Denote the circle of radius [tex]a[/tex] by [tex]C[/tex]. [tex]C[/tex] is simple and closed, so by Green's theorem the line integral reduces to a double integral over the interior of [tex]C[/tex] (call it [tex]D[/tex]):
[tex]\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\int_C(6x^2y+2y^3+4e^x)\,\mathrm dx+(7e^{y^2}+54x)\,\mathrm dy[/tex]
[tex]=\displaystyle\iint_D\left(\frac{\partial(7e^{y^2}+54x)}{\partial x}-\frac{\partial(6x^2y+2y^3+4e^x)}{\partial y}\right)\,\mathrm dx\,\mathrm dy[/tex]
[tex]=\displaystyle\iint_D(54-6x^2-6y^2)\,\mathrm dx\,\mathrm dy[/tex]
[tex]D[/tex] is a circle of radius [tex]a[/tex], so we can write the double integral in polar coordinates as
[tex]\displaystyle\iint_D(54-6x^2-6y^2)\,\mathrm dx\,\mathrm dy=\int_0^{2\pi}\int_0^a(54-6r^2)r\,\mathrm dr\,\mathrm d\theta[/tex]
a. For [tex]a=1[/tex], we have
[tex]\displaystyle\int_0^{2\pi}\int_0^1(54-6r^2)r\,\mathrm dr\,\mathrm d\theta=2\pi\int_0^1(54r-6r^3)\,\mathrm dr=\boxed{51\pi}[/tex]
b. Let [tex]I(a)[/tex] denote the integral with unknown parameter [tex]a[/tex],
[tex]I(a)=12\pi\int_0^a(9r-r^3)\,\mathrm dr\,\mathrm d\theta[/tex]
By the fundamental theorem of calculus,
[tex]I'(a)=12\pi(9a-a^3)[/tex]
[tex]I(a)[/tex] has critical points when
[tex]12\pi(9a-a^3)=12\pi a(9-a^2)=0\implies a=0,a=\pm3[/tex]
If [tex]a=0[/tex], then line integral is 0, so we ignore that critical point. For the other two, we would find [tex]I(\pm3)=243\pi[/tex].
