# Calculus III problem

• Jul 28th 2010, 08:03 PM
SolitaryRaf
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.
• Jul 28th 2010, 08:19 PM
Vlasev
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!
• Jul 28th 2010, 08:32 PM
bondesan
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 $\displaystyle Q=3xy^2-x^3$.

$\displaystyle \frac{\partial Q}{\partial x} = 3y^2 - 3x^2$

$\displaystyle \frac{\partial P}{\partial y} = - 6x^2$

$\displaystyle \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} = 3y^2 + 3x^2$ (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" counter-clockwise - 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.
• Jul 28th 2010, 08:33 PM
Vlasev
It says counter-clockwise in the question!
• Jul 28th 2010, 08:39 PM
bondesan
Oh, my mistake, sorry. I'll correct it.
• Jul 28th 2010, 08:44 PM
SolitaryRaf
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.
• Jul 28th 2010, 08:47 PM
bondesan