CHAPTER 5 - Applications of Integrals

Section 5.1, page 371

Problem 31

First, we'll plot the region of interest. Since the equations are easily solved for y, we'll do that first:

>	eqn1:=4*x^2+y=4; eqn2:=x^4-y=1;

>	y1:=solve(eqn1,y); y2:=solve(eqn2,y);

And we'll solve for where the curves cross so we know where to plot:

>	solve({eqn1,eqn2},{x,y});

The third thing is complex, so looks like x should go from -1 to 1. We'll go a little farther, just in case:

>	plot({y1,y2},x=-2..2,color=blue,thickness=2);

Now we're ready to calculate the area. y1 is on top since it's the parabola that opens down:

>	area:=int(y1-y2,x=-1..1);

Section 5.3, page 385

Problem 9

First we'll graph the region and the solid:

>	with(plots,tubeplot):

>	plot(sqrt(cos(x)),x=0..Pi/2,color=blue,thickness=2);

>	tubeplot([x,0,0],x=0..Pi/2,radius=sqrt(cos(x)));

To calculate the volume, we just integrate  from 0 to :

>	volume:=int(Pi*(sqrt(cos(x))^2),x=0..Pi/2);