- #1

kostoglotov

- 234

- 6

## Homework Statement

I want to know if I got the answer correct and if my reasoning is sound. The text answers and solutions manual only gives answers/solutions for odd numbered problems.

Here is the problem:

And a direct link to the imgur page: http://i.imgur.com/Tko1xFh.png

## Homework Equations

[tex]A(S) = \int \int_D \left|\vec{r}_u\times\vec{r}_v\right|dA[/tex]

## The Attempt at a Solution

Instead of trying to come up with a parametric vector equation for the entire surface, I started subdividing the surface as much as possible using symmetry. I got it down to one sixteenth of the whole surface. Basically, just consider the surface in the first octant, and then divide that in two again along the line y = x.

By doing that I get the following:

[tex]A(S) = 16 \times \int\int_D\left|\vec{r}_x\times\vec{r}_y\right|dA[/tex]

where [itex]\vec{r} = \left< x,y,\sqrt{1-x^2}\right>[/itex] and [itex]D = \{(x,y)|0\leq x \leq 1, 0 \leq y \leq x\}[/itex]

so

[itex]\vec{r}_x = \langle 1,0,\frac{-x}{\sqrt{1-x^2}} \rangle[/itex] and [itex]\vec{r}_y = \langle 0,1,0 \rangle[/itex]

[tex]\vec{r}_x \times \vec{r}_y = \langle \frac{x}{\sqrt{1-x^2}},0,1 \rangle[/tex]

[tex]\left| \vec{r}_x \times \vec{r}_y \right| = \sqrt{1+\frac{x^2}{1-x^2}} = \frac{1}{1-x^2}[/tex]

Now

[tex]A(S) = 16 \times \int_0^1\int_0^x \frac{1}{1-x^2} dydx[/tex]

[tex]A(S) = 16 \times \int_0^1 \frac{x}{1-x^2} dx = -16 \times\left[\sqrt{1-x^2}\right]_0^1 = 16[/tex]