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}.

Leave a comment

Your email address will not be published. Required fields are marked *