Let C be the country of the square whose vertices are (0,0),(1,0),(0,1), and (1,1), evaluate a)...


Let C be the country of the square whose vertices are {eq}\displaystyle (0,0),(1,0),(0,1), {/eq} and {eq}\displaystyle (1,1), {/eq} evaluate

{eq}\displaystyle a) \oint x^3 dx+y^3 dy {/eq}

{eq}\displaystyle b) \oint y^3 dx+x^3 dy {/eq}

Green's Theorem:

Suppose that {eq}C {/eq} is a counterclockwise simple closed curve in the plane which encloses the region {eq}R {/eq}. Let {eq}P(x,y) {/eq} and {eq}Q(x,y) {/eq} be differentiable functions whose domains include all of {eq}R {/eq}. Then Green's theorem states that:

{eq}\displaystyle \oint_C \left[P(x,y) \, dx + Q(x,y) \, dy\right] = \iint_R \left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right) \, dA \, . {/eq}

Green's theorem can be thought of as a higher-dimensional version of the fundamental theorem of calculus. Just as the fundamental theorem of calculus relates an integral over an interval (a one-dimensional object) to the evaluation of a function at two points (a zero-dimensional object), Green's theorem relates an integral over a plane region (a two-dimensional object) to an integral over a curve (a one-dimensional object).

Answer and Explanation:

a) Let {eq}S {/eq} be the square bounded by the curve {eq}C {/eq}. Then applying Green's theorem gives

{eq}\begin{align*} \oint_C (x^3 \, dx + y^3 \, dy)&=\iint_S \left[\frac{\partial}{\partial x}(y^3)-\frac{\partial}{\partial y}(x^3)\right] \, dA&&\text{(by Green's theorem)}\\ &=\iint_S (0-0) \, dA&&\text{(evaluating partial derivatives)}\\ &=\iint_S 0 \, dA\\ &=0 \, . \end{align*} {/eq}

That is, {eq}\boxed{\oint_C (x^3 \, dx + y^3 \, dy)=0\, .} {/eq}

b) We can again use Green's theorem to evaluate this integral:

{eq}\begin{align*} \oint_C (y^3 \, dx + x^3 \, dy)&=\iint_S \left[\frac{\partial}{\partial x}(x^3)-\frac{\partial}{\partial y}(y^3)\right]\, dA&&\text{(by Green's theorem)}\\ &=\iint_S (3x^2-3y^2) \, dA&&\text{(evaluating partial derivatives).} \end{align*} {/eq}

Now, the square {eq}S {/eq} has the Cartesian coordinate bounds {eq}0 \le x \le 1 {/eq} and {eq}0 \le y \le 1 {/eq}. So we can integrate over {eq}S {/eq} as follows:

{eq}\begin{align*} \iint_S (3x^2-3y^2) \, dA&=\int_{x=0}^1\int_{y=0}^1 (3x^2-3y^2) \, dy \, dx&&\text{(writing the area integral as a Cartesian iterated integral)}\\ &=\int_{x=0}^1\Big(3x^2y-y^3\Big|_{y=0}^1 \, dx&&\text{(evaluating the }y\text{-integral)}\\ &=\int_{x=0}^1(3x^2-1^3-0+0) \, dx\\ &=\int_0^1 (3x^2-1) \, dx\\ &=\Big(x^3-x\Big|_0^1&&\text{(evaluating the }x\text{-integral)}\\ &=1-1-0+0\\ &=0 \, . \end{align*} {/eq}

In summary, we've computed that {eq}\boxed{\oint_C (y^3 \, dx + x^3 \, dy)=0 \, .} {/eq}

Learn more about this topic:

Double Integrals & Evaluation by Iterated Integrals

from GRE Math: Study Guide & Test Prep

Chapter 15 / Lesson 4

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