# Set {h}(x,y) = (2x sin y - e^{2x} + 4) {i} + (x^2 cos y + 2e^{3y} - y^2) {j}. Let C be the...

## Question:

Set {eq}\mathbf{h}(x,y) = (2x \sin y - e^{2x} + 4) \mathbf{i} + (x^2 \cos y + 2e^{3y} - y^2) \mathbf{j}. {/eq} Let {eq}C {/eq} be the curve {eq}\mathbf{r}(u) = 4 \sin u \mathbf{i} + 2 \cos u \mathbf{j}, \; 0 \leq u \leq 2 \pi {/eq}.

{eq}\int_{C} \mathbf{h}(\mathbf{r} \cdot \mathbf{d} \mathbf{r}) = ? {/eq}

## Green's Theorem:

We will use {eq}\oint F_{1}dx+F_{2}dy=\int \int \left ( \frac{\partial F_{2}}{\partial x}-\frac{\partial F_{1}}{\partial y} \right )dxdy {/eq} by solving the problem where we will find the partial derivatives with respect to x and y.

To solve the problem we will use Green's Theorem:

{eq}\oint F_{1}dx+F_{2}dy=\int \int \left ( \frac{\partial F_{2}}{\partial x}-\frac{\partial F_{1}}{\partial y} \right )dxdy {/eq}

Now let us find the partial derivatives:

{eq}\frac{\partial F_{2}}{\partial x}=2x\cos y\\ \frac{\partial F_{1}}{\partial y}=2x\cos y {/eq}

The integral will be:

{eq}=\int \int (2x\cos y-2x\cos y)dxdy\\ =0 {/eq} 