2 cylinders intersect, area of resulting parametric surface

[kostoglotov]

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

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

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:

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

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

so

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

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

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

Now

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

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

 

[LCKurtz]

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

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

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:

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

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

so

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

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

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

Shouldn't that be$$
\frac{1}{\sqrt{1-x^2}}$$

 

[kostoglotov]

Shouldn't that be$$
\frac{1}{\sqrt{1-x^2}}$$

Typo :) In my original work that typo isn't there.

 

[haruspex]

I haven't tried to follow your working (yet), but a mental calculation using a simpler method gives me 16 also.

 

[haruspex]

I haven't tried to follow your working (yet), but a mental calculation using a simpler method gives me 16 also.

My method:
Convert the xz plane to polar, measuring theta from lowest point.
Consider face on the positive x side.
Between $\theta$ and $\theta+d\theta$, there is a rectangle measuring $r.d\theta$ by $2r\sin(\theta)$. Integrate 0 to $\pi$ and multiply by 4.