1. Question about Green's Theorem

I am working on a take home quiz for my Calculus III class, and one of the problems says "Verify Green's Theorem for the line integral ∫xy dx + x² dy, where C is the triangle with vertices (0,0) (1,1) (2,0)."

Ok so this might be kind of a dumb question, but since it says "Verify" instead of "evaulate" am I supposed to do something different rather than find what the integral is equal to? I know how to evaluate this but I just am not sure if that means the same thing as verify. I don't want to misinterpret the problem and end getting a completely wrong answer.

2. The Green's Theorem says that:

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

In other words, it tells you that you can either calculate a double integral over the region R, or you can calculate a line integral over the contour wich encloses R. So to verify this theorem you must evaluate both and show that they're equal.

I hope it helps.

3. Thank you bondesan, I solved the double integral portion and got an answer of 1. However, I have no idea how to solve the ∫P dx + Q dy part. Could someone help out with this please?

4. You can use the definition

$\displaystyle \displaystyle{\int_C F\cdot dr}$

Where C is a picewise smooth curve described by the parametric equation $\displaystyle r(t) = x(t) \overset{\rightharpoonup }{i} + y(t) \overset{\rightharpoonup }{j}$ and F is a bidimensional vector field $\displaystyle F(x,y)=P(x,y)\overset{\rightharpoonup }{i} +Q(x,y)\overset{\rightharpoonup }{j}$

By doing $\displaystyle \dfrac{dr}{dt} = r'(t)$ we can rewrite the line integral as:

$\displaystyle \displaystyle{\int_C F(r(t))\cdot r'(t)~dt}$

Where we have a dot product. So for the problem you have $\displaystyle F(x,y) = xy \overset{\rightharpoonup }{i} + x^2 \overset{\rightharpoonup }{j}$ and $\displaystyle r(t)$ will depend, because you should take three line integrals over the three lines that closes the triangle, and each one will have a different parametrization.

So, let $\displaystyle P1=(0,0),~P2=(2,0),~P3=(1,1)$.

We can set the first line $\displaystyle L1$ as the one who links $\displaystyle P1$ to $\displaystyle P2$, $\displaystyle L2$ links $\displaystyle P2$ to $\displaystyle P3$, and $\displaystyle L3$ links $\displaystyle P3$ to $\displaystyle P0$. We followed a counter-clockwise direction.

Now we should parametrize $\displaystyle L1$, $\displaystyle L2$, $\displaystyle L3$.

For $\displaystyle L1$:

You can see that this line has $\displaystyle y=0$ and $\displaystyle 0 \leq x \leq 2$, so maybe the parametrization $\displaystyle r_1(t)=(2t,0)$ with $\displaystyle t\in[0,1]$ should be nice for us.

For $\displaystyle L2$:

This line begins at the $\displaystyle P2$ then go to $\displaystyle P3$, so we want that for some $\displaystyle r(t)$, when t=a we will be at $\displaystyle P2$and when t=b we will be at $\displaystyle P3$. You can use the formula $\displaystyle r(t)=(1-t)P2 + tP3$. It follows that $\displaystyle r_2(t)=(2-t,t)$ with $\displaystyle t\in[0,1]$.

For $\displaystyle L3$:

We can use the same formula again, so $\displaystyle r(t)=(1-t)P3 + tP0$ give us $\displaystyle r_3(t)=(1-t,1-t)$ with $\displaystyle t\in[0,1]$

Now that we have the three parametric equations for each line we can evaluate:

$\displaystyle \displaystyle{\int_{Li} F(r_i(t))\cdot r_i'(t)~dt}$ for $\displaystyle i = 1,2,3$

Finally, we sum each value to get the final answer.

I apologize if I did any mistake, I don't have any book around here by now and I had to remember the whole process - that's why it lacks some deep explanations. I did my best and I hope it helps. Maybe the other guys around here can add some thoughts about it.

5. Your integral should be $\displaystyle - \int \int_D ( Q_x - P_y )~dydx$ because the direction of the path is clockwise , in this case ...

Then it should be $\displaystyle - \int \int_D x dydx$

Note that the x-coordinates of the centroid of a region is defined as $\displaystyle \frac{ \int \int_D x dydx }{ \int \int_D dydx }$

Therefore , $\displaystyle - \int \int_D x dydx = - ( \text{ x-centroid } ) ~ \cdot ( \text{ area })$

$\displaystyle = - ( \frac{ 0 + 1 + 2 }{3} ) ( 1 ) = -1$