# Homework Help: Evaluate the Line Integral

1. Dec 5, 2011

### Bamboozled91

1. The problem statement, all variables and given/known data
Let C be the (positively oriented) boundary of the first quadrant of the unit disk. Use the definition of the line integral to find ∫(xy)dx+(x+y)dy

2. Relevant equations
x=rcos(x)
y=rsin(x)
dx=-sin(x)
dy=cos(y)
0≤ t ≤ ∏/2

3. The attempt at a solution
∫-cos(t)sin^2(t)+cos^2(t)+sin(t)cos(t) dt from 0 to ∏/2
Then I finished out the integral and was left with ∏/4-5/6 which is incorrect. Could it possibly have to do with the r or my limits of integration?

2. Dec 5, 2011

### Bamboozled91

BTW I can do this using Greens theorem

3. Dec 5, 2011

### Dick

No, the setup looks ok. I think you just did the integral wrong. I don't get the number you got as a answer.

4. Dec 5, 2011

### Bamboozled91

I used wolfram to see where I mesed up and when it gave me the integral it gave the answer pi/4-1/6 which is also incorrect. Also another weird thing is I did the integral again and I got the same thing wolfram did unfortunatley the back of the book disagrees it says the answer is pi/4-1/3. Which I believe because I can solve using greens so you got anything else

5. Dec 5, 2011

### Dick

You really need to show more of your work before we can help. Yes, Green's theorem gives you pi/4-1/3. So far you've only given the line integral over the arc. What do you get for the line integral over the x=0 and y=0 parts of the region?

6. Dec 6, 2011

### Bamboozled91

Yah sorry about not showing too much work but that would require a ton of typing and I am lazy. Other than that, I think I have solved it. I found that the line integral over x=0 would cancel out. As for y=0, I determined it to be -1/2 so when I add -1/2 to pi/4+1/6. I get pi/4-1/3 which is correct. Let me show my work at least for the part I just did x=0 and y=t and dx = 0 and dy = 1 so I get ∫t dt from 1 to 0. which gives a -1/2

7. Dec 6, 2011

### Dick

That's it all right.