Calculate ∫C(7(x2−y)i⃗ +3(y2+x)j⃗ )⋅dr⃗ if: (a) C is the circle (x−2)2+(y−3)2=9 oriented counterclockwise. ∫C(7(x2−y)i⃗ +3(y2+x)j⃗ )⋅dr⃗ = (b) C is the circle (x−a)2+(y−b)2=R2 in the xy-plane oriented counterclockwise. ∫C(7(x2−y)i⃗ +3(y2+x)j⃗ )⋅dr⃗ =

By Green's theorem,[tex]\displaystyle\int_C(7(x^2-y)\,\vec\imath+3(y^2+x)\,\vec\jmath)\cdot\mathrm d\vec r=\iint_D\left(\frac{\partial3(y^2+x)}{\partial x}-\frac{\partial7(x^2-y)}{\partial y}\right)\,\mathrm dx\,\mathrm dy[/tex][tex]\displaystyle=10\iint_D\mathrm dx\,\mathrm dy[/tex]where [tex]D[/tex] is the region bounded by the closed curve [tex]C[/tex]. The remaining integral is 10 times the area of [tex]D[/tex].Since [tex]D[/tex] is a circle in both cases, and we're given the equations for them right away, it's just a matter of determining the radius of each one and plugging it into the well-known formula for the area of a circle with radius [tex]r[/tex], [tex]\pi r^2[/tex].(a) [tex]C[/tex] is a circle with radius 3, so the line integral is [tex]10\pi(3^2)=\boxed{90\pi}[/tex].(b) [tex]C[/tex] is a circle with radius [tex]R[/tex], so the line integral is [tex]\boxed{10\pi R^2}[/tex].