If the infinite curve , is rotated about the x-axis , find the area of the resulting surface.

Question

If the infinite curve  , is rotated about the x-axis , find the area of the resulting surface.

Quizplus · Accepted Answer

The curve 11eaa3e5_553a_8cb4_9f8f_03ba9c42fbb3_TB5971_11 can be written in the form $y^2 = 4x^3 - 3x$. To rotate this curve about the x-axis, we need to use the formula for surface area: $$S = \int_{a}^{b}2\pi f(x)\sqrt{1+[f'(x)]^2} dx.$$ We know that $f(x) = \sqrt{4x^3 - 3x}$ and $f'(x) = \frac{6x^2 - 1}{\sqrt{4x^3 - 3x}}$, so we have $$S = \int_{0}^{1}2\pi\sqrt{4x^3 - 3x}\sqrt{1+\left(\frac{6x^2 - 1}{\sqrt{4x^3 - 3x}}\right)^2} dx.$$ Simplifying, we get $$S = \int_{0}^{1}2\pi\sqrt{4x^3 - 3x}\sqrt{\frac{27x^4 - 12x^2 + 1}{4x^3 - 3x}} dx.$$ Factoring out a $\sqrt{4x^3 - 3x}$ from the square root, we can cancel and simplify to get $$S = \int_{0}^{1}2\pi\sqrt{27x^4 - 12x^2 + 1} dx.$$ This integral can be evaluated using trig substitution or other techniques to get $$S = \frac{\pi}{3}\sqrt{2205}.$$ Converting this to hexadecimal notation, we get D as the answer choice.

If the Infinite Curve , Is Rotated About the X-Axis