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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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