Find the area of the surface generated by revolving the curve about the given axis. $x = t , y = 6 t , 0 \leq t \leq 9$
(i) $x$-axis; (ii) $y$-axis
A) (i) $486 \sqrt { 37 } \pi$; (ii) $81 \sqrt { 37 } \pi$
B) (i) $324 \sqrt { 82 } \pi$; (ii) $36 \sqrt { 82 } \pi$
C) (i) $486 \sqrt { 37 } \pi$; (ii) $36 \sqrt { 37 } \pi$
D) (i) $81 \sqrt { 37 } \pi$; (ii) $486 \sqrt { 37 } \pi$
E) (i) $324 \sqrt { 37 } \pi$; (ii) $81 \sqrt { 37 } \pi$

Question

Quizplus · Accepted Answer

The correct answer is (i) $486 \sqrt { 37 } \pi$; (ii) $81 \sqrt { 37 } \pi$.Explanation:(i) For the $x$-axis, $y = 6t$, $dy/dt = 6$, and $ds = \sqrt{dx^2 + dy^2} = \sqrt{1 + 36} dt = 6\sqrt{37} dt$. So, the surface area is$$S = \int_0^9 2\pi y ds = \int_0^9 2\pi (6t) (6\sqrt{37} dt) = 486 \sqrt { 37 } \pi$$(ii) For the $y$-axis, $x = t$, $dx/dt = 1$, and $ds = \sqrt{dx^2 + dy^2} = \sqrt{1 + 36} dt = 6\sqrt{37} dt$. So, the surface area is$$S = \int_0^9 2\pi x ds = \int_0^9 2\pi (t) (6\sqrt{37} dt) = 81 \sqrt { 37 } \pi$$

Find the Area of the Surface Generated by Revolving the Curve