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!