Question

Let vector F = (6 x^2 y + 2 y^3 + 4 e^x) i + (7 e^{y^2} + 54 x) j . Consider the line integral of vector F around the circle of radius a, centered at the origin and traversed counterclockwise.
(a) Find the line integral for a=1 .
(b) For which value of a is the line integral a maximum?

Answer 1

Denote the circle of radius $a$ by $C$. $C$ is simple and closed, so by Green's theorem the line integral reduces to a double integral over the interior of $C$ (call it $D$):

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\int_C(6x^2y+2y^3+4e^x)\,\mathrm dx+(7e^{y^2}+54x)\,\mathrm dy$

$=\displaystyle\iint_D\left(\frac{\partial(7e^{y^2}+54x)}{\partial x}-\frac{\partial(6x^2y+2y^3+4e^x)}{\partial y}\right)\,\mathrm dx\,\mathrm dy$

$=\displaystyle\iint_D(54-6x^2-6y^2)\,\mathrm dx\,\mathrm dy$

$D$ is a circle of radius $a$, so we can write the double integral in polar coordinates as

$\displaystyle\iint_D(54-6x^2-6y^2)\,\mathrm dx\,\mathrm dy=\int_0^{2\pi}\int_0^a(54-6r^2)r\,\mathrm dr\,\mathrm d\theta$

a. For $a=1$, we have

$\displaystyle\int_0^{2\pi}\int_0^1(54-6r^2)r\,\mathrm dr\,\mathrm d\theta=2\pi\int_0^1(54r-6r^3)\,\mathrm dr=\boxed{51\pi}$

b. Let $I(a)$ denote the integral with unknown parameter $a$,

$I(a)=12\pi\int_0^a(9r-r^3)\,\mathrm dr\,\mathrm d\theta$

By the fundamental theorem of calculus,

$I'(a)=12\pi(9a-a^3)$

$I(a)$ has critical points when

$12\pi(9a-a^3)=12\pi a(9-a^2)=0\implies a=0,a=\pm3$

If $a=0$, then line integral is 0, so we ignore that critical point. For the other two, we would find $I(\pm3)=243\pi$.