Section 16.3

answered 2014-07-30 19:20:51 -0600

Justin
1009 ●2 ●20 ●38

updated 2014-07-30 19:23:16 -0600

Since 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).