# Thread: This question is so hard!! any help??

1. ## This question is so hard!! any help??

Consider a function $f(x)$ where $f(x) \succeq 0$ for $a \preceq x \preceq b$.

Define the 'surface of revolution' to be the surface swept out by a complete revolution of the curve $f(x)$, $a \preceq x \preceq b$, about the x-axis.

This surface of revolution can be described using the parameterization $r(u,v)=[v, f(v)cos u, f(v)sin u]$, $a\preceq v \preceq b,$ $0 \preceq u \preceq 2\pi$.

Show that the area of this surface is given by $A=2\pi \int_a^b f(v) \sqrt{1+(f'(v))^2}dv$ where $f'(v)=\frac{df}{dv}$

2. Originally Posted by Ash_underpar
Consider a function $f(x)$ where $f(x) \succeq 0$ for $a \preceq x \preceq b$.

Define the 'surface of revolution' to be the surface swept out by a complete revolution of the curve $f(x)$, $a \preceq x \preceq b$, about the x-axis.

This surface of revolution can be described using the parameterization $r(u,v)=[v, f(v)cos u, f(v)sin u]$, $a\preceq v \preceq b,$ $0 \preceq u \preceq 2\pi$.

Show that the area of this surface is given by $A=2\pi \int_a^b f(v) \sqrt{1+(f'(v))^2}dv$ where $f'(v)=\frac{df}{dv}$
Hi
Please consider the following graph

The curvilinear abscissa ds is
$ds=\sqrt{dv^2+dy^2}$

$y=f(v)$
$dy=f'(v) dv$

$ds=\sqrt{dv^2+f'(v)^2dv^2}$
$ds=\sqrt{1+(f'(v))^2}dv$

Rotation of curvilinear abscissa ds around x axis with an angle u leads to an elementary surface
$dA=ds f(v)du$
f(v) du is the equivalent of $r d\theta$, the radius being f(v) and the angle being u
$dA=f(v)\sqrt{1+(f'(v))^2}dv\;du$

Surface swept out by a complete revolution of the curve $f(x)$ is
$A=\int_0^{2\pi} \int_a^b f(v) \sqrt{1+(f'(v))^2}dv \;du$

$A=2\pi \int_a^b f(v) \sqrt{1+(f'(v))^2}dv$