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Let $f(x,y) = x^2 y$ and let $C$ be curve $\vec{r}(t) = \langle e^{t^2},\ \sin(\pi t/2 )\rangle$ for $0\le t \le 1$

$$\int_C \nabla f \cdot d\vec{r}$$

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What have you tried so far? Do you know the gradient theorem? –  EuYu Nov 4 '12 at 21:39
    
I took the gradient of f and plug in r to obtain F to dot –  40Plot Nov 4 '12 at 21:44
    
Do you have access to the gradient theorem? –  EuYu Nov 4 '12 at 21:47
    
What is the gradient theorem? –  40Plot Nov 4 '12 at 21:53
    
Sometimes it's also called the Fundamental Theorem for Line Integrals. Here. –  EuYu Nov 4 '12 at 21:54

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Well, we have that $$\int_{C}\nabla f\cdot d\mathbf{r}=\int_{t_{i}}^{t_{f}}\nabla f(\mathbf{r}(t))\cdot\frac{d\mathbf{r}}{dt}dt=\int_{t_{i}}^{t_{f}}\frac{df(\mathbf{r}(t))}{dt}dt$$ or $$\int_{C}\nabla f\cdot d\mathbf{r}=f(\mathbf{r}(t_{f}))-f(\mathbf{r}(t_{i}))$$ So, in this particular case $f(\mathbf{r})\equiv f(x,y)=x^2y$; $\mathbf{r}(t)\equiv (x(t),y(t))=(e^{t^2},\sin \frac{\pi t}{2})$; and $t_i=0$, $t_f=1$; $$\int_{C}\nabla f\cdot d\mathbf{r}=f(\mathbf{r}(1))-f(\mathbf{r}(0))=f(e,1)-f(1,0)$$ or $$\int_{C}\nabla f\cdot d\mathbf{r}=e^2\cdot1-1^2\cdot 0=e^2$$

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The Gradient Theorem says that for a differentiable function $f$, $$ f(x_1)-f(x_0)=\int_{\gamma(x_0,x_1)}\nabla f(r)\cdot\mathrm{d}r $$ where $\gamma(x_0,x_1)$ is a path from $x_0$ to $x_1$.

This is a generalization of the Fundamental Theorem of Calculus.

Applying to $f(x,y)=x^2y$ and $x_0=\langle1,0\rangle$ and $x_1=\langle e,1\rangle$, we get $f(x_1)-f(x_0)=e^2$.

Of course, your assignment might be to compute the integral so as to verify this result. In that case you need to use $$ \begin{align} \nabla f(x,y) &=\langle2xy,x^2\rangle\\ &=\langle2e^{t^2}\sin(\pi t/2),e^{2t^2}\rangle \end{align} $$ and $$ \mathrm{d}r=\langle2te^{t^2},\pi/2\,\cos(\pi t/2)\rangle\mathrm{d}t $$

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