Let $c:I^2\rightarrow\mathbb{R}^3$ be the singular $2$-cube given by $$c(s,t)=(\frac{1}{2}s^2,st,\frac{1}{2}t^2)$$ Let $$\omega=xy^2dz$$
Questions:
i) Compute $c^*\omega$
ii) Compute $c^*d\omega$
iii) Compute $\int_cd\omega$
iv) Without using Stokes theorem, compute $\int_{\partial c}\omega$ (Hence verifying Stokes theorem)
Answers:
(Parts (i)-(iii) checked and are correct)
i) I have gotten: $c^*\omega=\frac{1}{2}s^4t^2dt$
ii) $c^*d\omega=d(\frac{1}{2}s^4t^2dt)=2s^3t^2ds\wedge dt$
iii) $\int_cd\omega=2\int_0^1(\int_0^1s^3t^2ds)dt=2\frac{1}{4}\frac{1}{3}=\frac{1}{6}$
iv) I am using the formula $\int_{\partial c}\omega=\sum_{j=1}^2\sum_{\alpha=0,1}(-1)^{j+\alpha}\int_0^1c^*_{(j,\alpha)}\omega$
I have computed that $c_{(1,0)}(t)=(0,0,\frac{1}{2}t^2),c_{(1,1)}(t)=(\frac{1}{2},t,\frac{1}{2}t^2),c_{(2,0)}(t)=(\frac{1}{2}t^2,0,0),c_{(2,1)}(t)=(\frac{1}{2}t^2,t,\frac{1}{2})$
So $c_{(1,0)}^*\omega=0,c_{(2,0)}^*\omega=0,c_{(2,1)}^*\omega=0$
Point of contention:
My calculation:
$c_{(1,1)}^*\omega$ is the only non-zero part for $\partial c$
And $c_{(1,1)}^*\omega=\omega(\frac{1}{2},t,\frac{1}{2}t^2)=\frac{1}{2}\cdot t^2\cdot d(\frac{1}{2}t^2)$ from the definition of $\omega=xy^2dz$
This gives $\omega(\frac{1}{2},t,\frac{1}{2}t^2)=\frac{1}{2}t^2\cdot tdt=\frac{1}{2}t^2\cdot tdt$
So I have gotten $c_{(1,1)}^*\omega=\frac{1}{2}t^3dt$
this is wrong since,
Answer given: $c_{(1,1)}^*\omega=\frac{1}{2}t^2dt$
and the next step is $$\int_{\partial c}\omega=\int_0^1\frac{1}{2}t^2dt=\frac{1}{6}$$
Which must be correct since this verifies Stokes theorem from part (iii)
And my answer must be wrong since it gives $\int_{\partial c}\omega=\frac{1}{8}$
The answer also has $c_{(1,1)}(t)=(\frac{1}{2},t,\frac{1}{2}t^2)$, but computes $c_{(1,1)}^*\omega$ as $\frac{1}{2}t^2dt$, whereas I get $\frac{1}{2}t^3dt$
Where am I going wrong with this calculation, it is probably a simple mistake, any help would be greatly appreciated.