# Calculate the surface area of the surface

• Jun 25th 2009, 06:19 AM
karpatzio
Calculate the surface area of the surface
Calculate the surface area of the part of the sphere x^2 + y^2 + z^2 = 1 that is contained in the lemniscata (x^2 + y^2)^2 = x^2 - y^2

I solved this quickly but i cant seem to get the same answer given in the book, And i would like to know which is the correct one. thanks in advance to anyone that helps.(Clapping)
• Jun 25th 2009, 08:06 AM
Jester
Quote:

Originally Posted by karpatzio
Calculate the surface area of the part of the sphere x^2 + y^2 + z^2 = 1 that is contained in the lemniscata (x^2 + y^2)^2 = x^2 - y^2

I solved this quickly but i cant seem to get the same answer given in the book, And i would like to know which is the correct one. thanks in advance to anyone that helps.(Clapping)

Show us what you did.
• Jun 26th 2009, 05:03 AM
karpatzio
the area element of a sphere given in x,y is $\sqrt{\dfrac{1}{1-x^2-y^2}}dxdy$
integrating it over the are of the lemniscata:

$\int\int \sqrt{\dfrac{1}{1-x^2-y^2}}dxdy$
$D$

when i change to polar coordinates
$x = r*cos(\phi)$
$y = r*sin(\phi)$

i get $4\displaystyle\int^{\pi/4}_{-\pi/4} \, d\phi\displaystyle\int^{\sqrt{cos(2\phi)}}_0 \dfrac{r}{1-r^2}\, dr = -4\displaystyle\int^{\pi/4}_{-\pi/4} \sqrt{1 - cos(2\phi)} - 1\, d\phi =$

$= 4*(\sqrt{2}*cot(\phi)*\sqrt{sin^2(\phi)} + \phi [\pi/4 , -\pi/4])$

eventually solving and getting the wrong answer, what am i doing wrong?