Use Green’s Theorem to evaluate the integral:

and C is the ellipse with equation $3x^2+4y^2=25$ traversed counterclockwise.

It is given that

Green Theorem states that:

Let R be a region in the plane with smooth boundary curve C oriented counterclockwise by

$\mathbf{r}\left(t\right)=⟨\left(x\right(t),y(t\left)\right)⟩\text{for}a\le t\le b$

According to given question

${\int }_C\mathbf{F}·d\mathbf{r}={\iint }_R\left(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right)dA.$

From given equations

$F_1(x,y)=e^x\cos y$

$F_2(x,y)=e^x\sin y$

$\frac{\partial F_2}{\partial x}=\frac{\partial }{\partial x}\left(e^x\sin y\right)$

$\Rightarrow \frac{\partial F_2}{\partial x}=e^x\sin y$

And

$\frac{\partial F_1}{\partial y}=\frac{\partial }{\partial y}\left(e^x\cos y\right)$

$\Rightarrow \frac{\partial F_1}{\partial y}=-e^x\sin y$

Using Green's Theorem, we get

${\int }_C\mathbf{F}·d\mathbf{r}={\iint }_R\left(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right)dA$

$={\iint }_R\left(e^x\sin y-\left(-e^x\sin y\right)\right)dA$

$={\iint }_{R}2e^x\sin ydA$

According to condition given in question,

$3x^2+4y^2=25$

$\Rightarrow y=\pm \frac{1}{2}\sqrt{25-3x^2}$

If $y=0$, equation becomes

$3x^2=25$

$x=\pm \frac{5\sqrt{3}}{3}$

Hence, region of integration is described as:

$R=\left\{(x,y)\mid -\frac{5\sqrt{3}}{3}\le x\le \frac{5\sqrt{3}}{3},-\frac{1}{2}\sqrt{25-3x^2}\le y\le \frac{1}{2}\sqrt{25-3x^2}\right\}$

Solving integral, we get

${\int }_C\mathbf{F}·d\mathbf{r}={\iint }_R\left(2e^x\sin y\right)dA$

role="math" localid="1653247009175" $={\int }_{-\frac{5\sqrt{3}}{3}}^{\frac{5\sqrt{3}}{3}}{\int }_{-\frac{1}{2}\sqrt{25-3x^2}}^{\frac{1}{2}\sqrt{25-3x^2}}\left(2e^x\sin y\right)dydx$

role="math" localid="1653247018268" $={\int }_{-\frac{5\sqrt{3}}{3}}^{\frac{5\sqrt{3}}{3}}\left[{\int }_{-\frac{1}{2}\sqrt{25-3x^2}}^{\frac{1}{2}\sqrt{25-3x^2}}\left(2e^x\sin y\right)dy\right]dx$

role="math" localid="1653247062123" $={\int }_{-\frac{5\sqrt{3}}{3}}^{\frac{5\sqrt{3}}{3}}2e^x[-\cos y{]}_{-\frac{1}{2}\sqrt{25-3x^2}}^{\frac{1}{2}\sqrt{25-3x^2}}dx$

$={\int }_{-\frac{5\sqrt{3}}{3}}^{\frac{5\sqrt{3}}{3}}2e^x\left[\left(-\cos \left(\frac{1}{2}\sqrt{25-3x^2}\right)\right)-\left(-\cos \left(-\frac{1}{2}\sqrt{25-3x^2}\right)\right)\right]dx$

$={\int }_{-\frac{5\sqrt{3}}{3}}^{\frac{5\sqrt{3}}{3}}2e^x\left[\left(-\cos \left(\frac{1}{2}\sqrt{25-3x^2}\right)\right)+\cos \left(-\frac{1}{2}\sqrt{25-3x^2}\right)\right]dx$

$={\int }_{-\frac{5\sqrt{3}}{3}}^{\frac{5\sqrt{3}}{3}}2e^x\left[\left(-\cos \left(\frac{1}{2}\sqrt{25-3x^2}\right)\right)+\cos \left(\frac{1}{2}\sqrt{25-3x^2}\right)\right]dx$

$={\int }_{-\frac{5\sqrt{3}}{3}}^{\frac{5\sqrt{3}}{3}}2e^x·0dx$

$=0$

Hence,${\int }_C\mathbf{F}·d\mathbf{r}=0$