Evaluating the surface integral $\iint_S {({x^2} + {y^2})} \,dS$ using spherical coordinates For the integral $$\iint\limits_S {({x^2} + {y^2})} \,dS\quad,\,S:{x^2} + {y^2} + {z^2} = 2z$$The
correct answer is $${{8\pi } \over 3}$$ I used Spherical coordinate system,it turns to $$\int_0^{2\pi } {d\theta \int_0^{{\pi  \over 2}} {({r^2}{{\sin }^2}\varphi } )({r^2}\sin \varphi )\,d\varphi } ,r = 2\cos \varphi $$Then use $r = 2\cos \varphi$,it turns to $$32\pi \int_0^{{\pi  \over 2}} {{{\sin }^3}\varphi {{\cos }^4}\varphi \,d\varphi }  = {{64} \over {35}}\pi $$Doesn't match the answer,I wonder where am I wrong.

As an extra, there is a way of computing this integral without ever integrating by using spherical coordinates, which all hinges on symmetry. Notice that the sphere could also be written as
$$x^2+y^2+(z-1)^2=1$$
Since the integrand doesn't depend on the location in $z$ at all, we could safely translate the sphere down and be assured that
$$\iint\limits_{x^2+y^2+(z-1)^2=1}x^2+y^2\:dS = \iint\limits_{x^2+y^2+z^2=1}x^2+y^2\:dS$$
holds. Second, notice that by rotational symmetry (or just pairwise swapping the variables AKA reflectional symmetry) on the second surface we have that
$$\iint\limits_{x^2+y^2+z^2=1}x^2\:dS = \iint\limits_{x^2+y^2+z^2=1}y^2\:dS = \iint\limits_{x^2+y^2+z^2=1}z^2\:dS \equiv I$$
which is a quantity we will denote by $I$. Then the beauty of being on a sphere tells us that
$$3I = \iint\limits_{x^2+y^2+z^2=1}x^2+y^2+z^2\:dS = \iint\limits_{x^2+y^2+z^2=1}1\:dS = 4\pi$$
Thus the answer to the problem is $2I$, or
$$\frac{8\pi}{3}$$
