notice that $\angle DST$ and $\angle DCT$ are both subtended by arc $DC$.
2. By the properties of angles in circles, we derive: $\angle DST + \angle DCT = \pi$.
3. Thus, we can conclude that: $\angle DST = \pi - \angle DCT$.

Using the result from part (ii), if $\angle DST + \angle DCT = \pi$, it implies that $R$, $S$, and $T$ must lie on a straight line. This follows from the fact that the angles add up to a linear pair, indicating collinearity.

To find the probability that the number exceeds 400, consider that:

1. The hundreds digit should be 4, 5, 6, 7, 8, or 9, giving us 6 choices.
2. The number of ways to select the remaining two digits from the other 8 cards: $\binom{8}{2} = 28$
3. Total successful outcomes = $6 * 28 = 168$.
4. Total possible outcomes = $\binom{9}{3} = 84$.
5. Therefore, the probability is: $P = \frac{168}{84} = 2$.

To find this probability:

1. Any selection of 3 digits can be arranged in $3! = 6$ ways.
2. However, only 1 of those arrangements will be in descending order.
3. Thus, the probability is: $P = \frac{1}{6}$.