Use Green's Theorem to evaluate the line integral I = \int_C 3xe^{-2x} \, dx + 2(x^4 +...


Use Green's Theorem to evaluate the line integral

{eq}I = \int_C 3xe^{-2x} \, dx + 2(x^4 + 2x^2y^2)\, dy {/eq}

along the positively oriented boundary of the region between the circles

{eq}x^2 + y^2 = 1 , \quad x^2 + y^2 = 16 {/eq}

Green's Theorem:

Suppose that {eq}P(x,y) {/eq} and {eq}Q(x,y) {/eq} are real-valued functions whose domains are contained in the plane. Let {eq}R {/eq} be a region in the plane which is within the domains of both {eq}P {/eq} and {eq}Q {/eq}, which is bounded by the simple closed curves {eq}C_1,C_2,\dots,C_n {/eq}. If the curves {eq}C_1,C_2,\dots,C_n {/eq} are all oriented so that the interior of {eq}R {/eq} lies to their left when they are traversed in the positive direction, and {eq}C=C_1\cup C_2\cup \dots\cup C_n {/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) \, . {/eq}

Answer and Explanation:

The region {eq}R {/eq} which is bounded by the two circles {eq}x^2+y^2=1 {/eq} and {eq}x^2+y^2=16 {/eq} has the polar coordinate bounds {eq}1 \le r \le 4 {/eq} and {eq}0 \le \theta \le 2\pi {/eq}. So we can use Green's theorem to evaluate the given integral, as follows:

$$\begin{align*} \oint_C \left[3xe^{-2x} \, dx + 2(x^4+2x^2y^2) \, dy\right]&=\iint_R \left[\frac{\partial}{\partial x}(2(x^4+2x^2y^2))-\frac{\partial}{\partial y}(3xe^{-2x})\right] \, dA&&\text{(by Green's theorem)}\\ &=\iint_R \left[(8x^3+8xy^2) - 0\right] \, dA&&\text{(evaluating partial derivatives)}\\ &=\iint_R (8x^3+8xy^2) \, dA\\ &=\iint_R 8x(x^2+y^2) \, dA&&\text{(factoring)}\\ &=\iint_R (8r\cos \theta)(r^2) \, dA&&\text{(rewriting the integrand in polar coordinates)}\\ &=\iint_R 8r^3 \cos \theta \, dA\\ &=\int_{r=1}^4 \int_{\theta=0}^{2\pi} (8r^3 \cos \theta)(r \, d\theta \, dr)&&\text{(writing the area integral as an iterated integral in polar coordinates)}\\ &=\int_{r=1}^4 \int_{\theta=0}^{2\pi} 8r^4 \cos \theta \, d\theta \, dr\\ &=\int_{r=1}^4 \Big(8r^4\sin \theta\Big|_{\theta=0}^{2\pi} \, dr&&\text{(evaluating the }\theta\text{-integral)}\\ &=\int_{r=1}^4 (8r^4\sin(2\pi)-8r^4\sin 0) \, dr\\ &=\int_{r=1}^4 (0-0) \, dr\\ &=0 \, . \end{align*} $$

That is, it follows from Green's theorem that {eq}\boxed{\oint_C \left[3xe^{-2x} \, dx+ 2(x^4+2x^2y^2) \, dy\right]=0\,.} {/eq}

Learn more about this topic:

Double Integrals & Evaluation by Iterated Integrals

from GRE Math: Study Guide & Test Prep

Chapter 15 / Lesson 4

Related to this Question

Explore our homework questions and answers library