# Calculate flux across a curve. Would like my solution checked please

• Nov 21st 2010, 03:14 PM
downthesun01
Calculate flux across a curve. Would like my solution checked please
$F=(x^{2}+y^{3})i+(2xy)j$

Curve $C:x^{2}+y^{2}=9$

$x=3cos(t)$, $y=3sin(t)$; $0\leq t \leq 2\pi$

So then:

$F=(9cos^{2}(t)+27sin^{3}(t))i+(18cos(t)sin(t))j$

$M=9cos^{2}(t)+27sin^{3}(t)$

$N=18cos(t)sin(t)$

$dx=-3sin(t)$

$dy=3cos(t)$

$Flux=\displaystyle \int^{2\pi}_{0}(Mdy-Ndx)dt$

$Flux=\displaystyle \int^{2\pi}_{0}(27cos^{3}(t)+81sin^{3}(t)cos(t)-54sin^{2}(t)cos(t))dt$

So, then I just did the integral part by part

$\displaystyle 27 \int^{2\pi}_{0}cos^{3}(t)dt$

$\displaystyle \frac{81}{4} \int^{2\pi}_{0}(cos(t)+cos(3t))dt$

$\displaystyle \frac{81}{4}[sin(t)]|^{2\pi}_{0}+\frac{81}{12}[sin(3t)]|^{2\pi}_{0}=>\frac{81}{4}(0-0)+\frac{81}{12}(0-0)=0$

$\displaystyle \int^{2\pi}_{0}81sin^{3}(t)dt$

$u=sin(t)$ $du=cos(t)dt$

$\displaystyle 81 \int^{2\pi}_{0}u^{3}du$

$\frac{81}{4}[sin^{4}(t)]|^{2\pi}_{0}=>\frac{81}{4}(0-0)=0$

$-54 \displaystyle\int^{2\pi}_{0}sin^{2}(t)cos(t))dt$

$u=sin(t)$ $du=cos(t)dt$

$-54 \displaystyle\int^{2\pi}_{0}u^{2}(t)du$

$\frac{-54}{3}[sin^{3}(t)]|^{2\pi}_{0}=>\frac{-54}{3}(0-0)=0$

So, that means that:

$Flux=\displaystyle \int^{2\pi}_{0}(27cos^{3}(t)+81sin^{3}(t)cos(t)-54sin^{2}(t)cos(t))dt=0$

Did, I do the problem wrong or something? I'd hate to think I did all of this work for the answer to just be 0...
• Nov 21st 2010, 04:10 PM
spruancejr
You are correct here. Integrals measure area across plane(s) and a curve doesn't have any area to it. Therefore it makes sense that the answer would be zero.