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October 1, 2014

October 1, 2014

Posted by **Elyse** on Thursday, June 6, 2013 at 3:10pm.

y^(2)dx+xy dy

where C is the boundary of the region lying between the graphs of y=0,

y=sqrt(x), and x=9

- Calculus -
**Steve**, Thursday, June 6, 2013 at 4:22pmUsing the definition, we have

P = y^2

Q = xy

and the integral becomes

∫[0,9]∫[0,√x] (∂Q/∂x - ∂P/∂y) dy dx

= ∫[0,9]∫[0,√x] (y - 2y) dy dx

= ∫[0,9]∫[0,√x] -y dy dx

= ∫[0,9] -x/2 dx

= -x^2/4 [0,9]

= -81/4

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