Compute $\int_{\gamma} \nabla f \cdot d\mathbf{x}$

Question

Compute $\int_{\gamma} \nabla f \cdot d\mathbf{x}$ for the following choices of $f$ and $\gamma$.
(a) $f(x,y) = x^2+y^2; \gamma:g(t) = (1+t^2, 1-t^2), -1 \le t \le2$

What I have tried: $$\int_{-1}^2(1+t^2)^2dx + \int_{-1}^2(1-t^2)^2dy$$

Where $dx = 2t$ and $dy = -2t$ So we have $$\int_{-1}^2(1+t^2)^2(2t)dt + \int_{-1}^2(1-t^2)^2(-2t)dt$$

however this outputs the wrong result. How do I proceed from here>?

Answer 1

Note that $f(x, y) = x^2 + y^2$, so the vector field $\nabla f$ is $$ \nabla f (x, y) = (2x, 2y).$$ By definition, \begin{align} \int_\gamma \nabla f \cdot dx &= \int_{-1}^2 2(1+ t^2) 2t dt + \int_{-1}^2 2 (1-t^2) (-2t) dt, \end{align} which gives $30$.

In general, when the vector fields in the gradient of a function, the line integral depends only on the end points but not the path: for any path $\gamma : [a, b]\to \mathbb R^2$,
$$\int_\gamma \nabla f \cdot dx = f(\gamma(b)) - f(\gamma(a)).$$

In our case $$f(\gamma(b)) - f(\gamma(a)) = f(\gamma(2)) - f(\gamma(-1)) =f(5, -3) - f(2,0) = 25 + 9 - 4 = 30.$$

Answer 2

For $f(x,y) = x^2+y^2$ we have
$$ \nabla f. X = 2(x^2+y^2)$$ Thus,
$\begin{array}{ccl} \displaystyle\int_{\gamma} \nabla f. X ds &=& 2 \displaystyle\int_{\gamma} (x^2+y^2) ds\\ &=& 2 \displaystyle\int_{-1}^{2} ((1+t^2)^2+(1-t^2)^2) \sqrt{8t^2} dt\\ &=& 2 \sqrt{8} \left( - \displaystyle\int_{-1}^{0} ((1+t^2)^2+(1-t^2)^2) t dt +\displaystyle\int_{0}^{2} ((1+t^2)^2+(1-t^2)^2) t dt \right) \\ &=& 8\sqrt{2} \left(-\displaystyle\int_{-1}^{0} t^5+tdt+\displaystyle\int_{0}^{2}t^5+tdt \right) \\ &=&96\sqrt{2}. \end{array}$