http://calc3.askbot.com/questions/Open source question and answer forum written in Python and DjangoenCopyright Askbot, 2010-2011.Wed, 30 Jul 2014 19:20:51 -0500http://calc3.askbot.com/question/197/section-163/Number 8 in section 16.3 asks, Evaluate $\int (10x^4=2xy^3)dx-3x^2y^2dy$ where $C$ is the part of the curve $x^5-5x^2y^2-7x^2=0$ from $(0,0)$ to $(3,2)$. Can someone explain how to solve something like this?Wed, 30 Jul 2014 07:15:50 -0500http://calc3.askbot.com/question/197/section-163/http://calc3.askbot.com/question/197/section-163/?answer=206#post-id-206Since nobody seems to be answering, I'll try to lend a hand! This problem is just a cleverly cloaked version of the problems earlier in the homework (such as problems one through seven). The best most amazing part of The Fundamental Theorem of Line Integrals, is that *it doesn't matter what path the curve takes!* That is basically the most amazing thing ever since the regular Fundamental Theorem of Calculus. Seriously, it's awesome. Okay, with that enthusiastic introduction aside, I will solve this problem. I first will make sure that there exists a function $f$ such that $\nabla f = F$. Note that in this problem, $F = \langle P, Q \rangle$ where $P = 10x^4 - 2xy^3$ and $Q = -3x^2y^2$. If $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} $, then the mixed partials of $f$ are equal and $f$ should exist. $$\frac{\partial P}{\partial y} = -6xy^2$$ $$\frac{\partial Q}{\partial x} = -6xy^2$$ They are clearly equal and therefore we should be able to find $f$. I will integrate $P$ with respect to $x$ to gain some information about $f$ (although we will need to gather some more information after this). $$ f = \int 10x^4 - 2xy^3 \ dx = 2x^5 - x^2y^3 + C(y)$$ The missing information that we must now find is $C(y)$. Since differentiation is a "lossy" operation, we lose anything that is independent of $x$. In this case, we could lose some information about how $f$ depends on $y$. To find $C(y)$, we will solve for $C'(y)$ and antidifferentiate it to get what $C(y)$ is equal to. I believe that this is best illustrated by doing, rather than explaining, so: $$ \frac{\partial F}{\partial y} = -3x^2y^2 + C'(y) $$ Note that $ \frac{\partial F}{\partial y} $ necessarily *has* to equal $Q$, so that means: $$ -3x^2y^2 + C'(y) = -3x^2y^2 $$ Clearly $C'(y) = 0$ and therefore $C(y)$ equals $0$. So our function $f = 2x^5 - x^2y^3$. You can verify this by making sure that $\nabla f = F$ (it does). Now, we can utilize one of the most amazingly overpowered and totally groovy theorems! All we must do is plug in the endpoints to $f$ and we will obtain our final answer! The endpoints, in this case, are $(0, 0)$ and $(3, 2)$. Note that we start at $(0, 0)$ and go to $(3, 2)$. If we went the other way around, the sign of our answer would be flipped. $$ f(3, 2) - f(0, 0) = \left(2(3)^5 - (3)^2(2)^3\right) - \left(2(0)^5 - (0)^2(0)^3\right) $$ $$ = \left(2(243) - (9)(8)\right) $$ $$ = \left(486 - 72 \right) $$ $$ = 414 $$ Tada! Our answer is $414$ (the book agrees).Wed, 30 Jul 2014 19:20:51 -0500http://calc3.askbot.com/question/197/section-163/?answer=206#post-id-206