Question

Use Green's Theorem to evaluate the following line integrals. Unless stated otherwise, assume all curves are oriented counterclockwise. \(\oint\left(2 x+e^{y^{2}}\right) d y-\left(4 y^{2}+e^{x^{2}}\right) d x,\) where \(C\) is the boundary of the square with vertices \((0,0),(1,0),(1,1),\) and (0,1)

Short answer

Based on the given information and the application of Green's Theorem, the line integral is equal to 6.

Step-by-step solution

Find the partial derivatives

We need to find the partial derivatives \(\frac{\partial Q}{\partial x}\) and \(\frac{\partial P}{\partial y}\): \(\frac{\partial P}{\partial y} = \frac{\partial}{\partial y}\left(-\left(4 y^{2}+e^{x^{2}}\right)\right) = -8y\) \(\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}\left(\left(2 x+e^{y^{2}}\right)\right) = 2\)

Apply Green's Theorem

Using the results from step 1, we can apply Green's Theorem: \(\oint_{C}(P d x+Q d y)=\iint_{D}\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right) d A= \iint_{D}(2-(-8y)) d A = \iint_{D}(2+8y) d A\)

Set up the double integral

Now we set up the double integral over the region \(D\), which is the square with vertices at \((0,0),(1,0),(1,1),\) and \((0,1)\). We can express the integral in terms of \(x\) and \(y\) as follows: \(\iint_{D}(2+8y) d A = \int_{0}^{1}\int_{0}^{1}(2+8y) dy\, dx\)

Evaluate the double integral

Now we evaluate the double integral: \(\int_{0}^{1}\int_{0}^{1}(2+8y) dy\, dx = \int_{0}^{1}\left[2y+4y^2\right]_{0}^{1} dx\) \(=\int_{0}^{1}(2+4) dx = \int_{0}^{1}6 dx\) \([6x]_{0}^{1} = 6(1) - 6(0) = 6\). Thus, the given line integral is equal to 6.

Key concepts

Line Integrals

Line integrals are a type of integral where you sum a function along a curve or path. They are useful in various fields like physics and engineering to measure things such as work done by a force field. To better understand, picture a path on a map, and you want to compute the total elevation gain along that path. In a mathematical sense, line integrals help with this type of calculation.

There are two primary uses of line integrals. One is to integrate and find the total effect across a curve, such as finding the circulation around a path due to a vector field. The second use is finding the scalar field effect, such as finding the work done by a force on particles moving along a path.

With Green's Theorem, line integrals can be converted into double integrals which can simplify calculations, especially when the curve encloses a simple region like a square or a circle. In our exercise, Green's Theorem helps change an often complex line integral into a more straightforward problem of calculating a double integral over a straightforward region.

Double Integrals

Double integrals allow you to calculate the volume under a surface in a two-dimensional region. Picture a sheet with varying thickness laid across a flat plane. The thickness at each point on the sheet is given by a function of two variables, say, \(x\) and \(y\).

Double integrals give you a way to integrate over an entire area, summing up little infinitesimal volumes throughout the region. This is how you can find total volume, mass, or charge distribution over some area. The process involves integrating the function twice: once with respect to \(y\), and then with respect to \(x\) (or vice versa), over the limits of the region you're interested in.

In the current exercise, the double integral was used over an easy-to-define region - a square with vertices \(0,0\), \(1,0\), \(1,1\), and \(0,1\). By setting up the integral boundaries correctly, Green's Theorem makes it possible to turn the problem of evaluating a line integral into a more straightforward task of solving a double integral.

Partial Derivatives

Partial derivatives help you understand how a multi-variable function changes when you tweak one variable while keeping the others constant. They are like a spotlight, revealing how much a single variable contributes to the overall change of a function.

If you have a function involving two variables, say \(f(x, y)\), the partial derivatives \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\) show the rates of change relative to \(x\) and \(y\) respectively. The partial derivative with respect to \(x\) treats \(y\) as a constant and vice versa.

In Green's Theorem, these partial derivatives play a crucial role. They are used to ensure that a line integral, taken counterclockwise around a closed curve, can be transformed into a double integral over the region enclosed by the curve. For the exercise, finding the partial derivatives \(\frac{\partial Q}{\partial x}\) and \(\frac{\partial P}{\partial y}\) was essential to correctly apply Green's Theorem and move from a line integral into a double integral format.