Example 2: Exercise #22 (p. 706)

We begin by plotting the region defined by the given polar curves:

> P1 := polarplot( -5*sec(theta), theta=3*Pi/4..4*Pi/3 ): P1;

> P2 := polarplot( [r,3*Pi/4,r=0..10], color=BLUE ): P2;

> P3 := polarplot( [r,4*Pi/3,r=0..10], color=GREEN ): P3;

> display( [P1,P2,P3] );

>

Observe that this triangular region is vertically simple. To find the top and bottom curves:

> eq1;

> eq2a := isolate( eval( eq1, theta=3*Pi/4 ), y );

> eq2b := isolate( eval( eq1, theta=4*Pi/3 ), y );

>

That is, this region can be described in Cartesian coordinates as:

<= <=

-5 <= <= 0

>

To convert the integrand, recall that there is a factor of r that must be associated with dA. That is,

> eq4 := r*f = r^3 * sin(theta)^2;

> eq5 := isolate( eq4, f );

> eval( eq5, {r=sqrt(x^2+y^2), theta=arctan(y/x)} );

> simplify( % );

>

The given integral is therefore equivalent to

> Int( Int( y^2, y=sqrt(3)*x..-x ), x=-5..0 );

> value( % );

> evalf( % );

>