3-Ball. 3−manifold in $\mathbb{R}^3$

The $3-$ball ${B_R}^3 = \{(u,v,w) \in \mathbb{R}^3 | u^2+v^2+w^2 \le R^2\}$ is a $3-$manifold in $\mathbb{R}^3$; orient it naturally and give $${S_R}^2 = \partial {B_R}^3 = \{ (u,v,w)\in \mathbb{R}^3 | u^2+v^2+w^2 = R^2\}$$ the induced orientation. Assume that $\omega$ is a $2-form$ defined in $\mathbb{R}^2 \setminus \{0\}$ such that $$\int_{{S_R}^2} \omega = a+\dfrac{b}{R}$$for each $R>0$,

a) Given $0<c<d$, let $M$ be the $3-$manifold in $\mathbb{R}^3$ consisting of all $x$ with $c\le ||x|| \le d$, oriented naturally. Find $\int_{M} d\omega$.

b) If $d\omega =0$, what can you say about $a$ and $b$?

c) If $\omega = d\eta$ for some $\eta$ in $\mathbb{R}^3 \setminus \{0\}$, what can you say about $a$ and $b$?

*Munkres. Chapter 7. Paragraph 37. Problem 5.

Just finished part a)

Need help with b) and c)

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I think if you solved part a) with Stokes' theorem, then part b) and c) follows easily from the result and/or the method in part a). So, could you please show me your work on part a)? –  23rd Dec 20 '12 at 15:54
I got $b(\frac{1}{d}-\frac{1}{c})$ Does this make b) b=0 and c) a=0 and b=0 ? –  John Lennon Dec 20 '12 at 23:21
Yes, for part b), $b=0$ follows from part a); for part c), $a,b=0$ follows from Stokes' theorem. Could you please post an answer by yourself? –  23rd Dec 21 '12 at 16:06

a) $\int_M d\omega=\int_{\partial M}\omega=(\int_{S_d^2}-\int_{S_c^2}) \omega = (a+\frac{b}{d})-(a+\frac{b}{c})=b(\frac{1}{d}-\frac{1}{c})$
b) If $d\omega=0$, $0=\int_M d\omega=b(\frac{1}{d}-\frac{1}{c})$ $\Rightarrow b=0$
c) If $\omega=d\eta$, $\int_{S_R^2} \omega=\int_{S_R^2} d\eta = \int_{\partial S_R^2} \eta = \int_{\emptyset} \eta = 0$ $\Rightarrow a=b=0$