# Thread: Calculate the surface area of the surface

1. ## 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.

2. 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.
Show us what you did.

3. 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?