Calculus III problem

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July 28th 2010, 09:03 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.
July 28th 2010, 09:19 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!
July 28th 2010, 09:32 PM
bondesan

1 Attachment(s)

I think we should go for Green's Theorem.

$\displaystyle{\int_{\partial R} P dx + Q dy = \int\int_{R}\left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)dxdy}$

So, lets do it:

Let have $P = \sin x - 6x^2y$ and $Q=3xy^2-x^3$ .

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

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

$\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} = 3y^2 + 3x^2$ (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 $\int\int_{R} 3y^2 + 3x^2~dxdy$ . 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:

$x = r cos t$ and $y=r sin t$ .

Our angle will vary from 0 to $\pi/2 + \pi/4 = 3\pi/4$ 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 $dx dy = r dr dt$ . Finally, the integral becomes:

$\displaystyle{\int\int_{R} 3 (y^2 + x^2)~dxdy = \int_{0}^{3\pi/4}\int_{0}^{1} 3 r^2\cdot r~drdt$ because $x^2+y^2 = r^2$ by our change.
July 28th 2010, 09:33 PM
Vlasev

It says counter-clockwise in the question!
July 28th 2010, 09:39 PM
bondesan

Oh, my mistake, sorry. I'll correct it.
July 28th 2010, 09:44 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.
July 28th 2010, 09:47 PM
bondesan

Already done, I guess you can follow now.