Question

Evaluate each integral in the simplest way possible. $$\oint \mathbf{F}\cdot d\mathbf{r}\text{ around the circle }{x}^2+y^2+2x=0,\text{ where }\mathbf{F}=y\mathbf{i}-x\mathbf{j}$$

Short answer

$-2\pi$

Step-by-step solution

Rewrite the circle equation

Start by rewriting the given equation of the circle. The equation is given as:$$x^2+y^2+2x=0$$Complete the square for the terms involving x.$$x^2+2x+y^2=0$$$$(x+1{)}^2-1+y^2=0$$$$(x+1{)}^2+y^2=1$$So, the circle can be rewritten as centered at (-1,0) with radius 1.

Set up the integral using Green's Theorem

Green's Theorem states that for a positively oriented, simple closed curve C enclosing a region D:$${\oint }_C\mathbf{F}\cdot d\mathbf{r}={\iint }_D(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})dA$$Here, $\mathbf{F}=y\mathbf{i}-x\mathbf{j}$ which means $P=y$ and $Q=-x$.

Compute the partial derivatives

Compute the necessary partial derivatives:$$\frac{\partial Q}{\partial x}=\frac{\partial (-x)}{\partial x}=-1$$$$\frac{\partial P}{\partial y}=\frac{\partial (y)}{\partial y}=1$$

Set up the double integral

Substitute the partial derivatives into Green's Theorem:$${\oint }_C\mathbf{F}\cdot d\mathbf{r}={\iint }_D(-1-1)dA$$$${\oint }_C\mathbf{F}\cdot d\mathbf{r}={\iint }_D-2dA$$

Evaluate the double integral

The region D is the interior of the circle centered at (-1,0) with radius 1. The area of this region is given by $\pi r^2$, where $r=1$:$${\oint }_C\mathbf{F}\cdot d\mathbf{r}={\iint }_D-2dA$$$$=-2\times \text{Area of circle}$$$$=-2\times \pi \times (1^2)$$$$=-2\pi$$

Key concepts

vector calculus

Vector calculus is a branch of mathematics that deals with vector fields and differentiable functions. This branch is crucial for understanding various physical phenomena like fluid flow, electromagnetism, and more. In vector calculus, we use vectors to describe quantities that have both magnitude and direction. A vector field associates a vector with every point in a space. For instance, $\mathbf{\text{F}}=y\mathbf{\text{i}}-x\mathbf{\text{j}}$ is a vector field that assigns a vector to each point $(x,y)$. In our problem, vector calculus helps us analyze the behaviors of fields and forces acting over regions in a coordinate space.

line integrals

Line integrals combine calculus and vector fields to evaluate a function along a curve. They are used to measure the total effect of a vector field along a specific path or curve. The standard notation for a line integral along a curve $\mathbf{\text{C}}$ is: $\text{[}onumber{\oint }_{\mathbf{\text{C}}}\mathbf{\text{F}}\bullet d\mathbf{\text{r}}\text{]}$. This integral sums up the dot product of a vector field and a differential element vector along the path. It essentially computes the work done by a force field along a path. In our exercise, we use Green’s Theorem to simplify the line integral around a circle. Instead of computing it directly, we relate it to a double integral over the area enclosed by the path.

double integrals

Double integrals extend single-variable integrals to functions of two variables over a region in the plane. They are fundamental in calculating areas, volumes, and other quantities involving two-dimensional regions. The notation for a double integral over a region $D$ is: $\text{[}onumber{\iint }_{D}f(x,y)dA\text{]}$. In our exercise, we convert a line integral around a circle to a double integral over the disk enclosed by the circle using Green's Theorem. Specifically, we compute $\text{[}onumber{\iint }_D(-2)dA\text{]}$, where the negative constant $-2$ indicates uniform distribution over the entire area.

area of a circle

The area of a circle is a key geometric concept often required in physics, engineering, and mathematics problems. The formula for the area of a circle with radius $r$ is: $\text{[}onumber\text{Area}=\pi r^2\text{]}$. In our problem, when applying Green’s Theorem, we need to calculate the double integral over a region that's a circle with radius 1, centered at $-1,0$. The area of this circle is $\text{[}onumber\pi \times (1{)}^2=\pi \text{]}$. Multiplying this with the constant from our integral gives the final result $\text{-}2\text{textbackslash}\text{textbackslashpi}$. This highlight shows how geometric properties play into solving calculus problems.