# 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

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.

• How did you calculate $\dfrac{\partial Q}{\partial x}$? – Umberto P. Nov 16 '18 at 17:37
• You seem to be missing a term. – Doug M Nov 16 '18 at 17:39
• @UmbertoP. Yes, I had an error in that but I already corrected it and I got to where I show in the question, what else can I do? Use polar coordinates? – Nash Nov 16 '18 at 17:41

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.

• Excellent answer, could you see if what I did in the question is right? Thank you. – Nash Nov 16 '18 at 17:52

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.