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:For (i), $x = t$ and $y = 6 t$, so $dy/dx = 6$. The surface area is$$S = 2\pi \int_0^9 t \sqrt{1 + (6)^2} \ dt = 2\pi \int_0^9 t \sqrt{37} \ dt = 2\pi \left[ \frac{1}{2} \sqrt{37} t^2 \right]_0^9 = 486 \sqrt { 37 } \pi$$For (ii), $x = t$ and $y = 6 t$, so $dx/dy = 1/6$. The surface area is$$S = 2\pi \int_0^{54} y \sqrt{1 + (1/6)^2} \ dy = 2\pi \int_0^{54} y \sqrt{37/36} \ dy = 2\pi \left[ \frac{1}{2} \sqrt{37/36} y^2 \right]_0^{54} = 81 \sqrt { 37 } \pi$$

Find the Area of the Surface Generated by Revolving the Curve