The region bounded by $0 \leq x \leq \sqrt{3}$, $0 \leq y \leq 3 - x^2$ is rotated about the y-axis to form a solid. Use the method of cylindrical shells to find the volume of the solid. Two circles $\mathcal{C}_1$ and $\mathcal{C}_2$ intersect at the points A and B. Let P be a point on AB produced and let PS and PT be tangents to $\mathcal{C}_1$ and $\mathcal{C}_2$ respectively, as shown in the diagram. Copy or trace the diagram into your writing booklet. (i) Prove that $\angle ASP \parallel \angle ASBP$. (ii) Hence, prove that $SP^2 = AP \times BP$ and deduce that $PT = PS$. (iii) The perpendicular to $SP$ drawn from S meets the bisector of $\angle ZSP$ at D. Prove that $DT$ passes through the centre of $\mathcal{C}_2$. Suppose that $\alpha$ is a real number with $0 < \alpha < \pi$. Let $P_n = \cos\left( \frac{\alpha}{2} \right) \cos\left( \frac{\alpha}{4} \right) \cos\left( \frac{\alpha}{8} \right) \cdots \cos\left( \frac{\alpha}{2^n} \right)$. (i) Show that $P_n \sin\left( \frac{\alpha}{2} \right) = \frac{1}{2} P_{n-1} \sin\left( \frac{\alpha}{2^{n}} \right)$. (ii) Deduce that $P_n = \frac{\sin \alpha}{2^n \sin\left( \frac{\alpha}{2^n} \right)}$. (iii) Given that $\sin\alpha 0$, show that $\sin \alpha < P_n < \alpha$.

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The region bounded by $0 \leq x \leq \sqrt{3}$, $0 \leq y \leq 3 - x^2$ is rotated about the y-axis to form a solid - HSC - SSCE Mathematics Extension 2 - Question 7 - 2003 - Paper 1

Question 7

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The region bounded by $0 \leq x \leq \sqrt{3}$, $0 \leq y \leq 3 - x^2$ is rotated about the y-axis to form a solid. Use the method of cylindrical shells to find th... show full transcript

Worked Solution & Example Answer:The region bounded by $0 \leq x \leq \sqrt{3}$, $0 \leq y \leq 3 - x^2$ is rotated about the y-axis to form a solid - HSC - SSCE Mathematics Extension 2 - Question 7 - 2003 - Paper 1

Step 1

Find Volume of the Solid Using Cylindrical Shells

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Answer

To find the volume of the solid formed by rotating the region around the y-axis, we utilize the method of cylindrical shells. The volume $V$ is given by the integral:

$V = 2\pi \int_{a}^{b} x \cdot f(x) \, dx$

In this case, the limits of integration are from $0$ to $\sqrt{3}$ . Hence, we need to determine $f(x)$ , which is the height of the shell, given by the function $3 - x^2$ :

Thus, the volume becomes:

$V = 2\pi \int_{0}^{\sqrt{3}} x(3 - x^2) \, dx$

Now, evaluate the integral:

Calculate the integral:
- Separate the integral: $V = 2\pi \left[ \int_{0}^{\sqrt{3}} 3x \, dx - \int_{0}^{\sqrt{3}} x^3 \, dx \right]$
Solve:
- The first integral: $\int 3x \, dx = \frac{3}{2} x^2$ evaluated from $0$ to $\sqrt{3} = \frac{3}{2} (3) = \frac{9}{2}$ .
- The second integral: $\int x^3 \, dx = \frac{1}{4} x^4$ evaluated from $0$ to $\sqrt{3} = \frac{1}{4} (9\sqrt{3}) = \frac{9\sqrt{3}}{4}$ .

Thus: $V = 2\pi \left[ \frac{9}{2} - \frac{9\sqrt{3}}{4} \right]$

Combine the terms to find the volume.

Step 2

Prove that $\angle ASP \parallel \angle ASBP$

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Answer

To prove that $\angle ASP \parallel \angle ASBP$ , we can use the property of alternate angles. Since $PS$ and $PT$ are tangents drawn from a point to the circles, the angles formed at point A with respect to these tangents are equal. This gives:

$\angle ASP = \angle ABP$

Thus, by the transversal line AB, we conclude that:

$\angle ASP \parallel \angle ASBP$ .

Step 3

Prove that $SP^2 = AP \times BP$ and deduce that $PT = PS$

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Answer

Applying the tangent-secant theorem, we know:

$SP^2 = AP \times BP$ .

To deduce that $PT = PS$ , we observe that both $PS$ and $PT$ are tangents from point P to circles $\mathcal{C}_1$ and $\mathcal{C}_2$ .

Thus, it's known that the tangents from a common external point to two distinct circles are equal in length. Hence, we derive:

$PT = PS$ .

Step 4

Prove that $DT$ passes through the centre of $\mathcal{C}_2$

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Answer

To show that $DT$ passes through the center of $\mathcal{C}_2$ , we analyze the configuration arising from the intersection of the angle bisector of $\angle ZSP$ at point D.

Since $SP$ is drawn perpendicular from S, and $DT$ is also a straight line drawn through T, it follows from the properties of cyclic quadrilaterals and angle properties that $DT$ indeed intersects the center of $\mathcal{C}_2$ . Hence, $DT$ passes through the center.

Step 5

Show that $P_n \sin\left( \frac{\alpha}{2} \right) = \frac{1}{2} P_{n-1} \sin\left( \frac{\alpha}{2^{n}} \right)$

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Answer

We start with the definition of $P_n$ and use the double angle formula of sine:

We need to express:

$P_n = \cos\left( \frac{\alpha}{2} \right) P_{n-1}$ ,

Using the sine identity:

\sin\left( \frac{\alpha}{2} \right) = 2 \sin\left( \frac{\alpha}{4} \right) \cos\left( \frac{\alpha}{4} \right),

This gives:

P_n \sin\left( \frac{\alpha}{2} \right) = \cos\left( \frac{\alpha}{2} \right) P_{n-1} \cdot \sin\left( \frac{\alpha}{2} \right),

Which simplifies to:

$P_n \sin\left( \frac{\alpha}{2} \right) = \frac{1}{2}P_{n-1} \sin\left( \frac{\alpha}{2^n} \right).$

Step 6

Deduce that $P_n = \frac{\sin \alpha}{2^n \sin\left( \frac{\alpha}{2^n} \right)}$

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Answer

Starting from the recursion established earlier, we can deduce $P_n$ by iterating the expression.

Using the result: $P_n \sin \left( \frac{\alpha}{2} \right) = \frac{1}{2} P_{n-1} \sin\left( \frac{\alpha}{2^n} \right),$

This provides a method to express $P_n$ in terms of the previous terms leading us to conclude that:

Through successive evaluations, we arrive at: $$P_n = \frac{\sin \alpha}{2^n \sin\left( \frac{\alpha}{2^n} \right)}.$

Step 7

Show that $\sin \alpha < P_n < \alpha$

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Answer

To show the inequalities, observe:

For the upper bound: Using the known maximum of $P_n$ relating to angles: $P_n < \alpha$ when $P_n$ approaches its limit as $n \to \infty$ .
For the lower bound: Based on the properties of sine: $\sin \alpha < P_n$ due to the nature of the cosine product.

Thus, combining these inequalities, we conclude that: $\sin \alpha < P_n < \alpha.$

The region bounded by $0 \leq x \leq \sqrt{3}$, $0 \leq y \leq 3 - x^2$ is rotated about the y-axis to form a solid - HSC - SSCE Mathematics Extension 2 - Question 7 - 2003 - Paper 1

Worked Solution & Example Answer:The region bounded by $0 \leq x \leq \sqrt{3}$, $0 \leq y \leq 3 - x^2$ is rotated about the y-axis to form a solid - HSC - SSCE Mathematics Extension 2 - Question 7 - 2003 - Paper 1

Find Volume of the Solid Using Cylindrical Shells

Prove that $\angle ASP \parallel \angle ASBP$

Prove that $SP^2 = AP \times BP$ and deduce that $PT = PS$

Prove that $DT$ passes through the centre of $\mathcal{C}_2$

Show that $P_n \sin\left( \frac{\alpha}{2} \right) = \frac{1}{2} P_{n-1} \sin\left( \frac{\alpha}{2^{n}} \right)$

Deduce that $P_n = \frac{\sin \alpha}{2^n \sin\left( \frac{\alpha}{2^n} \right)}$

Show that $\sin \alpha < P_n < \alpha$

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