How should I find the boundaries for the following surface $S:$
$S$ is the surface with equations $z = x^2+y^2 $ and $z\leq 1.$
I can find $x^2=1-y^2$ and take the square root to find the boundaries for $y,$ but how should I find the boundaries for $x?$
Thanks
The trace of the paraboloid $$ z = x^2+y^2$$ in plane $ z= 1$ projects onto the circle $$ x^2+y^2 =1 $$ in the $xy $ plane, and the portion of the paraboloid that lies below the plane $ z= 1 $ projects onto the region $ R $ that is enclosed by the circle. Thus, it follows from definition of surface area of surfaces of the form $ z = f(x,y)$ that $$ |S| =\iint_{(R)}\sqrt{f'_{|x}(x,y)^2 + f'_{|y}(x,y)^2+1}dA = \iint_{(R)}\sqrt{ 4x^2 +4y^2 +1}dA. $$ The expresion $4x^2 +4y^2 +1 = 4(x^2+y^2)+1 $ in the integrand suggests that we evaluate the integral in polar coordinates. We substitute $ x =r\cos(\phi),\ \ y = r\sin(\phi)$ in the integrand, replace $dA = rdrd\phi $, and find the limits of integration by expressing the region $ R $ in polar coordinates. This yields $$ |S| = \int_{0}^{2\pi}\int_{0}^{1}\sqrt{4r^2+1}rdrd\phi = \int_{0}^{2\pi}\left[ \frac{1}{12}(4r^2 +1)^{\frac{3}{2}}\right]_{0}^{1}d\phi = \int_{0}^{2\pi}\frac{1}{12}(5\sqrt{5}-1)d\phi = \frac{1}{6}\pi(5\sqrt{5}-1).$$
