Copyright

Closed integral x^2y^2 dx + xy dy, C consists of arc of the parabola y = x^2 from (0, 0) to (1,...

Question:

Consider the integral {eq}\displaystyle \oint_C x^2y^2\ dx + xy\ dy {/eq}, where {eq}C {/eq} consists of the arc of the parabola {eq}y = x^2 {/eq} from {eq}(0,\ 0) {/eq} to {eq}(1,\ 1) {/eq} followed by the line segment from {eq}(1,\ 1) {/eq} to {eq}(0,\ 1) {/eq} and the line segment from {eq}(0,\ 1) {/eq} to {eq}(0,\ 0) {/eq}.

Evaluate the line integral by two methods:

(a) Directly.

(b) By applying Green's Theorem.

Green's Theorem:

Let {eq}C {/eq} be an oriented simple closed curve which is the oriented boundary of the region {eq}R {/eq}. Suppose that {eq}\vec{F}=\left<P,Q\right> {/eq} is a vector field whose domain contains the entire region {eq}R {/eq}. Then Green's theorem is the following integral equality:

{eq}\displaystyle \oint_C \vec{F} \cdot ds = \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. Where the fundamental theorem of calculus applies to intervals on the real line, Green's theorem applies to regions in the plane.

Answer and Explanation:

(a) Let {eq}C_1 {/eq} be the parabolic arc from {eq}(0,0) {/eq} to {eq}(1,1) {/eq}; let {eq}C_2 {/eq} be the line segment from {eq}(1,1) {/eq} to {eq}(0,1) {/eq}; and let {eq}C_3 {/eq} be the line segment from {eq}(0,1) {/eq} to {eq}(0,0) {/eq}. Then

{eq}\displaystyle \oint_C (x^2y^2 \, dx + xy \, dy)=\int_{C_1} (x^2y^2 \, dx + xy \, dy)+\int_{C_2} (x^2y^2 \, dx + xy \, dy)+\int_{C_3} (x^2y^2 \, dx + xy \, dy) \, . {/eq}

We'll evaluate each of these line integrals separately.

First, parameterize the parabolic arc {eq}C_1 {/eq} by the function {eq}r(t)=(t,t^2) {/eq} for {eq}0 \le t \le 1 {/eq}. Then {eq}r'(t)=\left<1,2t\right> {/eq}, and so

{eq}\begin{align*} \int_{C_1} (x^2y^2 \, dx + xy \, dy) &=\int_0^1 \left[t^2(t^2)^2(1)+t(t^2)(2t)\right] \, dt&\text{(by the definition of the line integral)}\\ &=\int_0^1 (t^6+2t^4) \, dt\\ &=\left(\frac{1}{7}t^7+\frac{2}{5}t^5\right|_0^1&\text{(evaluating the integral)}\\ &=\frac{1}{7}+\frac{2}{5}\\ &=\frac{19}{35} \, . \end{align*} {/eq}

Next, parameterize the line segment {eq}C_2 {/eq} by the function {eq}r(t)=(-t,1) {/eq} for {eq}-1 \le t \le 0 {/eq}. Then {eq}r'(t)=\left<-1,0\right> {/eq}, and so:

{eq}\begin{align*} \int_{C_1} (x^2y^2 \, dx + xy \, dy) &=\int_{-1}^0 \left[(-t)^2(1)^2(-1)+(-t)(1)(0)\right] \, dt&\text{(by the definition of the line integral)}\\ &=\int_{-1}^0 -t^2 \, dt\\ &=\left(-\frac{1}{3}t^3\right|_{-1}^0&\text{(evaluating the integral)}\\ &=-\frac{1}{3} \, . \end{align*} {/eq}

Finally, on the line segment {eq}C_3 {/eq}, the coordinate {eq}x {/eq} is always zero and so the line element {eq}x^2y^2 \, dx + xy \, dy=0y^2 \, dx + 0y \, dy {/eq} is always zero. It follows that {eq}\int_{C_3} (x^2y^2\, dx + xy \,dy)=0 {/eq}.

Putting all this together, we have:

{eq}\begin{align*} \oint_C (x^2y^2 \, dx + xy \, dy)&=\int_{C_1} (x^2y^2 \, dx + xy \, dy)+\int_{C_2} (x^2y^2 \, dx + xy \, dy)+\int_{C_3} (x^2y^2 \, dx + xy \, dy) \\ &=\frac{19}{35}-\frac{1}{3}+0\\ &=\frac{22}{105} \, . \end{align*} {/eq}

(b) The curve {eq}C {/eq} is drawn below:

the curve C described in the problem statement

As you can see, the region {eq}R {/eq} bounded by {eq}C {/eq} lies above the parabola {eq}y=x^2 {/eq} and below the line {eq}y=1 {/eq} for {eq}0 \le x \le 1 {/eq}. So the Cartesian coordinate bounds on {eq}R {/eq} are {eq}0 \le x \le 1 {/eq} and {eq}x^2 \le y \le 1 {/eq}.

Now, we can evaluate the integral via Green's theorem as follows:

{eq}\begin{align*} \oint_C (x^2y^2 \, dx + xy \, dy)&=\iint_R \left(\frac{\partial}{\partial x}(xy)-\frac{\partial}{\partial y}(x^2y^2)\right) \, dA&\text{(by Green's theorem)}\\ &=\iint_R (y-2x^2y) \, dA&\text{(evaluating the partial derivatives)}\\ &=\int_{x=0}^1 \int_{y=x^2}^1 (y-2x^2 y) \, dy \, dx&\text{(rewriting the area integral as an iterated integral in Cartesian coordinates)}\\ &=\int_{x=0}^1 \left(\frac{y^2}{2}-x^2y^2\right|_{y=x^2}^1 \, dx&\text{(evaluating the }y\text{-integral)}\\ &=\int_{x=0}^1 \left(\frac{1}{2}-x^2-\frac{x^4}{2}+x^6\right) \, dx\\ &=\left(\frac{1}{2}x-\frac{1}{3}x^3-\frac{1}{10}x^5+\frac{1}{7}x^7\right|_0^1&\text{(evaluating the }x\text{-integral)}\\ &=\frac{1}{2}-\frac{1}{3}-\frac{1}{10}+\frac{1}{7}\\ &=\boxed{\frac{22}{105}} \, . \end{align*} {/eq}

Both methods of evaluating the line integral give the result {eq}\dfrac{22}{105} {/eq}, verifying Green's theorem for this integral.


Learn more about this topic:

Loading...
Line Integrals: How to Integrate Functions Over Paths

from GRE Math: Study Guide & Test Prep

Chapter 15 / Lesson 2
933

Related to this Question

Explore our homework questions and answers library