Discussion
Let \(C\) be the positively oriented boundary of the domain bounded by the curves \(y = 2x^{2}\) and \(y^{2} = 4x\). Then the value of the line integral
\[ \oint_{C} (2y^{2} + 2xy + 4y)dx + (x^{2} + 4xy + 8x)dy \]
is equal to
Answer: A
The line integral equals \(\frac{8}{3}\).
By Green's theorem with \(M = 2y^{2}+2xy+4y\) and \(N = x^{2}+4xy+8x\):
\[ \begin{aligned} \frac{\partial N}{\partial x}-\frac{\partial M}{\partial y} &= (2x+4y+8)-(4y+2x+4) \\ &= 4 \end{aligned} \]
So the integral is 4 times the enclosed area. The curves \(y = 2x^{2}\) and \(y = 2\sqrt{x}\) meet at x = 0 and x = 1, with the square-root curve on top:
\[ \begin{aligned} A &= \int_{0}^{1}(2\sqrt{x}-2x^{2})\,dx = \frac{4}{3}-\frac{2}{3} = \frac{2}{3} \\ \oint &= 4 \times \frac{2}{3} = \frac{8}{3} \end{aligned} \]
2/3 is the area alone, without the factor 4. 4/3 and 1/3 come from using only one of the curves or halving the integrand.
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