Question

Find the circulation and flux of the fields: $\vec{F}_1 = 2x\vec{i} - 3y\vec{j}$; $\vec{F}_2 = 2x\vec{i} + (x - y)\vec{j}$ around and across the curve $\vec{r}(t) = (a\cos t)\vec{i} + b\sin t\vec{j}$; $0 \le t \le 2\pi$.

          Find the circulation and flux of the fields:
$\vec{F}_1 = 2x\vec{i} - 3y\vec{j}$;
$\vec{F}_2 = 2x\vec{i} + (x - y)\vec{j}$
around and across the curve
$\vec{r}(t) = (a\cos t)\vec{i} + b\sin t\vec{j}$;
$0 \le t \le 2\pi$.
        
Find the circulation and flux of the fields:
F⃗1 = 2xi⃗ - 3yj⃗;
F⃗2 = 2xi⃗ + (x - y)j⃗
around and across the curve
r⃗(t) = (acos t)i⃗ + bsin tj⃗;
0 ≤ t ≤ 2π.

Added by Christopher M.

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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Find the circulation and flux of the fields Fi = 2xi - 3yj; F2=2xi+(x -y)j around and across the curve r(t)=(acost)i+bsintj; 7>7>0
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Transcript

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00:01 Hello everyone, let's solve the given question.
00:02 So, as according to the question we are having f is equals to x minus y i cap plus 7 y minus x j cap x is equals to 0 x is equals to 7 y is equals to 0 y is equals to 7.
00:21 So, now using green theorem integration p dx plus s dy is equals to double integration r del q over del x minus del p over del y d a.
00:48 So, as to be moving further according to the question we are having a p is equals to x minus y q is equals to 7 y minus x.
00:58 So, the del q over del x is equals to minus 1 and del p over del y is equals to minus 1.
01:07 So, therefore we get integration r del q minus del x minus del p over del y d a is equals to integration minus 1 minus 1 d a.
01:23 So, as to be moving further we get it equals to the integration r 0 d a...
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