Multi-Variable, Line Integrals

[cloudmaster]

Hi! I am currently in a Multi-Variable (Calculus 3) class and we are on the unit with line and surface integrals. I am a bit confused with the following problem:

Let C be the closed path in the xy plane consisting of the four line segments and a half circle with points (6,3), (10,3), (10,7), (10,9), (6,9), and back to (6,3).

**The shape is essentially a box, with a semi-circle cutting into the rectangle from (10,3) to (10,7). Below is a sketch of it **



Evaluate: Integral (y/x+lny)dx + (ln(xy) +x/y -3x)dy

The first thing I tried was Green's Theorem, since this is a closed integral in a counterclock-wise direction. I said P= (y/x+lny) and Q= (ln(xy) +x/y -3x)
Thus, P(y)= 1/y + 1/y and Q(x)= 1/x +1/y -3

Then since Greens uses the double integral of Q(x) - P(y), I attempted to evaluate the integral over 6<x<10 and 3<y<9 (The bounds of the rectangle).

My plan is to first evaluate over the entire rectangle, and then evaluate over the semi-circle and then subtract the semi-circle from the rectangle.

I was wondering particularly:
Am I approaching this problem correctly? (With Greens?)
If I can use greens, how do I take the semi-circle into account?

I am actually super confused and have been staring at this problem for a while...
Thank-you to any thoughts/help!

 

[HallsofIvy]

Yes, you can use Green's theorem here. To integrate over the semi-circle, with center at (10, 8), radius 2, you could do it directly: integrate with x going from 9 to 10 and, for each x, y from \(\displaystyle 8- \sqrt{4- (x- 10)^2}\) to \(\displaystyle 8+ \sqrt{4- (x- 10)^2}\).

Or shift to modified "polar coordinates". Since the center of the circle is at (10, 8) rather than (0, 0), let x= 10+ rcos(t), y= 8+ rsin(t). Integrate with r from 0 to 2 and t from \(\displaystyle \pi/2\) to \(\displaystyle 3\pi/2\).

Once you have the integral over that half circle, subtract, as you said, from the integral over the entire rectangle.

It doesn't look like it would be too hard to do the line integral directly rather than using Green's theorem. On the straight line from (6, 3) to (10, 3), y is constant. x goes from 6 to 10, y= 3 so the integral is \(\displaystyle \int_6^{10} 3/x+ ln(3) dx\). Around the semi-circle, x= 10- 2 sin(t), y= 4- cos(t) with t going from 0 to \(\displaystyle \pi\).
For the line from (10, 7) to (10, 9), x= 10, y goes from 7 to 9 with dx= 0. For the line from (10, 9) to (6, 9), y= 9 and dy= 0 integrating over x from 10 to 6. From (6, 9) to (6, 3), x= 6 so dx= 0 and you integrate over y from 9 to 3.

 

[cloudmaster]

Thanks! In the end, I decided to find the rectangle first, and then subtract the semi-circle.