Ok, I just found out, I wrote it on polar form and integrated 0≤theta≤pi/2, 0≤r≤2a*sin(theta) 4*integral(r*dr*dtheta), and the answer is s*sqrt(2)*a^2*pi.
Hi! I'm trying to find the area of the cone x^2+y^2=z^2 inside x^2+(y-a)^2=a^2 -- but I believe I'm stuck... :/
I parametrize the cone as r=u*i+v*j+sqrt(u^2+v^2)k, find dr/du=i+u/sqrt(u^2+v^2)*k and dr/dv=j+v/sqrt(u^2+v^2)k, then the cross product dr/du*dr/dv=-u/sqrt(u^2+v^2)*i-v/sqrt(u^2+v^2)*j +k and the area element dS=norm(dr/du*dr/dv)du*dv=sqrt(2)*du*dv.
How shall I proceed now? I simply can't think of any good way...
Thanks!