Question

Use Green's Theorem to evaluate

integrate C (F · dr). (Check the orientation of the curve before applying the theorem.)

F(x, y) = vector ( sqrt(x)+4y^3, 4x^2 + sqrt(y) )


C consists of the arc of the curve y = sin(x) from (0, 0) to (pi, 0) and the line segment from (pi, 0) to (0, 0)

Answer 1

By Green's theorem, the line integral of F along C is equal to the integral of the curl of F (two-dimensional curl, that is) over the region bounded by C, where C is a generic path that is oriented counterclockwise. However, our C run clockwise, so we multiply the following by -1.

`\displaystyle \int_C \vec F \cdot d\vec r = -\iint_R (\partial(4x^2+\sqrt y))/(\partial x) - (\partial(\sqrt x+4y^3))/(\partial y) \, dx \, dy`

where R is the set

`R = \left\{(x, y) : 0 \le x \le \pi \text{ and } 0 \le y \le \sin(x)\right\}`

Compute the double integral:

`\displaystyle -\iint_R (\partial(4x^2+\sqrt y))/(\partial x) - (\partial(\sqrt x+4y^3))/(\partial y) \, dx \, dy = -\int_0^\pi \int_0^(\sin(x)) (8x - 12y^2) \, dy \, dx`

Integrating with respect to y is trivial:

`\displaystyle \cdots = - \int_0^\pi (8x\sin(x) - 4\sin^3(x)) \, dx`

Integrating by parts with

u = x ⇒ du = dx

dv = sin(x) dx ⇒ v = -cos(x)

gives

`\displaystyle \int x \sin(x) \, dx = -x \cos(x) + \int \cos(x) \, dx = -x\cos(x) + \sin(x)`

while in the other integral, we have by substitution

`\displaystyle \int \sin^3(x) \, dx = \int \sin(x) (1-\cos^2(x)) \, dx \\= - \int (1-\cos^2(x)) d(\cos(x)) \\ = -\cos(x) + \frac13 \cos^3(x)`

Then our last integral evaluates to

`\displaystyle \cdots = - \left(8\pi - 4 + \frac43\right) + \left(4 - \frac43\right) = \boxed{\frac{16}3-8\pi}`