Figure 1 shows the curve with equation y = \sqrt{\frac{2x}{3x^{2} + 4}} , \quad x > 0 The finite region S, shown shaded in Figure 1, is bounded by the curve, the x-axis and the line x = 2. The region S is rotated 360° about the x-axis. Use integration to find the exact value of the volume of the solid generated, giving your answer in the form k ln a, where k and a are constants.

A-Level Edexcel Maths Pure Applications of Differentiation: Figure 1 shows the curve with eq

Figure 1 shows the curve with equation y = \sqrt{\frac{2x}{3x^{2} + 4}} , \quad x > 0 The finite region S, shown shaded in Figure 1, is bounded by the curve, the x-axis and the line x = 2 - Edexcel - A-Level Maths Pure - Question 5 - 2012 - Paper 8

Question 5

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Figure 1 shows the curve with equation y = \sqrt{\frac{2x}{3x^{2} + 4}} , \quad x > 0 The finite region S, shown shaded in Figure 1, is bounded by the curve, the x... show full transcript

Worked Solution & Example Answer:Figure 1 shows the curve with equation y = \sqrt{\frac{2x}{3x^{2} + 4}} , \quad x > 0 The finite region S, shown shaded in Figure 1, is bounded by the curve, the x-axis and the line x = 2 - Edexcel - A-Level Maths Pure - Question 5 - 2012 - Paper 8

Step 1

Use integration to find the volume of the solid generated

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Answer

To find the volume of the solid generated when the region S is rotated about the x-axis, we can use the formula:

$V = \pi \int_{0}^{2} y^{2} \, dx$

Substituting the expression for y, we have:

$V = \pi \int_{0}^{2} \left( \sqrt{\frac{2x}{3x^{2} + 4}} \right)^{2} \, dx$

This simplifies to:

$V = \pi \int_{0}^{2} \frac{2x}{3x^{2} + 4} \, dx$

Next, we perform integration by parts or a suitable substitution to evaluate this integral.

Step 2

Evaluate the integral

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Answer

Using the substitution method, let:

$u = 3x^{2} + 4 \implies du = 6x \, dx$

Thus, we need to adjust our limits. When x = 0, u = 4; when x = 2, u = 16. Rewriting the integral, we get:

$V = \pi \int_{4}^{16} \frac{1}{6} \frac{2}{u} \, du$

This can be simplified to:

$V = \frac{\pi}{3} \left[ \ln{u} \right]_{4}^{16} = \frac{\pi}{3} \left( \ln{16} - \ln{4} \right)$

Since ( \ln{16} = 4 \ln{2} ) and ( \ln{4} = 2 \ln{2} ), the expression becomes:

$V = \frac{\pi}{3} (4\ln{2} - 2\ln{2}) = \frac{2\pi}{3} \ln{2}$

Step 3

Final answer in the required form

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Answer

Thus, the volume can be expressed as:

$V = \frac{2\pi}{3} \ln{2}$

Comparing with the required form ( k \ln{a} ), we can identify:

( k = \frac{2\pi}{3} )
( a = 2 )

Figure 1 shows the curve with equation y = \sqrt{\frac{2x}{3x^{2} + 4}} , \quad x > 0 The finite region S, shown shaded in Figure 1, is bounded by the curve, the x-axis and the line x = 2 - Edexcel - A-Level Maths Pure - Question 5 - 2012 - Paper 8

Worked Solution & Example Answer:Figure 1 shows the curve with equation y = \sqrt{\frac{2x}{3x^{2} + 4}} , \quad x > 0 The finite region S, shown shaded in Figure 1, is bounded by the curve, the x-axis and the line x = 2 - Edexcel - A-Level Maths Pure - Question 5 - 2012 - Paper 8

Use integration to find the volume of the solid generated

Evaluate the integral

Final answer in the required form

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