Surface area integral

**qwesl** · March 1st 2010, 12:27 PM

If you revolve the function $y=x^3$ , $1\leq x\leq 2$ around the x-axis how do you calculate the surface area by using the parametrization
$x=t$
$y=r(t)\cos\theta$
$z=r(t)\sin\theta$
for a suitable $r(t)$

**HallsofIvy** · March 1st 2010, 12:50 PM

Originally Posted by qwesl

If you revolve the function $y=x^3$ , $1\leq x\leq 2$ around the x-axis how do you calculate the surface area by using the parametrization
$x=t$
$y=r(t)\cos\theta$
$z=r(t)\sin\theta$
for a suitable $r(t)$

When you have a surface defined in terms of two parameters, x= f(u,v), y= g(u,v), z= h(u,v), you can write each point as the vector equation, $\vec{r}(u,v)= f(u,v)\vec{i}+ g(u,v)\vec{j}+ h(u,v)\vec{k}$ . The two partial derivatives, with respect to u and v are $\vec{r}_u= f_u\vec{i}+ g_u\vec{j}+ h_u\vec{k}$ and $\vec{t}= f_v\vec{i}+ g_v\vec{j}+ h_v\vec{k}$ . The "fundamental vector product" for the surface is $\vec{r}_u\times\vec{r}_v$ .

Finally, the "differential of surface area is the magnitude of that product time drdt: $\left|\vec{r}_u\times\vec{r}_v\right|dudv$

Here, since this is rotated around the x-axis, x= t, as you say, and so r(t), perpendicular to the x, axis, is $r(t)= t^3$ . $x= t$ , $y= t^3cos(\theta)$ , and $z= t^3sin(\theta)$ . $\vec{r}(t,\theta)= t\vec{i}+ t^3cos(\theta)\vec{j}+ t^3 sin(\theta)\vec{k}$ .

Thread: Surface area integral

LinkBack

Thread Tools

Display