Attached are a question and the matlab code. The code gives the total surface area as 639.4828, but the figure is not showing up. Also, the question says to use surface integral to calculate the...

Attached are a question and the matlab code. The code gives the total surface area as 639.4828, but the figure is not showing up.

Also, the question says to use surface integral to calculate the surface area, not matlab.

Please help me on this. Thanks.

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Expert Answers
sbernabel eNotes educator| Certified Educator

i, j, k are unit vectors.

r(u,v)=(b+a*cos(u))*cos(v) i + (b+a*cos(u))*sin(v) j + a*sin(u) k

0<=u,v<=2pi and a=1/4 and b=1

 

The surface area is:

A=double integral of magnitude of (r_u x r_v) dA

r_u is the partial derivative of r(u,v) with respect to u and r_v with respect to v:

r_u(u,v)= -a*sin(u)*cos(v) i -a*sin(u)*sin(v) j +a*cos(u) k

r_v(u,v)= -(b+a*cos(u))*sin(v) i +(b+a*cos(u))*cos(v) j

 

Compute the cross product:

r_u x r_v = -a*cos(u)*(b+a*cos(u))*cos(v) i - a*cos(u)*(b+a*cos(u))*sin(v) j - a*(b+a*cos(u))*sin(u) k

 

Compute the magnitude:

magnitude(r_u x r_v)= sqrt(a^4*(cos(u))^2+2*a^3*b*cos(u)+a^2*b^2)= a*(b+a*cos(u))

 

Lastly, evaluate the integral:

integral(integral(a*(b+a*cos(u)),v,0,2*pi),u,0,2*pi) = 4*pi^2 * a * b

With a=1/4 and b=1 the surface area is pi^2.

 

The Matlab code for the plot is:

clear all
syms u v
a=1/4; b=1;
x=(b+a*cos(u))*cos(v);
y=(b+a*cos(u))*sin(v);
z=a*sin(u);
ezsurf(x,y,z,[0,2*pi,0,2*pi])

 

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