Over a curve $C$ the line integral $$\int \limits_C [-(x^2)y – 3x + 2y]dx + [4(y^2)x-2x]dy.$$
Applying Green's Theorem -> Over a region $R$
$$\iint \limits_R [x^2 + 4y^2 – 4]dx dy$$
Answer: On the boundary of this double integral over region $R$ we would like the value to be zero. On the inside of the region we would like the value to be negative.
Why isn't there a case where the curve encompasses a region with both positive and negative values, and the negative values have a much greater value? I don't understand the explanation of the answer. Can you explain?
Best Answer
The values of the two integrals are equal. The integrand of the double integral is negative only when $x^2 + 4y^2 \leq 4$, which is the interior of an ellipse centered at the origin (with axes parallel to the coordinate axes). So all of the negative values of the integrand are in this compact region. If we put the path along that ellipse (with positive orientation), we include all the negative contributions and exclude all the positive contributions. This is necessarily the lowest value attainable by the double integral and hence by the path integral.