91影视

It is given that

Green Theorem states that:

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

$\mathbf{r}\left(t\right)=鉄�\left(x\right(t),y(t\left)\right)鉄�\text{for}a鈮�t鈮�b$

According to given question

${鈭�}_C\mathbf{F}路d\mathbf{r}={鈭�}_R\left(\frac{鈭�F_2}{鈭�x}-\frac{鈭�F_1}{鈭�y}\right)dA.$

From given equations

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

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

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

$鈬�\frac{鈭�F_2}{鈭�x}=e^x\sin y$

And

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

$鈬�\frac{鈭�F_1}{鈭�y}=-e^x\sin y$

Using Green's Theorem, we get

${鈭�}_C\mathbf{F}路d\mathbf{r}={鈭�}_R\left(\frac{鈭�F_2}{鈭�x}-\frac{鈭�F_1}{鈭�y}\right)dA$

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

$={鈭�}_{R}2e^x\sin ydA$

According to condition given in question,

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

$鈬�y=卤\frac{1}{2}\sqrt{25-3x^2}$

If $y=0$, equation becomes

$3x^2=25$

$x=卤\frac{5\sqrt{3}}{3}$

Hence, region of integration is described as:

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

Solving integral, we get

${鈭�}_C\mathbf{F}路d\mathbf{r}={鈭�}_R\left(2e^x\sin y\right)dA$

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

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

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

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

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

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

$={鈭�}_{-\frac{5\sqrt{3}}{3}}^{\frac{5\sqrt{3}}{3}}2e^x路0dx$

$=0$

Hence,${鈭�}_C\mathbf{F}路d\mathbf{r}=0$