# Let f(x,y) = \sin x \cos y + e^{2xy} and let C be the path in the xy plane that follows the arc...

## Question:

Let {eq}f(x,y) = \sin x \cos y + e^{2xy} {/eq} and let C be the path in the xy plane that follows the arc of {eq}y = \sin x {/eq} from {eq}(\frac \pi 2, 1) {/eq} to {eq}(\pi,0) {/eq}. Then:

(a) Find the gradient of {eq}f {/eq}, that is, find {eq}\vec F = \nabla f {/eq} .

(b) Explain why Green's Theorem cannot be applied to find {eq}\int_C \vec F\, d \vec r {/eq}

(c) Use a different method to find {eq}\int_C \vec F\, d \vec r {/eq}.

## Fundamental Theorem of Line Integrals:

The Fundamental Theorem of Line Integrals (FTLI) states that the line integral of a conservative vector field is independent by the choice of the path.

If a vector field {eq}\displaystyle \mathbf{F}(x,y)=\langle P(x,y), Q(x,y)\rangle {/eq} is conservative, i.e. {eq}\displaystyle \frac{\partial P}{\partial y}= \frac{\partial Q}{\partial x} {/eq}, then there is a potential function {eq}\displaystyle f(x,y) {/eq} such that {eq}\displaystyle \mathbf{F}(x,y)=\nabla f(x,y) {/eq} and {eq}\displaystyle \int_{\mathcal{C}} \mathbf{F}\cdot d\mathbf{r} =f(x(b),y(b))-f(x(a),y(a)), {/eq} where {eq}\displaystyle \mathcal{C}: x=x(t), y=y(t), a\leq t\leq b. {/eq}

## Answer and Explanation:

(a) The gradient of {eq}\displaystyle f(x,y)=\sin x\cos y+e^{2xy} {/eq} is given by the vector whose components are the first partial derivatives, as below

{eq}\displaystyle \begin{align} \mathbf{F}(x,y)=\nabla f(x,y)&= \left\langle \frac{\partial }{\partial x}\left( \sin x\cos y+e^{2xy}\right), \frac{\partial }{\partial y}\left( \sin x\cos y+e^{2xy}\right)\right\rangle \\ &= \boxed{\left\langle \cos x\cos y+2ye^{2xy}, -\sin x\sin y+2xe^{2xy}\right\rangle}. \end{align} {/eq}

b. To evaluate the line integral {eq}\displaystyle \int_{\mathcal{C}} \mathbf{F}\cdot d\mathbf{r}, {/eq} where {eq}\displaystyle \mathbf{F} {/eq} is the gradient vector obtained in a) and {eq}\displaystyle \mathcal{C} {/eq} is the curve {eq}\displaystyle y=\sin x, \frac{\pi}{2}\leq x\leq \pi, {/eq}

we need to use the direct computation of the line integral, since we cannot apply Green's Theorem

{eq}\displaystyle \boxed{\text{ because the curve is not a closed curve}}. {/eq}

c. To evaluate the line integral {eq}\displaystyle \int_{\mathcal{C}} \mathbf{F}\cdot d\mathbf{r}, {/eq} where {eq}\displaystyle \mathcal{C} {/eq} is the curve {eq}\displaystyle y=\sin x, \frac{\pi}{2}\leq x\leq \pi, {/eq} and

{eq}\displaystyle \mathbf{F}(x,y)=\nabla f(x,y) \implies \text{ conservative vector field} {/eq}

we will apply the Fundamental Theorem of Line Integrals,

that states that the line integral of a conservative vector field is independent of the choice of the curve and

{eq}\displaystyle \begin{align} \int_{\mathcal{C}} \mathbf{F}\cdot d\mathbf{r}&=\int_{\mathcal{C}} \nabla f(x,y)\cdot d\mathbf{r}=f(\pi, 0)-f\left(\frac{\pi}{2},1\right)\\ &=\sin \pi \cos 0+1-\sin\left(\frac{\pi}{2}\right)\cos 1+e^{\pi}\\ &=\boxed{1+e^{\pi}-\cos 1}. \end{align} {/eq}