The curve with equation $y = 3 \, ext{sin} \left( \frac{x}{2} \right)$, $0 \leq x \leq 2\pi$, 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 Differentiation: The curve with equation $y = 3 \

The curve with equation $y = 3 \, ext{sin} \left( \frac{x}{2} \right)$, $0 \leq x \leq 2\pi$, is shown in Figure 1 - Edexcel - A-Level Maths Pure - Question 4 - 2006 - Paper 6

Question 4

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

Worked Solution & Example Answer:The curve with equation $y = 3 \, ext{sin} \left( \frac{x}{2} \right)$, $0 \leq x \leq 2\pi$, is shown in Figure 1 - Edexcel - A-Level Maths Pure - Question 4 - 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 evaluate the integral:

$\text{Area} = \int_{0}^{2\pi} 3 \, \text{sin} \left( \frac{x}{2} \right) \, dx$

To solve this integral, we first apply integration:

Let ( u = \frac{x}{2} )
Then, ( du = \frac{1}{2} dx ) or ( dx = 2 , du )
Changing the limits:
- When ( x = 0, u = 0 )
- When ( x = 2\pi, u = \pi )
Thus, the integral becomes:

$\text{Area} = 6 \int_{0}^{\pi} \text{sin}(u) \, du$

Now, integrating:

= 6 \left[ 1 + 1 \right] = 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 formed by rotating the shaded region about the x-axis, we use the formula for the volume:

$V = \pi \int_{a}^{b} y^2 \, dx$ , where ( y = 3 , \text{sin} \left( \frac{x}{2} \right) ) and we know the limits are from ( 0 ) to ( 2\pi ).

Substituting for y,

$V = \pi \int_{0}^{2\pi} \left( 3 \, \text{sin} \left( \frac{x}{2} \right) \right)^2 \, dx$

This can be simplified to:

$V = 9\pi \int_{0}^{2\pi} \text{sin}^2 \left( \frac{x}{2} \right) \, dx$

Using the identity ( \text{sin}^2(a) = \frac{1 - \text{cos}(2a)}{2} ), we rewrite:

$V = \frac{9\pi}{2} \int_{0}^{2\pi} \left( 1 - \text{cos}(x) \right) \, dx$

Now calculate:
- The integral of ( 1 ) over ( [0, 2\pi] ) is ( 2\pi ), and the integral of ( \text{cos}(x) ) over the same interval is 0:
$V = \frac{9\pi}{2} \left[ 2\pi - 0 \right] = 9\pi^2$

Thus, the volume of the solid generated is ( 9\pi^2 \approx 88.8264 ).

The curve with equation $y = 3 \, ext{sin} \left( \frac{x}{2} \right)$, $0 \leq x \leq 2\pi$, is shown in Figure 1 - Edexcel - A-Level Maths Pure - Question 4 - 2006 - Paper 6

Worked Solution & Example Answer:The curve with equation $y = 3 \, ext{sin} \left( \frac{x}{2} \right)$, $0 \leq x \leq 2\pi$, is shown in Figure 1 - Edexcel - A-Level Maths Pure - Question 4 - 2006 - Paper 6

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

Find the volume of the solid generated.

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