Question

Consider the following region R and the vector field F. a. Compute the​ two-dimensional curl of the vector field. b. Evaluate both integrals in​ Green's Theorem and check for consistency. Bold Upper F equals left angle negative x comma negative y right angle​; Upper R equals StartSet (x comma y ): x squared plus y squared less than or equals 5 EndSet a. The​ two-dimensional curl is 0. ​(Type an exact​ answer.) b. Set up the integral over the region R. Write the integral using polar​ coordinates, with r as the radius and theta as the angle. Integral from 0 to nothing Integral from 0 to nothing (nothing )r font size decreased by 3 dr font size decreased by 3 d theta ​(Type exact​ answers.)

Answer 1

Looks like we're given

`\vec F(x,y)=\langle-x,-y\rangle`

which in three dimensions could be expressed as

`\vec F(x,y)=\langle-x,-y,0\rangle`

and this has curl

`\mathrm{curl}\vec F=\langle0_y-(-y)_z,-(0_x-(-x)_z),(-y)_x-(-x)_y\rangle=\langle0,0,0\rangle`

which confirms the two-dimensional curl is 0.

It also looks like the region
`R` is the disk
`x^2+y^2\le5`. Green's theorem says the integral of
`\vec F` along the boundary of
`R` is equal to the integral of the two-dimensional curl of
`\vec F` over the interior of
`R`:

`\displaystyle\int_(\partial R)\vec F\cdot\mathrm d\vec r=\iint_R\mathrm{curl}\vec F\,\mathrm dA`

which we know to be 0, since the curl itself is 0. To verify this, we can parameterize the boundary of
`R` by

`\vec r(t)=\langle\sqrt5\cos t,\sqrt5\sin t\rangle\implies\vec r'(t)=\langle-\sqrt5\sin t,\sqrt5\cos t\rangle`

`\implies\mathrm d\vec r=\vec r'(t)\,\mathrm dt=\sqrt5\langle-\sin t,\cos t\rangle\,\mathrm dt`

with
`0\le t\le2\pi`. Then

`\displaystyle\int_(\partial R)\vec F\cdot\mathrm d\vec r=\int_0^(2\pi)\langle-\sqrt5\cos t,-\sqrt5\sin t\rangle\cdot\langle-\sqrt5\sin t,\sqrt5\cos t\rangle\,\mathrm dt`

`=\displaystyle5\int_0^(2\pi)(\sin t\cos t-\sin t\cos t)\,\mathrm dt=0`