If the infinite curve $y = e ^ { - x } , x \geq 0$, is rotated about the $x$-axis, find the area of the resulting surface. Select the correct answer.
A) $\pi [ \sqrt { 2 } + \ln ( 1 + \sqrt { 2 } ) ]$
B) $\frac { \pi } { 6 } [ \sqrt { 3 } + \ln ( 1 + \sqrt { 3 } ) ]$
C) $\frac { \pi } { 3 } [ \sqrt { 2 } + \ln ( 1 + \sqrt { 2 } ) ]$
D) $\frac { \pi } { 4 } [ \sqrt { 2 } + \ln ( 1 + \sqrt { 2 } ) ]$
E) $\frac { \pi } { 6 } [ \sqrt { 2 } + \ln ( 1 + \sqrt { 2 } ) ]$

Question

Quizplus · Accepted Answer

The correct answer is A because the surface area of the resulting surface is given by the integral $2\pi \int_0^\infty y \sqrt{1 + (\frac{dy}{dx})^2} dx$, which evaluates to $\pi [ \sqrt { 2 } + \ln ( 1 + \sqrt { 2 } ) ]$.

If the Infinite Curve y=e−x,x≥0y = e ^ { - x } , x \geq 0y=e−x,x≥0

If the Infinite Curve $y = e ^ { - x } , x \geq 0$