# Recitation 14

Section 13.3 Problem 8: Determine whether or not $F$ is a conservative vector field. If it is, find a function $f$ such that $F = \nabla f$. $F(x,y)=(2xy+y^{-2})i+(x^2-2xy^{-3})j, y>0$.

Solution: Since $\partial 2xy+y^{-2} / \partial y = 2x-2y^{-3} = \partial x^2-2xy^{-3} / \partial y$, $F$ is conservative. To figure out $f$ such that $f_x(x,y,z)=2xy+y^{-2}, f_y(x,y,z)=x^2-2xy^{-3}$. Take the first equation and integrate it with respect to $x$. We get $f(x,y,z)=x^2y+xy^{-2}+a(y)$. Differentiate it with respect to $y$. We obtain $f_y(x,y,z)=x^2-2xy^{-3}+a'(y)$. Compare it with $f_y(x,y,z)=x^2-2xy^{-3}$. We get $a'(y)=0$. Therefore, $a(y)=constant$. Hence $f(x,y,z)=x^2y+xy^{-2}+constant$.

Section 13.3 Problem 13: (a) Find a function $f$ such that $F = \nabla f$ and (b) use part (a) to evaluate $\int_C F \cdot dr$ along the given curve $C$. $F(x,y,z)=yzi+xzj+(xy+2z)k$, $C$ is the line segment from $(1, 0, -2)$ to $(4, 6, 3)$.

Solution: For part (a), by observation, we can verify that $f=xyz+z^2+c$ is the ‘anti-gradient’ of $F$. For part (b), the line integral is simply $f(4,6,3)-f(1,0,-2)=77$.

Section 13.3 Problem 28: Determine whether or not the given set is (a) open, (b) connected, and (c) simply-connected. $\{(x,y) | 1<|x|<2\}$.

Answer: The region is open, but not connected. It is not simply-connected because it is not connected.

Section 13.4 Problem 1: Evaluate the line integral by two methods: (a) directly and (b) using Green’s Theorem. $\oint_C(x-y)dx+(x+y)dy$, $C$ is the circle with center the origin and radius 2.

Section 13.4 Problem 8: Use Green’s Theorem to evaluate the line integral along the given positively oriented curve. $\int_C y^4dx+2xy^3dy$, $C$ is the ellipse $x^2+2y^2=2$.

Solution: Using Green’s Theorem, $\int_C y^4dx+2xy^3dy = \iint_E 2y^3-4y^3 = -2\iint_E y^3$, where $E$ is the interior part of the ellipse. Notice that $E$ has $x$-axis symmetry. The double integral is zero.

Section 13.4 Problem 17: Use Green’s Theorem to find the work done by the force $F(x, y) = x(x + y) i + xy^2 j$ in moving a particle from the origin along the $x$-axis to $(1, 0)$, then along the line segment to $(0, 1)$, and then back to the origin along the $y$-axis.

Comment: Denote the triangular region by $T$. By Green’s Theorem, $\int_C F\cdot dr = \iint_T (y^2-x) dA$. Then turn this double integral into an intreated integral.

Section 13.6 Problem 15: Find a parametric representation for the surface. The plane through the origin that contains the vectors $i-j$ and $j-k$.

Solution: Given one point $P$ on the plane and two vectors $u, v$ that lie in the plane, the parametric representation for the plane is simply $r(s,t)=P+us+vt$, where $s, t$ are two parameters. For this problem, $P=(0,0,0), u=(1,-1,0), v=(0,1,-1)$. Therefore, $x=s, y=-s+t, z=-t$.

Section 13.6 Problem 17: Find a parametric representation for the surface. The part of the hyperboloid $4x^2 - 4y^2 - z^2 = 4$ that lies in front of the $yz$-plane.

Solution: Let $y=s, z=t$, i.e., let $y,z$ be the parameters. Then $4x^2 = 4+4s^2+t^2$. Since $x\geq 0$, $x=\sqrt{1+s^2+t^2/4}$.