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(B) The formula for the curved surface area of a solid generated by revolving a curve $y = f(x)$ about the $x$-axis from $x = a$ to $x = b$ is given b

Vedclass · Accepted Answer

$7\pi \sqrt{2}$

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(B) The formula for the curved surface area of a solid generated by revolving a curve $y = f(x)$ about the $x$-axis from $x = a$ to $x = b$ is given by:$S = \int_{a}^{b} 2\pi y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx$Given the line $y = x + 1$,we differentiate with respect to $x$:$\frac{dy}{dx} = 1$Substituting the values $a = 2$,$b = 3$,$y = x + 1$,and $\frac{dy}{dx} = 1$ into the formula:$S = \int_{2}^{3} 2\pi (x + 1) \sqrt{1 + (1)^2} dx$$S = \int_{2}^{3} 2\pi (x + 1) \sqrt{2} dx$$S = 2\sqrt{2}\pi \int_{2}^{3} (x + 1) dx$Evaluating the integral:$S = 2\sqrt{2}\pi \left[ \frac{(x + 1)^2}{2} \right]_{2}^{3}$$S = \sqrt{2}\pi \left[ (3 + 1)^2 - (2 + 1)^2 \right]$$S = \sqrt{2}\pi [16 - 9]$$S = 7\sqrt{2}\pi = 7\pi \sqrt{2}$

The part of the straight line $y = x + 1$ between $x = 2$ and $x = 3$ is revolved about the $x$-axis. The curved surface area of the solid thus generated is

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