Evaluating line integrals given multiple paths

Question

Evaluate $$ \int_c (\ln(1+x^5)+1/2 y^2)dx+xydy $$ for the following sub-paths that together make up the closed path C.

a.) $y=x^2$ from $(-1,1)$ to $(1,1)$,

b.) $x=1$ from $(1,1)$ to $(1,3)$,

c.) $x^2+y^2=10$, from $(1,3)$ to $(-1,3)$

d.) $x=-1$ from $(-1,3)$ to $(-1,1)$.

I have my calc final coming up and I'm 90% certain this type of question will be on it. It's from my old quiz and I had no idea how to do it.. too many things, just super confused. what I DO know is that this is a conservative function, which in turn means that the path does not matter (which cuts all the work down tremendously) ...

What I don't know, is where I start.. and where I end. I don't have the slightest clue on how to do this problem. Please help!

Answer 1

There are two ways you can go by (but in the long run...)

you can perameterize each of the contours, and do 4 integrals.

Or you can apply Green's theorem.

over a closed contour

$\oint P(x,y) dx + Q(x,y) dy = \iint (\frac {\partial Q}{\partial x} - \frac{\partial P}{\partial y}) \;dA$

With $P$ and $Q$ as above $\frac {\partial Q}{\partial x} - \frac{\partial P}{\partial y} = y-y = 0$ you have a conservative force.

A third way that I hadn't considered when I strated.

Answer 2

I think that something is wrong with your exercise.

The vector field is conservative in the half-plane $D:=\{(x,y):x>-1\}$ with a scalar potential $$U(x,y)=\int\ln(1+x^5)dx+\frac{xy^2}{2}.$$ If $C$ is a closed curve contained in $D$ then the integral along $C$ is zero.

In your case $C$ is NOT contained in $D$. In part d) $x=-1$ and $\log(1+x^5)$ is NOT defined there because $1+(-1)^5=0$. This means that you can't evaluate it (or apply Green's Theorem).