I have a function $$f(x,y)=1/2x^2y-2x+1/6y^3-5/2 y$$ and I need to find $\int_{d1}\nabla f\cdot dr$ with $d_1$ being orientated from the second quadrant towards the fourth quadrant.
$d_1$ will be the straight edge of D which is $$d_1 = \{(x,y): x^2 +y^2\le 9 \mathrm{\ and\ } x+y=0\}$$
Ok so D is just the whole circle which means the line will intersect at $3\pi/4$ and $-\pi/4$ and I know we are going to be integrating from top left to bottom right (quadrant 2 – 4) would this mean we are going counter clockwise around the circle??
Furthermore I did manage to calculate the gradient which came out to be $∇f=<xy-2, 1/2$ $x^2-y^2/2$ $-5/2>$ so how exactly would I go about setting up this integral and solving it with polar coordinates and am I correct in assuming it will be going counter clockwise?
Best Answer
The question says $d_1$ is the straight edge of $D$ which is $x+y = 0$ and $x^2 + y^2 \leq 9$. That means straight line segment from $A \ (-\frac{3}{\sqrt2}, \frac{3}{\sqrt2})$ to $B \ (\frac{3}{\sqrt2}, -\frac{3}{\sqrt2})$.
So parametrize the line segment as $ \ \vec {r}(t) = (t, -t), -\frac{3}{\sqrt2} \leq t \leq \frac{3}{\sqrt2}$.
Now please note that you are finding line integral of $\vec F$ which is a conservative vector field as $\vec F = \nabla f$.
So your line integral will simply be $f(B) - f(A)$