The spherical surface $x^2+y^2+z^2=R^2$ has a charge with the density
$$\rho(x,y,z)=\sqrt{R^2-x^2-y^2} \quad C/m^2.$$
Determine the total charge of the sphere.
So I have two different parametrizations that I've attempted and my answer is very close to the correct one, which is $2\pi R^3$. Can someone check my two approaches to see where I make the mistakes?
Method 1
The surface can be parametrized by
$$\left\{ \begin{array}{rcr} x & = & R\sin{\theta}\cos{\varphi} \\ y & = & R\sin{\theta\sin{\varphi}} \\ z & = & R\cos{\theta} \end{array} \right. \implies D:\left\{ \begin{array}{rcr} 0\leq \theta \leq \pi\\ 0\leq \varphi \leq 2\pi \\ \end{array} \right.$$
A positionvector is $\vec{r}(\theta,\varphi)=R(\sin{\theta}\cos{\varphi},\sin{\theta\sin{\varphi}},\cos{\theta})$, thus the magnitude of the normalvector is
$$dS=||\vec{n}||d\theta d\varphi=||\vec{r}'_\theta\times\vec{r}'_\varphi||d\theta d\varphi=R^2\sin{\theta} \ d\theta d\varphi.$$
Putting things together I get
$$I=\iint_Y\rho \ dS=\iint_DR^2\sin{\theta}\sqrt{R^2-x^2-y^2} \ d\theta d\varphi = \\ = \iint_DR^3\cos{\theta}\sin{\theta} \ d\theta d\varphi = 2\pi R^3\int_0^\pi \cos{\theta}\sin{\theta} \ d\theta d\varphi = 0.$$
However, if the upper bound for $\theta$ is $\pi/2$, the answer becomes correct. But I believe my bounds are correct since that is standard for spherical coordinates. What is wrong?
Method 2
Just leaving the variables as they are I get the position vector expressed only in the variables $x$ and $y$ by
$$\vec{r}(x,y)=(x,y,\sqrt{R^2-x^2-y^2}),$$
so a the magnitude of the normalvector is
$$dS=||\vec{n}||dxdy=||\vec{r}'_x\times\vec{r}'_y||dx dy=\frac{R}{\sqrt{R^2-x^2-y^2}} \ dxdy$$
and I get that
$$I=\iint_Y \rho \ dS=\iint_D R r \ drd\theta =\int_0^{2\pi}\int_0^RRr \ drd\theta=\pi R^3,$$
which is half of the correct answer.
In method 2, setting $z=\sqrt{R^2-x^2-y^2}$ only gives you the upper hemisphere. When you add in the integral over the lower hemisphere with $z=-\sqrt{R^2-x^2-y^2}$, you'll double the answer to get your missing factor of $2$.
In method 1 you again have an issue with the signs of square roots: $\sqrt{R^2-x^2-y^2}$ is $R|\cos\theta|$, not $R\cos\theta$. Correcting this, you get the right answer. (Note that contrary to what you said, integrating from $0$ to $\pi/2$ does not give the right answer: it is still off by a factor of $2$, though it gives you the same answer as you got in method 2.)