Computing surface area and volume of a unit sphere When solving some exercises, I forgot the formula for the surface area of the unit sphere. However, I remember that the length of the perimeter of a circle of radius $r$ is $2 \pi r$. So I figured I'll just compute
$$2 \int_0^1 2 \pi r \, \mathrm dr,$$
starting to integrate from the north pole of a unit sphere up to its center and as that area occurs twice, I added a factor of $2$. However, this integral yields $2 \pi$ and not the desired $4 \pi$. Where did I forget another factor of $2$?
I tried doing the same computation with the volume of the sphere since its surface area will just be the derivative of the volume. However, again
$$2 \int_0^1 \pi r^2 \, \mathrm dr = \frac{2}{3} \pi,$$
missing a factor of $2$. What am I doing wrong?

Hint about the surface:
Consider the surface defined by $$r=\cos u\sin v~\mathbf{i}+\sin u\cos v~\mathbf{j}+\cos v~\mathbf{k},~~0\leq u\leq 2\pi,~0\leq v\leq \pi/2$$ so if we set $x=\cos u\sin v,~y=\sin u\cos v,~z=\cos v$ then $x^2+y^2+z^2=1$ and this is clearly your unit sphere. Now we can use the following well-known formula. I hope you don't mind I am giving you this not pretty one:
$$S=\iint_S ds,~ds=\sqrt{\left(\frac{\partial(y,z)}{\partial(u,v)}\right)^2+\left(\frac{\partial(z,x)}{\partial(u,v)}\right)^2+\left(\frac{\partial(x,y)}{\partial(u,v)}\right)^2}$$
