Calculate $\int_CPdx+Qdy$ where $P(x,y)=xe^{-y^2}$ and $Q(x,y)=-x^2ye^{-y^2}+\frac{1}{x^2+y^2+1}$, $C$ is the boundary of the square determined

Question

Calculate $\int_CPdx+Qdy$ where $P(x,y)=xe^{-y^2}$ and $Q(x,y)=-x^2ye^{-y^2}+\frac{1}{x^2+y^2+1}$, $C$ is the boundary of the square determined by the inequalities $-a\leq x\leq a$, $-a\leq y\leq a$, oriented positively.

I have thought to do the following:

Using Green's theorem we get to that

$\int_CPdx+Qdy=\int\int_D(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})dA= \int\int_D(-2xye^{-y^2}-\frac{2x}{(x^2+y^2+1)^2}+2xye^{-y^2})dA=\int\int_D-\frac{2x}{(x^2+y^2+1)^2}dA=\int_{-a}^{a}\int_{-a}^{a}-\frac{2x}{(x^2+y^2+1)^2}dxdy=-\int_{-a}^{a}\int_{a^2+y^2+1}^{a^2+y^2+1}u^{-2}dudy=-\int_{-a}^{a}0dy=0$

Is this fine? Thank you.

Answer 1

Since your region $D$ is symmetric about the $x$-axis, any integrable function $f$ satisfying $f(x,y) = -f(-x,y)$ will satisfy $$\int_D f(x,y) \, dA = 0$$ by symmetry.

Answer 2

Something you might want to consider....

$G(x,y) = (xe^{-y^2}, -x^2 e^{-y^2})$ is a conservative field.

i.e. $(xe^{-y^2}, -x^2 e^{-y^2}) = \nabla (\frac12 x^2 e^{-y^2})$ and $\nabla \times (xe^{-y^2}, -x^2 e^{-y^2}) = 0$

But, you are not obligated to use Greens theorem / stokes therm. You can just strip out the conservative portion of your field.

$F(x,y) = G(x,y) + (0, \frac {1}{x^2+y^2 + 1})$

$\oint F(x,y)\cdot dr = \oint G(x,y) \cdot dr + \oint(0, \frac {1}{x^2+y^2 + 1})\cdot dr$

$\oint G(x,y) \cdot dr = 0$ because it is a conservative field.

Leaving:

$\oint(0, \frac {1}{x^2+y^2 + 1})\cdot dr$

It is up to you whether you think it is easier to integrate this or

$\iint \frac {-2x}{(x^2+y^2 + 1)^2} \ dA$

They will be equal.

If you chose to stick with the contour integral, the the impact of horizontal movement around the contour is zero.

And since the square is centered at the origin, we go up one side of the square, and down the other and get two identical integrals, in opposite directions, that cancel each other out.