Specialist2024 NHT Exam 2 Section A · Q10CAS active1 mark

Surface area of revolution of y = 1/x

Consider the surface of revolution formed by rotating the curve $y = \frac{1}{x}$ , $1 \le x \le a$ about the $x$ -axis.

The area of this surface is given by

1 mark

2\pi \int_1^a x\sqrt{1+x^2}\,dx

2\pi \int_1^a \frac{\sqrt{1+x^4}}{x}\,dx

2\pi \int_1^a \frac{\sqrt{1+x^4}}{x^3}\,dx

2\pi \int_1^a \left(1+\frac{1}{x^2}\right)dx

2\pi \int_1^a \left(\frac{1+x^4}{x^5}\right)dx

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