m103ex-5-10.mw

Applications of Integrals

(Thomas-Finney Section 5.1, page 372)

Problem 31

Find the area of the region enclosed by the curves 4*x^2+y = 4 and x^4-y = 1 .

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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;

eqn1 := 4*x^2+y = 4

eqn2 := x^4-y = 1

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

y1 := -4*x^2+4

y2 := x^4-1

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

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

{x = -1, y = 0}, {x = 1, y = 0}, {x = RootOf(_Z^2+5, label = _L1), y = 24}

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);

[Plot]

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);

area := 104/15

Section 5.3, page 385 Problem 9

Find the volume of the solid generated by revolving the region bounded by

        y = sqrt(cos(x)) ,  for x in [0, Pi/2] ,     y = 0 ,   x = 0

about the x axis.

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First we'll graph the region and the solid:

> with(plots,tubeplot):

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

[Plot]

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

[Plot]

To calculate the volume, we just integrate Pi*y^2 from 0 to Pi/2 :

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

volume := Pi