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 + 3x$$y^2$$j
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 [itex] \frac{-r^4}{4} [/tex]. No negative.

You'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 :uhh: