Solutions to MATLAB #4!
1/5
>> clear all;
>> close all;
>> % Problem 1
>> clear all;
>> syms s t real;
>> % Problem 1a
>> % Using cylindrical coordinates, the hyperboloid is written as:
>>%
z^2 = r^2 - 1
>> % We know that r>0 so we can solve for r explicitly:
>> %
r = sqrt(z^2 + 1)
>> % We can substitute the expression of r in terms of z into the standard
>> % cylindrical coordinates transformation:
>> %
x = sqrt(z^2 + 1)*cos(theta)
>> %
y = sqrt(z^2 + 1)*sin(theta)
>> %
z = z
>> % In addition to the -2<=z<=2 bounds we also have 0<=theta<=2*pi.
>> figure;
>> ezmesh(sqrt(s^2+1)*cos(t),sqrt(s^2+1)*sin(t),s,[-2 2 0 2*pi]);
2
2
x = cos(t) sqrt(s +1.0), y = sin(t) sqrt(s +1.0), z = s
2
1.5
1
z
0.5
0
−0.5
−1
−1.5
−2
3
2
4
1
2
0
0
−1
−2
−2
y
>>
>>
>>
>>
>>
>>
−3
−4
x
% Problem 1b
% The surface of revolution is given by:
%
r(s,t) = (s, cos(s)*cos(t), cos(s)*sin(t))
% for -pi/2<=s<=pi/2 and 0<=t<=2*pi
figure;
ezmesh(s,cos(s)*cos(t),cos(s)*sin(t),[-pi/2 pi/2 0 2*pi]);
Solutions to MATLAB #4!
2/5
x = s, y = cos(s) cos(t), z = cos(s) sin(t)
1
z
0.5
0
−0.5
−1
1
2
0.5
1
0
0
−0.5
y
>>
>>
>>
>>
>>
>>
>>
>>
>>
>>
>>
−1
−1
% Problem 2
clear all;
syms x y real;
S1(x,y)=(2+cos(y))*cos(x);
S2(x,y)=(2+cos(y))*sin(x);
S3(x,y)=sin(y);
S=[S1,S2,S3];
% Problem 2a
figure;
ezmesh(S1,S2,S3,[0 2*pi 0 2*pi]);
axis equal;
−2
x
Solutions to MATLAB #4!
3/5
x = cos(x) (cos(y)+2.0), y = sin(x) (cos(y)+2.0), z = sin(y)
z
0.5
0
−0.5
2
3
1
2
0
1
0
−1
−2
y
>>
>>
>>
>>
>>
>>
−1
−2
x
% Problem 2b
Sx=diff(S,x);
Sy=diff(S,y);
Sx_cross_Sy=cross(Sx(x,y),Sy(x,y));
norm_Sx_cross_Sy=sqrt(dot(Sx_cross_Sy,Sx_cross_Sy));
I_2b=int(int(norm_Sx_cross_Sy,x,0,2*pi),y,0,2*pi)
I_2b =
8*pi^2
% Problem 3
[X,Y]=meshgrid(-10:10,-5:5);
M=X./(X.^2+Y.^2);
N=1./(X.^2+Y.^2);
figure;
quiver(X,Y,M,N);
Solutions to MATLAB #4!
4/5
6
4
2
0
−2
−4
−6
−15
−10
−5
0
% Problem 4
clear all;
% Define the curve
syms t real;
r1=cos(t);
r2=-2*t;
r3=sin(3*t);
r=[r1,r2,r3];
a=0;
b=pi;
% Define the vector field
syms x y z real;
M(x,y,z)=exp(y);
N(x,y,z)=x^2*y;
P(x,y,z)=x*z^3;
F=[M,N,P];
% Setup and evaluate the integral
I_4=int(dot(F(r1,r2,r2),diff(r,t)),t,a,b)
I_4 =
- exp(-2*pi)/5 - (41*pi^2)/4 - 1/5
5
10
15
Solutions to MATLAB #4!
5/5
% Problem 5
clear all;
syms t real;
r1=sin(2*t);
r2=sin(t);
r=[r1,r2];
a=0;
b=pi;
% Problem 5a
figure;
ezplot(r1,r2,[a,b]);
x = sin(2 t), y = sin(t)
1.2
1
0.8
y
0.6
0.4
0.2
0
−0.2
−0.8
−0.6
−0.4
−0.2
0
x
0.2
0.4
0.6
0.8
% The region is the diamond like area bounded by the curve
% Problem 5b
% By integrating the vector field, F, over the closed curve C, we can see
% by Green's theorem that the result will be the area of the region.
%
F = x*j
syms x y real;
% Define the vector field
F(x,y)=[0,x];
% Setup and evaluate the integral
area=int(dot(F(r1,r2),diff(r,t)),t,a,b)
area =
4/3