Area Between Curves

> with(plots):

(page 434) If f and g are continuous and for all x [a,b] then the area A of the region bounded by y = f(x), y = g(x), x = a and x = b is = the "integral of the 'top' function minus the 'bottom' function."

e.g. Find the region bounded by and .

> f:=x+1:g:=9-x^2:plot({f,g},x=-6..6,y=-10..10,colour=black);

> limits:=solve(f=g);

> Area:=Int((g-f),x=limits[2]..limits[1]);

> value(%);

> evalf(%,3);

=> The bounded area is ~ 31.6 sq. units.

e.g Find the area bounded by and .

> eq1:=y=x-1;eq2:=y^2=2*x+6;

> implicitplot({eq1,eq2},x=-6..8,y=-4..6,colour=black,numpoints=1000);

> solve({eq1,eq2});

There is a problem in that as you move from left to right the top and bottom functions change! From x = -3 to -1, is the top function and is the bottom function.From x = -1 to x = 5, is the top function but is the new bottom function and this must be taken into account when finding the area! See below.

> Top:=sqrt(2*x+6);Bottom1:=-Top;Bottom2:=x-1;

> Area:=Int(Top-Bottom1,x=-3..-1)+Int(Top-Bottom2,x=-1..5);

> value(Area)*`square units`;

Integration "along the y-axis"

The last e.g. was a lot of work because of the nature of the region. If you look at it with respect to the y-axis though it becomes much simpler. As you move along the y-axis from bottom to top there is a single (and simple) rightmost function of y (namely ) and a single (and simple) leftmost function of y (namely ). It follows that this would be a much smpler problem if we integrate with respect to y , moving from "down" to "up" (i.e. from the negative to positive y direction). See below, keeping in mind that we know from before the y coordinates of the intersection points.

> right:=y+1;left:=(y^2-6)/2;

> Area2:=Int(right-left,y=-2..4);

> value(Area2)*`square units`;

You can see we get the same answer and it's much less work!

In general : If and are continuous functions and for all y [c,d] then the area bounded by f and g between y = c and y = d is .

More e.g.'s: Find the area of the region bounded by the given curves.

A) and

> f:=x^2:g:=x^3:plot({f,g},x=-2..2,y=-2..2,colour=black);

> solve(f=g);

> A:=Int(f-g,x=0..1);value(A)*`sq units`;

B) , .

> eq1:=y^2=x;eq2:=x-2*y=3;implicitplot({eq1,eq2},x=-1..10,y=-4..4,colour=black);

> solve({eq1,eq2});

> right:=3+2*y;left:=y^2;

> A:=Int(right-left,y=-1..3);value(A)*`sq units`;

# 47, pg 439: Find the values of c so that the region enclosed by and is 576.

> top:=c^2-x^2;bottom:=x^2-c^2;

> solve(top=bottom);

> area:=Int(top-bottom,x=-c..c);area:=value(area);

> solve(area=576,c);

=> c can be 6 or -6.