From the book |
pgs. 273-275 # 2,8,10, pgs 279-280, # 2 4
pgs 286-287 #2 (skip plotting task), 4,6, pgs 292-293 #2,4 |
Also do: |
1) Take a parameterized curve C: (f(u),g(u)), where g(u)>0 in the x-y plane and
revolve it about the x-axis in space to create a surface (called a 'surface of
revolution'). a) Show that X(u,v) = (f(u), g(u) cos(v), g(u) sin(v)) parameterizes the surface where v is between 0 and 2 pi. b) Sketch the surface of revolution in the case, where f(u)=u and g(u)=1 and also in the case, where f(u)=cos(u) and g(u)=sin(u)+2. 2) Find a parameterization for the surface obtained by revolving the curve y=exp(x) about the x-axis. 3) Find a parameterization for the surface, where x-y^{2}+z^{4} y^{4}=0. 4) Find a parameterization for the y>0 part of the surface, where y^{2}-((x^{4}+z^{4})^{8}+1)=0. 5) Find the area of the part of the plane x+y+z=1, where x,y,z are all between 0 and 1. 6) Integrate the function f(x,y,z)=x-y over the part of the plane x+2y+3z=6, where x,y lie between 0 and 1. |
Section 5.3 |
2. The proof is almost the same as the one given in the book (page 269) but
x and y are reversed. 8. This vector field is the vector field of a vortex. a) With the parametrisation r(t)=(cos(t),sin(t)) or the circle, the vector field v=(P,Q) on the circle is v(r(t))=(-sin(t),cos(t)). The velocity vector is r'(t)=(-sin(t),cos(t)) and the line integral is the integral of (-sin(t),cos(t)) . (-sin(t),cos(t))=1 from t=0 to t=2 which is 2. b) No. Almost however: The vector field is a gradient field in the region where the point (0,0) is taken out: the potential is f(x,y) = arctan(y/x). In the whole plane, it not a conservative field because there, the line integral around a closed path is zero. This problem illustrates a condition in Green's theorem requiring the vector field to be continuously differentiable in the region under consideration. In this example, the vector field is not differentiable at the origin. c) The answer is the same as in b) because in the region between the circle and the square, the vector field is conservative. By Green's theorem, the integrals around the outer and inner boundaries are the same. d) If a path does not contain the origin, then the integral is zero because such a path is contained in the region where the field has a potential. 10. This is a useful formula for the area of an arbitrary polygon! Green's theorem says that the area is the line integral of the vector field v(x,y)=(0,x) over the polygon with the edges P_{k}=(x_{k},y_{k}). The path from P_{k} to P_{k+1} is r(t)=(x_{k},y_{k}) + t [(x_{k+1},y_{k+1})-(x_{k},y_{k})] has the velocity vector r'(t)=[(x_{k+1},y_{k+1})-(x_{k},y_{k})]. The dot product of the vector v(r(t))=(0,x_{k}+t ( x_{k+1} - x_{k} )) with r'(t) = (y_{k+1})-y_{k}) (x_{k}+t ( x_{k+1} - x_{k} )). Integrating this from t=0 to t=1 gives 1/2 ( x_{k+1} + x_{k} ) (y_{k+1}-y_{k}). The line integral over all vertices is the sum from k=1 to k=n of these expressions. This is the formula we needed to verify. |
Section 5.4 |
2a) 2^{1/2}, b) 2^{1/2},
c) 14^{1/2}, d) 14^{1/2} 4a) Parameterize as a graph X(u,v) = (u,v,(1-u^{2}-v^{2})^{1/2}), where (u,v) is in the unit disk. 4b) Parameterize X(u,v) = (sin(u) cos(v), sin(u) sin(v),cos(u)), where v runs from 0 to 2 and v from 0 to /2. |
Section 5.5 |
2. a) -2, b) 0, c) -2 2^{1/2},
d) The tangent plane has the equation x=1 4. The surface area is 3 2^{1/2}. 6. With D = |X_{u} x X_{v}| = ((1+f_{u}^{2}+f_{v}^{2})^{1/2}, the angle a between the vector n = X_{u} x X_{v}/|X_{u} x X_{v}|= (-f_{u},-f_{v},1)/D and z=(0,0,1) satisfies cos(a) = 1/D(u,v) = n . z or D=sec(a). |
Section 5.6 |
2. a) -/2, b) 0, c) /2 4) a) 2 , b) 2, c) , d) 0 |
Solutions to additional problems: |
1) A point on the circle is given by specifying a point on the curve
together with the angle in the (y-z) plane from the y axis where z = 0.
The point in the plane is given by the parameter u and the angle
by the parameter v. Here, you see that when v=0, the coordinates of
the point is (g(u),f(u),0) so it lies on the original curve. The first surface is the cylinder z^{2} + y^{2} = 1, the second surface is a torus (donut). 2) A parametrization sends (u,v) with positive u to (ln(u) cos(v),u,ln(u) sin(v)). 3) A possible parametrisation sends (u,v) to (u^{2}-v^{4}u^{4},u,v). 4) A parametrization sends (u,v) in the unit square to (u,(1+u^{4}+v^{4})^{8})^{1/2}, v). 5) The answer 3^{1/2}/2 is the integral of |X_{u} x X_{v}| = |(1,1,1)| = 3^{1/2} over the triangular region where u is between 0 and 1 and where v is less than u. 6) A parametrization which sends (u,v) in the unit square to (6-2u-3v, u, v) satisfies |X_{u}x X_{v}| = 14^{1/2}. Integrating |X_{u} x X_{v}| (y-x) =(14)^{1/2} (6-3u-3v) over this square gives 3*(14)^{1/2} |
Last update, 12/15/2000 |