how to calculate $\int_{C}(2x^2+1)e^{x^2+y^2}dx+(2xy)e^{x^2+y^2}dy$ using Green theorem

Question

Compute $\int_{C}(2x^2+1)e^{x^2+y^2}dx+(2xy)e^{x^2+y^2}dy$ where $C$ connects $(1,0)$ to $(0,1)$ by a straight line segment.

I tried to use green theorem since $Q_x = P_y$ so $\int \vec F_\dot{}\vec dr=0$

i need to close $C$ so i build a triangle with vertices $(0,0)$ , $(1,0)$ ,$(0,1)$ now i got a complicated integral in the line parametrization :

$y = 0 ~~~ dy=0$

$x = t ~~~ dx=dt$

$0\leq t \leq 1$

$\int_0^1 (2t^2 + 1)e^{t^2}dt$ , since $dy=0$ only the left side stays

any hints how i can build an easy enclosier ?

Answer 1

Since $e^{x^2+y^2}$ is a radial function, it is better to choose $D:(\cos t,\sin t), 0\le t\le \frac{\pi}{2}$ and use the fact $$\begin{eqnarray} \int_{C}(2x^2+1)e^{x^2+y^2}dx+(2xy)e^{x^2+y^2}dy&=&\int_{D}(2x^2+1)e^{x^2+y^2}dx+(2xy)e^{x^2+y^2}dy\\ &=&e\int_D (2x^2+1)dx + 2xydy\\ &=&e\left[\left(\frac{2}{3}x^3+x\right)\Big|^0_1+\frac{2}{3}\right]\\ &=&-e. \end{eqnarray}$$

Answer 2

Seeing $x^2+y^2$ in the expression hints at circles and polar coordinates. I would suggest connecting the two given points with the quarter of the circle $x^2+y^2=1$ in the first quadrant. Then you can parametrize it as $x=\cos t$, $y=\sin t$, for $0\le t\le\dfrac{\pi}{2}$.

Answer 3

Since $\int x e^{x^2 + y^2} d(y^2) = x e^{x^2 + y^2} + C(x)$, we can see that $\phi(x, y) = x e^{x^2 + y^2}$ is a potential for $\mathbf F$. Then, by the gradient theorem, $$I = \int_C (\nabla \phi) \cdot d\mathbf r = \phi(0, 1) - \phi(1, 0) = -e.$$