Discussion
Let C be the circle \((x-1)^{2}+y^{2}=25\) oriented counterclockwise. Then, the value of the line integral
\[ \oint_{C} \left[(x^{5}-3y)\,dx + (-2x+e^{y^{2}})\,dy\right] \]
is
Answer: D
The line integral equals \(25\pi\).
By Green's theorem, with \(M = x^{5}-3y\) and \(N = -2x+e^{y^{2}}\):
\[ \begin{aligned} \oint_{C}(M\,dx+N\,dy) &= \iint_{D}\left(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\right)dA \\ &= \iint_{D}(-2+3)\,dA = \text{area of } D \end{aligned} \]
The circle has radius 5 (its centre at (1, 0) does not matter), so the area is \(\pi \times 5^{2} = 25\pi\).
The awkward terms \(x^{5}\) and \(e^{y^{2}}\) drop out because each depends only on the variable it is integrated against. Taking the integrand as -2 - 3 or 3 alone gives multiples of \(25\pi\) not offered; 10, 15 and 20 times \(\pi\) do not match the area.
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