Expert: Abe Mantell - 1/6/2010

Question
hi

1- how can i find the volume by useing a triple integral of the solid enclosed by the paraboloid x=5y^2+5z^2 and x=5

i've done this:

5=5y^2+5z^2
r^2=1

0<=x<=5
0<=r<=1
0<=@<=2pi

is that right?
i get the volume=0 and it is wrong

2- how can i find the volume of of the solid that lies within the sphere  x^2+y^2+z^2=36 above the xy- plane, and outside the cone z=7sqrt(x^2+y^2)

your help is very much appericated

Answer
Hello Sara,

1. I gather you want to do the integral in cylindrical coordinates.  Thus,
.  The integral should be:
.  int(int(int(r, z = 5*r^2 .. 5), r = 0 .. 1), @ = 0 .. 2*Pi)=5*pi/2 units^3

2. V=(volume of the upper hemisphere)-(volume of solid between the sphere & cone)
.  =(1/2)((4/3)*pi*6^3)-int(int(int(r, z = 7*r .. sqrt(36-r^2)), r = 0 .. sqrt(18)), @ = 0 .. 2*Pi)
.  =144*pi-(144*Pi-288*sqrt(2)*Pi)=288*sqrt(2)*Pi units^3

I hope this helps.

Abe