- #1
hbomb
- 58
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I'm having trouble on a line integral.
Assuming that the closed curve C is taken in the counterclockwise sense. Use Green's Theorem.
[tex] \int_C F\bullet dR [/tex]
where F=([tex]x^2 + y^2[/tex])i + 3x[tex]y^2[/tex]j
and C is the circle
[tex]x^2 + y^2 = 9[/tex]
This is what I have done so far...
[tex]\int_0^{2\Pi} \int_0^3 \-r^2 rdrd\theta[/tex]
[tex]\int_0^{2\Pi} \int_0^3 \-r^3 drd\theta[/tex]
[tex]\int_0^{2\Pi} \frac{-r^4}{4} \\]_0^3d\theta[/tex]
[tex]\int_0^{2\Pi} \frac{-81}{4} d\theta[/tex]
[tex]\frac{-81}{4} \theta\\]_0^{2\Pi}[/tex]
[tex]\frac{-81\Pi}{2}[/tex]
The book gives the answer as [tex]\frac{243\Pi}{4}[/tex]
I have no idea where I went wrong.
Assuming that the closed curve C is taken in the counterclockwise sense. Use Green's Theorem.
[tex] \int_C F\bullet dR [/tex]
where F=([tex]x^2 + y^2[/tex])i + 3x[tex]y^2[/tex]j
and C is the circle
[tex]x^2 + y^2 = 9[/tex]
This is what I have done so far...
[tex]\int_0^{2\Pi} \int_0^3 \-r^2 rdrd\theta[/tex]
[tex]\int_0^{2\Pi} \int_0^3 \-r^3 drd\theta[/tex]
[tex]\int_0^{2\Pi} \frac{-r^4}{4} \\]_0^3d\theta[/tex]
[tex]\int_0^{2\Pi} \frac{-81}{4} d\theta[/tex]
[tex]\frac{-81}{4} \theta\\]_0^{2\Pi}[/tex]
[tex]\frac{-81\Pi}{2}[/tex]
The book gives the answer as [tex]\frac{243\Pi}{4}[/tex]
I have no idea where I went wrong.