# Surface rotation

• Jun 3rd 2009, 04:22 PM
Apprentice123
Surface rotation
A surface S is obtained by rotation, around the axis OZ, the arc of parabola $\displaystyle \alpha : z = \frac{x^2}{2}, y = 0$ with $\displaystyle 0 \leq x \leq 1$. Determine the area of the surface obtained
• Jun 4th 2009, 01:11 PM
Apprentice123
How do I calculate surface rotation? Must use the matrix?
• Jun 4th 2009, 02:00 PM
Jester
Quote:

Originally Posted by Apprentice123
How do I calculate surface rotation? Must use the matrix?

It's a standard formula in Calc II $\displaystyle SA = 2 \pi \int_a^b y \sqrt{1+y'^2}\,dx$
• Jun 4th 2009, 02:04 PM
Apprentice123
Quote:

Originally Posted by Danny
It's a standard formula in Calc II $\displaystyle SA = 2 \pi \int_a^b y \sqrt{1+y'^2}\,dx$

I do not know this formula.
Could you explain better?
• Jun 4th 2009, 02:57 PM
Jester
Quote:

Originally Posted by Apprentice123
I do not know this formula.

Could you explain better?

Are you familar with the formula's for volumes of revolution?

I need a starting place for our discussion.
• Jun 4th 2009, 05:59 PM
Spec
Or you can recognize that the rotation will produce an elliptical paraboloid $\displaystyle z=\frac{x^2+y^2}{2}$ which can be parametrized as $\displaystyle \sigma(u,v)=\left(u,v, \frac{u^2+v^2}{2}\right)$

A switch to polar coordinates will then basically produce the same integral as Danny showed you.

$\displaystyle \iint_D \left| \vec {\sigma'_u}\times\vec {\sigma'_v}\right|dudv=\{\text{Polar coordinates}\}=\int \int_0^{2\pi}\sqrt{r^2+1} \cdot rd\varphi dr=$ $\displaystyle 2\pi\int r\sqrt{r^2+1}dr$
• Jun 5th 2009, 01:07 PM
Apprentice123
Quote:

Originally Posted by Spec
Or you can recognize that the rotation will produce an elliptical paraboloid $\displaystyle z=\frac{x^2+y^2}{2}$ which can be parametrized as $\displaystyle \sigma(u,v)=\left(u,v, \frac{u^2+v^2}{2}\right)$

A switch to polar coordinates will then basically produce the same integral as Danny showed you.

$\displaystyle \iint_D \left| \vec {\sigma'_u}\times\vec {\sigma'_v}\right|dudv=\{\text{Polar coordinates}\}=\int \int_0^{2\pi}\sqrt{r^2+1} \cdot rd\varphi dr=$ $\displaystyle 2\pi\int r\sqrt{r^2+1}dr$

Ok. Thanks
• Jun 5th 2009, 04:56 PM
Apprentice123
Quote:

Originally Posted by Spec
Or you can recognize that the rotation will produce an elliptical paraboloid $\displaystyle z=\frac{x^2+y^2}{2}$ which can be parametrized as $\displaystyle \sigma(u,v)=\left(u,v, \frac{u^2+v^2}{2}\right)$

A switch to polar coordinates will then basically produce the same integral as Danny showed you.

$\displaystyle \iint_D \left| \vec {\sigma'_u}\times\vec {\sigma'_v}\right|dudv=\{\text{Polar coordinates}\}=\int \int_0^{2\pi}\sqrt{r^2+1} \cdot rd\varphi dr=$ $\displaystyle 2\pi\int r\sqrt{r^2+1}dr$

as you know that rotation will produce an elliptical paraboloid $\displaystyle z=\frac{x^2+y^2}{2}$