Question

Find the counterclockwise circulation and outward flux of the field Fequals7 xy Bold i plus 2 y squared Bold j around and over the boundary of the region C enclosed by the curves yequalsx squared and yequalsx in the first quadrant.

Answer 1

Answer:

The counterclockwise circulation is $\frac{7}{12}$ and the outward flux is $\frac{11}{15}$

Step-by-step explanation:

We are given the field $F(x,y) = (7xy,2y^2)$. A picture of the region and the path we are considering is attached. Recalll the following theorems.

Given a field of the form F(x,y)=(f(x,y),g(x,y) with f,g having continous partial derivates, C is a closed path counterclockwise oriented, R is the region enclosed by C and n is the normal vector pointing outwards of the path C. Then

$\oint_C F\cdot dr =\iint_R \frac{\partial f}{dy}- \frac{\partial g}{dx} dA$(this one calculates the counterclockwise circulation)

$\oint_C F\cdot n ds =\iint_R (\frac{\partial f}{dx}+ \frac{\partial g}{dy} dA$ (This one calculates the outward flux)

Then, recall that in our case f(x,y) = 7xy, g(x,y)=2y^2[/tex]. Then

$\frac{\partial f}{dx} = 7y,\frac{\partial f}{dy} = 7x$

$\frac{\partial g}{dx}=0, \frac{\partial g}{dy} = 4y$.

Note that we just need to describe our region R. The region R lies between the parabola y=x^2 and the line y=x. Thus, one way to describe the region is as follows $0\leq x \leq 1, x^2\leq y \leq x$. Then, using the previous results, we get that

$\oint_C F\cdot dr =\int_{0}^{1}\int_{x^2}^{x}7x-0dydx = 7\int_{0}^1x(x-x^2)dx = 7 \left.(\frac{x^3}{3}-\frac{x^4}{4})\right|_{0}^1 = 7(\frac{1}{3}-\frac{1}{4}) = \frac{7}{12}$ (circulation)

$\oint_C F\cdot n ds=\int_{0}^{1}\int_{x^2}^{x}7y+4ydydx = \frac{11}{2}\int_{0}^1\left.y^2\right_{x^2}^{x}dx = \frac{11}{2}\int_{0}^{1}x^2-x^4 dx = \frac{11}{2}\left(\frac{x^3}{3}-\frac{x^5}{5})\right|_{0}^{1}=\frac{11}{2}(\frac{1}{3}-\frac{1}{5})=\frac{11}{15}$(flux)