The curve with equation $y = 3 \, \sin \left( \frac{x}{2} \right)$, $0 \leq x \leq 2\pi$, is shown in Figure 1. The finite region enclosed by the curve and the x-axis is shaded. (a) Find, by integration, the area of the shaded region. This region is rotated through $2\pi$ radians about the x-axis. (b) Find the volume of the solid generated.

A-Level Edexcel Maths Pure Modelling with Functions: The curve with equation $y = 3 \

The curve with equation $y = 3 \, \sin \left( \frac{x}{2} \right)$, $0 \leq x \leq 2\pi$, is shown in Figure 1 - Edexcel - A-Level Maths Pure - Question 3 - 2006 - Paper 6

Question 3

$The-curve-with-equation-$y-=-3-\,-\sin-\left(-\frac{x}{2}-\right)$,-$0-\leq-x-\leq-2\pi$,-is-shown-in-Figure-1-Edexcel-A-Level Maths Pure-Question 3-2006-Paper 6.png$

The curve with equation $y = 3 \, \sin \left( \frac{x}{2} \right)$, $0 \leq x \leq 2\pi$, is shown in Figure 1. The finite region enclosed by the curve and the x-axi... show full transcript

Worked Solution & Example Answer:The curve with equation $y = 3 \, \sin \left( \frac{x}{2} \right)$, $0 \leq x \leq 2\pi$, is shown in Figure 1 - Edexcel - A-Level Maths Pure - Question 3 - 2006 - Paper 6

Step 1

Find, by integration, the area of the shaded region.

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Answer

To find the area of the shaded region, we need to calculate the integral of the curve from $0$ to $2\pi$ :

$\text{Area} = \int_0^{2\pi} 3 \sin \left( \frac{x}{2} \right) \, dx$

We will first find the integral:

Let:

\Rightarrow du = \frac{1}{2}dx \\ \Rightarrow dx = 2du\\ \text{Changing limits:} \quad x = 0 \Rightarrow u = 0 \quad ext{and} \quad x = 2\pi \Rightarrow u = \pi$$ Thus: $$\text{Area} = \int_0^{\pi} 6 \sin(u) \, du$$ Integrating gives: $$= -6 \cos(u) \bigg|_0^{\pi} = -6 \left[ \cos(\pi) - \cos(0) \right]$$ Substituting values, we find: $$= -6[-(-1)-1] = -6[-(-2)] = 12$$ Therefore, the area of the shaded region is $12$.

Step 2

Find the volume of the solid generated.

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Answer

To find the volume of the solid generated by revolving the region around the x-axis, we use the disk method:

$V = \pi \int_0^{2\pi} (3 \sin \left( \frac{x}{2} \right))^2 \, dx$

This simplifies to:

$= 9\pi \int_0^{2\pi} \sin^2 \left( \frac{x}{2} \right) \, dx$

Using the half-angle identity, we can express:

$\sin^2 \left( \frac{x}{2} \right) = \frac{1 - \cos(x)}{2}$

Thus:

$V = \frac{9\pi}{2} \int_0^{2\pi} (1 - \cos(x)) \, dx$

Calculating the integral:

$= \frac{9\pi}{2} \left[ x - \sin(x) \right]_0^{2\pi}$

Substituting the limits gives:

$= \frac{9\pi}{2} \left[ (2\pi - 0) - (0 - 0) \right] = 9\pi^2$

Finally, this results in:

$V = 9 \cdot \frac{22}{7} \approx 88.8264$

Therefore, the volume of the solid generated is approximately $88.8264$ .

The curve with equation $y = 3 \, \sin \left( \frac{x}{2} \right)$, $0 \leq x \leq 2\pi$, is shown in Figure 1 - Edexcel - A-Level Maths Pure - Question 3 - 2006 - Paper 6

Worked Solution & Example Answer:The curve with equation $y = 3 \, \sin \left( \frac{x}{2} \right)$, $0 \leq x \leq 2\pi$, is shown in Figure 1 - Edexcel - A-Level Maths Pure - Question 3 - 2006 - Paper 6

Find, by integration, the area of the shaded region.

Find the volume of the solid generated.

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