CHAPTER 13 - Multiple Integrals

Section 13.1, page 1010

Problem 33

The problem is

>	Int(Int(x^2*exp(x*y),x=y..1),y=0..1);

Maple can do this:

>

value(%);

But let's sketch the region of integration ("for every y between 0 and 1, x goes from the line x=y to the line x=1").

>	plot({[y,y,y=0..1],[1,y,y=0..1]},color=blue,thickness=2);

Another way to describe this triangle is "for every x from 0 to 1, y goes from the line y=0 to the line y=x", and so our integral is equivalent to:

>	Int(Int(x^2*exp(x*y),y=0..x),x=0..1);

>

value(%);

Problem 68

To get Maple to evaluate integrals numerically:

>	evalf(Int(Int(exp(-x^2-y^2),x=0..1),y=0..1));

Section 13.3, page 1024

Problem 35

First let's plot the base - inside the cardioid and outside the circle:

>	r1:=1; r2:=1+cos(t);

>	plot({[r1*cos(t),r1*sin(t),t=0..2*Pi],[r2*cos(t),r2*sin(t),t=0..2*Pi]},color=blue,thickness=2,scaling=constrained);

We can see that t goes from  to . The height over any point is equal to x=r cos(t). So we can calculate the volume:

>	vol:=int(int(r*cos(t)*r,r=r1..r2),t=-Pi/2..Pi/2);

Problem 44

The curves are y=0 (that's just ), y=x/2 (that's ), and x=1 (that's ). The domain is a triangle, we don't really need to plot it, right?

To do the integral in polar, set:

>	x:=r*cos(theta); y:=r*sin(theta);

>	Int(Int(simplify(x/(x^2+y^2)*r),r=0..1/cos(theta)),theta=0..arctan(1/2));

>

value(%);

Not so bad.

Section 13.6, page 1044

Problem 40

The region looks like an ice-cream cone:

>

restart;

>	with(plots,display3d):

>	A:=plot3d([rho*sin(Pi/4)*cos(theta),rho*sin(Pi/4)*sin(theta),rho*cos(Pi/4)],rho=0..3,theta=0..2*Pi):

>	B:=plot3d([3*sin(phi)*cos(theta),3*sin(phi)*sin(theta),3*cos(phi)],phi=0..Pi/4,theta=0..2*Pi):

>	display3d({A,B},scaling=constrained);

Now we can find the volume easily in spherical coordinates:

>	vol1:=int(int(int(rho^2*sin(phi),rho=0..3),phi=0..Pi/4),theta=0..2*Pi);

Or in cylindrical coordinates, the cone is r=z and the sphere is . So the volume is

>	int(int(int(r,z=r..sqrt(9-r^2)),r=0..3*sqrt(2)/2),theta=0..2*Pi);

How about that!