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

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

Question 6

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

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

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

Step 1

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

96%

114 rated

Answer

To find the area of the shaded region, we integrate the function from $0$ to $2\pi$ :

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

First, we calculate the integral:

Substitute $u = \frac{x}{2}$ $u = \frac{x}{2}$ , then $dx = 2 \, du$ $d x = 2 d u$ , and change the limits accordingly:
- When $x = 0$ , $u = 0$ .
- When $x = 2\pi$ , $u = \pi$ .

So,

$A = \int_0^{\pi} 3 \sin(u) \cdot 2 \, du = 6 \int_0^{\pi} \sin(u) \, du$

Now integrate $\sin(u)$ :

$6 [-\cos(u)]_0^{\pi} = 6 [(-\cos(\pi)) - (-\cos(0))] = 6 [1 + 1] = 6\cdot2 = 12.$

Thus, the area of the shaded region is $12$ square units.

Step 2

Find the volume of the solid generated.

99%

104 rated

Answer

To find the volume, we use the formula:

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

First, simplify the integrand:

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

Using the identity $\sin^2(u) = \frac{1 - \cos(2u)}{2}$ , we can rewrite the integral:

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

Now, calculate the integral:

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

Evaluate:

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

Finally, substituting the value of $\pi \approx 3.14$ , we find that:

$\approx 9 \cdot 3.14^2 \approx 88.8264.$

Thus, the volume of the solid generated is approximately $88.83$ cubic units.

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

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

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

Find the volume of the solid generated.

Join the A-Level students using SimpleStudy...