The Arc Length Formula

If f ' is continuous on [a, b], then the length of the curve NiMvJSJ5Ry0lImZHNiMlInhH on the interval is given by

NiMvJSJMRy0lJGludEc2JC0lJXNxcnRHNiMsJiIiIkYsKiQtJSVkaWZmRzYkJSJ5RyUieEciIiNGLC9GMjslImFHJSJiRw==

Alternately, if g ' is continuous on [c, d], then the length of the curve NiMvJSJ4Ry0lImdHNiMlInlH on the interval is given by

NiMvJSJMRy0lJGludEc2JC0lJXNxcnRHNiMsJiIiIkYsKiQtJSVkaWZmRzYkJSJ4RyUieUciIiNGLC9GMjslImNHJSJkRw==

The Surface Area Formula

If f ' is continuous on [a, b], then the surface area of the surface obtained by rotating the curve y = f(x) about the x-axis is

NiMqKCIiIyIiIiUjUGlHRiUtJSRpbnRHNiQqJiUieUdGJS0lJXNxcnRHNiMsJkYlRiUqJC0lJWRpZmZHNiRGKyUieEdGJEYlRiUvRjQ7JSJhRyUiYkdGJQ==

If x = g(y), on [c, d], then the surface area of the surface obtained by rotating the curve about the y-axis is

NiMqKCIiIyIiIiUjUGlHRiUtJSRpbnRHNiQqJiUieEdGJS0lJXNxcnRHNiMsJkYlRiUqJC0lJWRpZmZHNiRGKyUieUdGJEYlRiUvRis7JSJjRyUiZEdGJQ==

Example 1: Find the area and the perimeter of the ellipse NiMvLCYqJiUieEciIiMiIiohIiIiIiIqJiUieUdGJyIiJUYpRipGKg== and find the volume and the surface area of the solid created when the upper half of the ellipse is rotated about the x-axis.

restart;

with(plots):

f:=x^2/9+y^2/4=1;

implicitplot(f,x=-3..3,y=-2..2,scaling=constrained,thickness=2);

g:=solve(f,y);

gx:=diff(g[1],x);

A:=2*int(g[1],x=-3..3);

P:=2*Int(sqrt(1+gx^2),x=-3..3);

evalf(P);

V:=Pi*Int(g[1]^2,x=-3..3);

S:=2*Pi*Int(g[1]*sqrt(1+gx^2),x=-3..3);

evalf(S);

Parametric curves and Polar Coordinates Example 2: Find the arc length of a parametric curve: x = cos(3t), y = sin(3t) on [0, 2Pi].

xt,yt:=cos(3*t),sin(3*t);

plot([xt,yt,t=0..2*Pi]);

Dxt:=diff(xt,t);Dyt:=diff(yt,t);

L:=Int(sqrt(Dxt^2+Dyt^2),t=0..2*Pi);value(%);

To see a plot of the torus, input the following statements. There is also a semitorus command, which allows one to view parts of the torus.

with(plottools):

a:=torus([3,0,0],1,3):

plots[display](a,scaling=constrained,axes=boxed);

Plot the function r = 2*sin(theta) on the interval [0, Pi].

with(plots):

r:=2*sin(t):

polarplot(r,t=0..Pi,scaling=constrained,thickness=2);

animatecurve([r,t,t=0..Pi],coords=polar,thickness=2,frames=50,scaling=constrained);#Click on the plot and use the toolbar buttons to animate.

Plot r = 2-2*sin(2*theta) and r = 1 and determine the points of intersection.

restart:with(plots):

f:=2-2*sin(2*t):g:=1:

polarplot({f,g},t=0..2*Pi,scaling=constrained,thickness=2);

s1:=solve(f=4);

p1:=evalf([s1,subs(t=s1,f)]);

The solve command only gives one value. fsolve must be used and limits on t must be used. The animatecurve command may be of some use.

s2:=fsolve(f=g,t=Pi/4..Pi);

p2:=[s2,evalf(subs(t=s2,f))];

s3:=fsolve(f=g,t=Pi..5*Pi/4);

p3:=[s3,evalf(subs(t=s3,f))];

Find the area enclosed by the graph of r = 3+3*sin(theta).

restart:with(plots):

r:=3+3*sin(t):

polarplot(r,t=0..2*Pi,scaling=constrained);

A:=evalf(.5*int(r^2,t=0..2*Pi));

Find the area of the region common to the graph of r = 3+2*cos(theta) and the graph of r = 2.

restart:with(plots):

r1:=3+2*cos(t):r2:=2:

polarplot({r1,r2},t=0..2*Pi,scaling=constrained);

s:=solve(r1=r2);

A:=2*(.5*Int(r2^2,t=0..s)+.5*Int(r1^2,t=s..Pi));

value(%);

evalf(5);

Exercise

1. Two particles travel along the curves: x1 = 3sin(t), y1=2cos(t); x2=-3+cos(t), y2=1+sin(t), t is from 0 to 2Pi.

a) Graph the paths of both particles.

b) Do the particles collide?

c) Do their paths intersect?

d) Describe what happens if the path of the second particle is given by: x2=3+cos(t), y2=1+sin(t).

2. Plot the following curves and the area that it encloses a) r = 1+2*sin(6*theta) b) r = 2sin(theta)+3*sin(9*theta).

3. Plot r = 2+cos(2*theta) and r = 2+2*cos(3*theta) and find their points of intersection at the same points in their cycles.

4. Find the area of the region inside the graph of r = -2*sin(theta) and outside the inner loop of r = 2-3*sin(theta).

5. Find the area of the region enclosed by the graphs of r = sin(theta) and r = sqrt(3)*cos(theta).