1
$\begingroup$

Consider the vector field $${\mathbf F}(x,y)=(e^x)(\sin y){\mathbf i}+(e^x)(\cos y){\mathbf j},$$ and the curve $C$ composed of the graph of $\sqrt{x}+\sqrt{y}=5$ followed by segment from $(25,0)$ to $(0,0)$. Evaluate the line integral $$\int_C {\mathbf F} \cdot d {\mathbf r}.$$ I noticed that ${\mathbf F}(x,y)$ is conservative, so I applied the Fundamental Theorem of Line Integral and got the potential function $$f(x,y)=(e^x)(\sin y),$$ matching my professor's answer key. I then did $f(0,0)-f(25,0)$ to get an answer of $0$, but in my professor's answer key he used $f(0,0)-f(0,25)$ to get an answer of $-\sin(25)$. Why is that?

$\endgroup$
1
$\begingroup$

\begin{align*} F(0,0)-F(0,25)&=e^{0}\sin(0)-e^0\sin(25) \\ &=1\cdot 0-1\cdot\sin(25)\\ &=-\sin(25) \end{align*}

Edit

The graph of $$\sqrt{x}+\sqrt{y}=5$$ is the portion of a parabola going from $(0,25)$ to $(25,0)$, which you must include as part of your integral. So the integral starts at $(0,25)$, traverses the parabola, then goes to $(0,0)$ along the $x-axis$. So the start point is $(0,25)$ and the end point is $(0,0)$.

$\endgroup$
  • $\begingroup$ You are not addressing the OP's question. He understands that $f(0,0) - f(0,25) = - \sin(25)$. He does not understand why that is what he's supposed to do. $\endgroup$ – Mark Fantini Dec 7 '14 at 18:38
  • $\begingroup$ Oh woops...ill edit. $\endgroup$ – illysial Dec 7 '14 at 18:38
  • $\begingroup$ @Mark_Fantini I have made the appropriate edit $\endgroup$ – illysial Dec 7 '14 at 18:44
  • 1
    $\begingroup$ I have reversed the downvote and upvoted. :) $\endgroup$ – Mark Fantini Dec 7 '14 at 18:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.