Using stokes' theorem 
$B=\{(x,y), x^2+y^2\le1\} $ is a closed ball and $S=\{(x,y,z), z=x^2+y^2, (x,y)\in B\} $ oriented so that $f:B\to S$ defined by $$f(x,y)=(x,y,x^2+y^2)$$
  is orientation preserving. Compute $$\int_{\partial S}\ coszdx+(xz+\tan y)dy+y^2z^3dz$$

My work: I tried to use Stokes theorem and wrote
$$\int_Sd(coszdx)+d((xz+\tan y)dy)+d(y^2z^3dz)= \sin zdz\land dx+zdx+xdz+2yz^3dz$$
I cannot proceed further. How should I use the information about $B$ and $S$ here? I am learning Stokes theorem now and it is hard to use it for me. Any help would be appreciated! Please give an answer instead of commenting how the notation is weird or something like that. If I could do it, I would
Thanks in advance!
 A: Something went wrong in your calculation of the exterior derivative.  When you differentiate a $1$-form, you are going to get a $2$-form (but some of your terms have a single differential).
If
$$
\omega = \cos z \, dx + (xz + \tan y) \, dy + y^2 z^3 \, dz
$$
then
$$
\begin{align}
\def\pa#1{\frac{\lower2pt\hbox{$\partial$}}{\partial #1}}%
\def\we{\wedge}%
d \omega &= \pa{z} \cos z \, dz \we dx + \pa{x} (xz) \, dx \we dy 
+ \pa{z} (xz) \, dz \we dy + \pa{y} (y^2z^3) \, dy \we dz \\
&= -\sin z \, dz \we dx + z \, dx \we dy 
+ x \, dz \we dy + 2yz^3 \, dy \we dz \\
&= z \, dx \we dy + \sin z \, dx \we dz + (2yz^3 - x) \, dy \we dz.
\end{align}
$$

Added later in response to comments.  In order to integrate the $2$-form over the surface $S$, we have to parametrize the surface in $2$ variables.  This is pretty straightforward, as $S$ is the graph of a function on the domain $B$:
$$
F(x,y) = (x, y, x^2 + y^2)
$$
Since $z = x^2 + y^2$,
$$
dz = d(x^2 + y^2) = 2x \, dx + 2y \, dy.
$$
Making these substitutions (really, calculating the pullback of the form $d\omega$ along $F$) yields
$$
\begin{align}
\def\pa#1{\frac{\lower2pt\hbox{$\partial$}}{\partial #1}}%
\def\we{\wedge}%
F^*d \omega &= (x^2 + y^2) \, dx \we dy 
+ \sin (x^2 + y^2) \, dx \we (2x \, dx + 2y \, dy) \\
& \qquad {}+ (2y(x^2 + y^2)^3 - x) \, dy \we (2x \, dx + 2y \, dy) \\
&= (x^2 + y^2) \, dx \we dy + 2y \sin(x^2 + y^2) \, dx \we dy 
+ 4xy(x^2 + y^2)^3 \, dy \we dx \\
&= \Bigl( (x^2 + y^2) - 4xy(x^2 + y^2)^3 + 2y \sin(x^2 + y^2) \Bigr) \, dx \we dy.
\end{align}
$$
Now, you calculate (remember that $S = F(B)$):
$$
\int_{\partial S} \omega = \int_S d\omega = \int_B F^*d\omega.
$$
It occurs to me that this integral is a candidate for a change to polar coordinates, so maybe we should use one more pullback to make the integration easier.  Define the rectangle 
$$
R = [0, 1] \times [0, 2\pi] = \bigl\{ (r, \theta) \mid 0 \le r \le 1 \text{ and } 0 \le \theta \le 2\pi \bigr\}
$$
and $P: R \to B$, given by
$$
P(r, \theta) = (r \cos \theta, r \sin \theta) = (x, y).
$$
Then,
$$
\left\{
\begin{align}
dx = d(r \cos \theta) &= \cos \theta \, dr - r \sin \theta \, d\theta \\
dy = d(r \sin \theta) &= \sin \theta \, dr + r \cos \theta \, d\theta.
\end{align}
\right.
$$
I'll leave it there.  You can substitute in these expressions to get a (hopefully) reasonable integral, completing the chain:
$$
\int_{\partial S} \omega = \int_S d\omega 
= \int_B F^*d\omega = \int_R P^*F^*d\omega.
$$

Note:  If this seems like awfully many steps just to calculate an integral, you're right!  Some of these steps could be combined and streamlined.  Notice, for example how $F \circ P$ maps the rectangle in $(r, \theta)$-coordinates into the paraboloid in $(x, y, z)$ coordinates and we could have calculated the pullback in one step.  If we avoid making any mistakes, then we would find that
$$
(F \circ P)^* = P^* \circ F^*.
$$
I've described the calculation on the right side of this equation, but it would be more efficient to do the calculation on the left.  Try it!
