# Surface area of revolution using a parametrization

• Mar 4th 2010, 08:11 PM
My Little Pony
Surface area of revolution using a parametrization
Hi, the question is attached below.

Attempt at solution

Surface area = 2*pi*r*ds = 2*pi*integral (from 1 to 2) of x^3*sqrt(1+9x^4)dx

I ended up with 199.4804797 as my answer.

However, I'm having trouble finding the area of revolution using the parametrization method. What is r(t)?
• Mar 5th 2010, 04:18 AM
HallsofIvy
Since x= t and $\displaystyle y= x^3$, $\displaystyle r(t)= t^3$ so the parametric equations are x= t, $\displaystyle y= t^3 cos(\theta)$, $\displaystyle z= t^3 sin(\theta)$

As a vector equation, that is $\displaystyle \vec{r}(t,\theta)= t\vec{i}+ t^3 cos(\theta)\vec{j}+ t^3 sin(\theta)\vec{k}$.

The derivatives of that function, with respect to t and $\displaystyle \theta$ are
$\displaystyle \vec{r}_t= \vec{i}+ 3t^2 cos(\theta)\vec{j}+ 3t^2 sin(\theta)\vec{k}$ and
$\displaystyle \vec{r}_\theta= -t^3 sin(\theta)\vec{j}+ t^3 cos(\theta)\vec{k}$

The cross product of those two derivatives (the "fundamental vector product" for this surface) is
$\displaystyle 3t^5\vec{i}- t^3 cos(\theta)\vec{j}- t^3 sin(\theta)\vec{k}$

Its length is $\displaystyle \sqrt{9t^4+ t^2}= t^2\sqrt{9+ t^2}$ so the "differential of surface area" for this surface is
$\displaystyle t^2\sqrt{9+ t^2} dtd\theta$

The surface area is given by
$\displaystyle \int_{\theta= 0}^{2\pi}\int_{t=1}^2 t^2\sqrt{9+ t^2}dtd\theta$