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March 4, 2015

March 4, 2015

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 -
**kuai**, Thursday, June 6, 2013 at 6:47pmP = y^2

Q = xy

[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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