Solving a Line Integral Using Green's Theorem

[hbomb]

I'm having trouble on a line integral.

Assuming that the closed curve C is taken in the counterclockwise sense. Use Green's Theorem.

\int_C F\bullet dR
where F=(x^2 + y^2)i + 3xy^2j
and C is the circle
x^2 + y^2 = 9

This is what I have done so far...

\int_0^{2\Pi} \int_0^3 \-r^2 rdrd\theta

\int_0^{2\Pi} \int_0^3 \-r^3 drd\theta

\int_0^{2\Pi} \frac{-r^4}{4} \\]_0^3d\theta

\int_0^{2\Pi} \frac{-81}{4} d\theta

\frac{-81}{4} \theta\\]_0^{2\Pi}

\frac{-81\Pi}{2}

The book gives the answer as \frac{243\Pi}{4}

I have no idea where I went wrong.

 

[whozum]

Theres another form of Green's theorem that might be more appropriate for this problem, try looking for it in your textbook.

Also the integral of r^3 is not \frac{-r^4}{4} [/tex]. No negative.<br /> <br /> You&#039;ve also managed to neglect the vector field your integrating over.

 

[Data]

Can you show the work as to how you got

\int_0^{2\Pi} \int_0^3 \-r^2 rdrd\theta

?

 

[quasar987]

You've done something strange. Applying Green theorem should yield the equality

\int_C F\bullet dR = \iint_D \left(\frac{\partial (x^2+y^2)}{\partial y} - \frac{\partial (3xy^2)}{\partial x} \right) dxdy

and that's not

\int_0^{2\Pi} \int_0^3 \-r^2 rdrd\theta

 

[hbomb]

Sorry guys, I was looking at another problem and I took the derivative of the wrong force. I redid this problem and I got the correct answer. Thanks for the attempted help though.

 

[quasar987]

Np.. happens to me all the time