# Green's Theorem well, sort of.

1. Jan 10, 2009

### dtl42

1. The problem statement, all variables and given/known data
Evaluate $$\displaystyle \int_C y^2dx + x^2dy$$ for the path C: the boundary of the region lying between the graphs of $$\displaystyle y=x$$ and $$\displaystyle y=\frac{x^2}{4}$$.

2. Relevant equations
The catch is that you can't use Green's Theorem.

3. The attempt at a solution
I think you can break C into two other curves, $$C_1:r(t)=t\textbf{i}+t\textbf{j}$$ for $$0 \leq t \leq 4$$ and $$C_2:r(t)=(8-t)\textbf{i}+\frac{(8-t)^2}{4}\textbf{j}$$ for $$4 \leq t \leq 8$$.

I believe my error is somewhere below:

$$2\int_{C_1}t^2dt+\frac{-1}{2} \int_{C_2}(8-t)^3((8-t)+1)dt \Rightarrow 2\int_0^4t^2dt+\frac{-1}{2} \int_4^8(8-t)^3((8-t)+1)dt$$

I get that expression equal to $$\frac{-1376}{15}$$, which I know is incorrect, as I used Green's Theorem, and got $$\frac{32}{15}$$ which also coincides with the answer key I have.

2. Jan 10, 2009

### Dick

Shouldn't your second integrand be something like (8-t)^4/16+(8-t)^3/2? With only a (-1) coefficient outside?

Last edited: Jan 10, 2009
3. Jan 10, 2009

### dtl42

Yes, that's what I had, but I factored it just a little bit in the above expression.

4. Jan 10, 2009

### Dick

Then you factored it wrong. The integral of 2*t^2 from 0 to 4 minus the integral of (8-t)^4/16+(8-t)^3/2 from 4 to 8 is -32/15.

5. Jan 10, 2009

### dtl42

I realized my mistake, I had been thinking that I needed to get positive 32/15, but it all depends on the direction of the curve, which in my case was clockwise, and is usually taken to be counterclockwise. Is that correct?

6. Jan 10, 2009

### Dick

Yes, but how did you get -1376/15?

7. Jan 10, 2009

### dtl42

I forgot a factor of 1/8 that should be in the second expression when I factored, thanks for the help!

8. Jan 10, 2009

### Dick

I thought it was something like that. You set up the problem just fine. Very welcome.