Solved: Evaluating F dot dr Integral for P = pi

[joemama69]

Homework Statement

P = pi

Evaluate \int F \cdotdr where c is the curve given by r(t) = (t+sin\pit)i + (2tcos\pit)j

F = (4x³y² - 2xy³) i + (2x⁴y - 3x²y² + 4y³)j

Homework Equations

The Attempt at a Solution

When I dot them I get an extremely long expression.

\int 4x³y²t - 4xy³t - 2xy³sinPt + 4x⁴yt + 2x⁴cosPt - 6x²y²t - 3x²y²cosPt + 8y³t + 4y³cosPt dt evaluated from t = 0 to to = 1


2x³y² - 2xy³ +2Pxy³cosPt + 2x⁴y + 2Px⁴ysinPt - 3x²y² - 3Px²y²sinPt + 4y³ + 4Py³sinPt

 

[quasar987]

I haven'T checked it you have doted correctly but once that is done, you have to replace all the x in there by t+sin(pi*t) and all the y by 2tcos(pi*t). Then simplify if possible and integrate...

 

[Shooting Star]

F = (4x³y² - 2xy³) i + (2x⁴y - 3x²y² + 4y³)j

Don't even think of doing it directly!

What is the curl of F? Once you spot that, use Green's theorem or some other property to get the result in one line.

 

[joemama69]

Ah hah, Is this right

\intF dot dr = \intcurl F dot dA = 0

Because

F = (4x³y² - 2xy³) i + (2x⁴y - 3x²y² + 4y³)j


curl F = (8x³y - 6xy² - 8x³y + 6xy²)k = 0