Question

F(x, y) = e−x + y2, e−y + x2 , c consists of the arc of the curve y = cos(x) from − ???? 2 , 0 to ???? 2 , 0 and the line segment from ???? 2 , 0 to − ???? 2 , 0.

Answer 1

Given that y = cos(x) makes up part of the boundary of C, I suspect you mean the given points to be (-π/2, 0) and (π/2, 0).

I also assume the given vector field is

`\vec F(x,y) = \left\langle e^(-x) + y^2, e^(-y) + x^2 \right\rangle`

Since
`\vec F` has no singularities on C or in its interior, Green's theorem applies:

`\displaystyle \int_C \vec F(x,y) \cdot d\vec r = \iiint_D (\partial(e^(-y)+x^2))/(\partial x) - (\partial(e^(-x)+y^2))/(\partial y) \, dA = 2 \iiint_D (x + y) \, dA`

where D is the interior of C, the region

`D = \left\{ (x, y) : -\frac\pi2 \le x \le \frac\pi2 \text{ and } 0 \le y \le \cos(x) \right\}`

The integral then reduces to

`\displaystyle 2 \iint_D (x + y) \, dA = 2 \int_(-\frac\pi2)^(\frac\pi2) \int_0^(\cos(x)) (x + y) \, dy \, dx`

`\displaystyle 2 \iint_D (x + y) \, dA = 2 \int_(-\frac\pi2)^(\frac\pi2) \left( x\cos(x) + \frac12 \cos^2(x) \right) \, dx`

`\displaystyle 2 \iint_D (x + y) \, dA = 2 \int_(-\frac\pi2)^(\frac\pi2) \left( x\cos(x) + \frac{1 + \cos(2x)}4 \right) \, dx`

`\displaystyle 2 \iint_D (x + y) \, dA = \frac12 \int_(-\frac\pi2)^(\frac\pi2) \left( 4x\cos(x) + 1 + \cos(2x) \right) \, dx`

Since 4x cos(x) is an odd function over the symmetric interval [-π/2, π/2], its contribution to the integral is 0, and the remaining integral is trivial.

`\displaystyle 2 \iint_D (x + y) \, dA = \frac12 \int_(-\frac\pi2)^(\frac\pi2) \left( 1 + \cos(2x) \right) \, dx = \boxed{\frac\pi2}`