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.)

Looks like we're given[tex]\vec F(x,y)=\langle-x,-y\rangle[/tex]which in three dimensions could be expressed as[tex]\vec F(x,y)=\langle-x,-y,0\rangle[/tex]and this has curl[tex]\mathrm{curl}\vec F=\langle0_y-(-y)_z,-(0_x-(-x)_z),(-y)_x-(-x)_y\rangle=\langle0,0,0\rangle[/tex]which confirms the two-dimensional curl is 0.It also looks like the region [tex]R[/tex] is the disk [tex]x^2+y^2\le5[/tex]. Green's theorem says the integral of [tex]\vec F[/tex] along the boundary of [tex]R[/tex] is equal to the integral of the two-dimensional curl of [tex]\vec F[/tex] over the interior of [tex]R[/tex]:[tex]\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\iint_R\mathrm{curl}\vec F\,\mathrm dA[/tex]which we know to be 0, since the curl itself is 0. To verify this, we can parameterize the boundary of [tex]R[/tex] by[tex]\vec r(t)=\langle\sqrt5\cos t,\sqrt5\sin t\rangle\implies\vec r'(t)=\langle-\sqrt5\sin t,\sqrt5\cos t\rangle[/tex][tex]\implies\mathrm d\vec r=\vec r'(t)\,\mathrm dt=\sqrt5\langle-\sin t,\cos t\rangle\,\mathrm dt[/tex]with [tex]0\le t\le2\pi[/tex]. Then[tex]\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[/tex][tex]=\displaystyle5\int_0^{2\pi}(\sin t\cos t-\sin t\cos t)\,\mathrm dt=0[/tex]