# Solve line integral using Green's thereom

Given $\oint_c x^2+y^2dx + 2xydy$ and C is the boundry R bounded by the graphs of $y=\sqrt{x}, y = 0, x = 4.$ Here is what I have so far : $$\oint_c x^2+y^2dx + 2xydy = \int\int_R (\frac{dn}{dx}-\frac{dm}{dy})$$ $$M(x,y) = x^2 + y^2, \space N(x,y)=2xy$$ $$\oint_c x^2+y^2dx + 2xydy = \int\int_R\ (\frac{d}{dx}(2xy) - \frac{d}{dy}x^2 + y^2 dydx )$$ $$= \int_0^4\int_0^{\sqrt{x}} (2y -2y)dydx$$

Would this just equal 0, or after applying the integral would it become y and then be evaluated at the points, as I know y would become 1 when using the derivative not 0. Is it that I set this up incorrectly ?

• The integrand is exact and you’re integrating over a closed curve, so... – amd Dec 10 '17 at 1:01
• soooo ? what does that mean the total length of this line can't be zero right ? – Doug Ray Dec 11 '17 at 21:19
• You’re not computing arc length. The line integral of an exact form over a closed curve is zero, so you’re getting the right answer. – amd Dec 11 '17 at 21:59