
Calculus III problem
Hi, so tomorrow's pretty much the last day of the summer semester for me and I have a homework assignment due tomorrow that's confusing me sooo much..
Im here asking for some help in direction on how to do these problems so that I can find the solution by myself... I would really really appreciate any help anyone could provide ...
I am not sure how to represent the domain as a function, and if anyone could point me in the right direction, I'm sure I could get the answer.
The Problem 
http://img843.imageshack.us/img843/3995/calc3.jpg
I would really really really appreciate any help at all. I need to know how to represent the domain as a function.

Don't think of the domain as one big function. Think of it as 3 parts as in the descrption:
C1: line from (0,0) to (1,0)
C2: circular arc from (1,0) to the line x=y. What is x in this case?
C3: line from (x,x) to (0,0)
If you don't know how to parametrize lines, look in your book somewhere around parametric equations of a line.
The circular arc you can just parametrize using sines and cosines.
I hope this helps!

1 Attachment(s)
I think we should go for Green's Theorem.
$\displaystyle \displaystyle{\int_{\partial R} P dx + Q dy = \int\int_{R}\left(\frac{\partial Q}{\partial x}  \frac{\partial P}{\partial y}\right)dxdy}$
So, lets do it:
Let have $\displaystyle P = \sin x  6x^2y $ and $\displaystyle Q=3xy^2x^3$.
$\displaystyle \frac{\partial Q}{\partial x} = 3y^2  3x^2 $
$\displaystyle \frac{\partial P}{\partial y} =  6x^2 $
$\displaystyle \frac{\partial Q}{\partial x}\frac{\partial P}{\partial y} = 3y^2 + 3x^2 $ (How convenient!!)
Now we have to determine the region bounded by the curve C. I've made a sketch:
Attachment 18371
You see that we're "walking" counterclockwise  thus we should have $\displaystyle \int\int_{R} 3y^2 + 3x^2~dxdy$. There would be a negative signal if we walked clockwise.
Well. We should determine the limits of integration. I mentioned above the convenience because this integral is asking so much that we change it for polar coordinates  therefore we can easily find the limits of integration (I think the sketch speaks by itself). So we put:
$\displaystyle x = r cos t$ and $\displaystyle y=r sin t$.
Our angle will vary from 0 to $\displaystyle \pi/2 + \pi/4 = 3\pi/4$ and the radius will vary from 0 to 1 (unit circle). This will lead us to:
Ah, before, we can't forget the Jacobian (I will not show this, but you should know how its done), so $\displaystyle dx dy = r dr dt$. Finally, the integral becomes:
$\displaystyle \displaystyle{\int\int_{R} 3 (y^2 + x^2)~dxdy = \int_{0}^{3\pi/4}\int_{0}^{1} 3 r^2\cdot r~drdt $ because $\displaystyle x^2+y^2 = r^2$ by our change.

It says counterclockwise in the question!

Oh, my mistake, sorry. I'll correct it.

Thank you so much for the quick replies. I appreciate it a looooot! I will try it out and see what I can come up with. And keep you guys updated.

Already done, I guess you can follow now.