When integrating a function over a cylindrical surface or cone, you will find that using rectangular coordinates may make it complicated. Also, when integrating a function over a spherical surface, you will find that rectangular coordinates are not conducive to happy integrating, so what is a person to do? We will now talk about how to integrate using both cylindrical and spherical coordinates. Hopefully, they will make integrating over the above named surfaces easier.

Here are the conversion formulas for cylindrical coordinates.

EXAMPLE 1:

 Find the volume of the right circular cylinder whose base is the circle r = 2sin q in the xy-plane and whose top lies in the plane z = 4 - y.

SOLUTION:

The first thing that we must do is to convert z = 4 - y into polar coordinates.

z = 4 - y ® z = 4 - r sin q .

r will go from 0 to 2sin q , and q will go from 0 to p .

EXAMPLE 2:

 Find the volume of the prism whose base is the triangle in the xy-plane bounded by the x-axis and the lines y = x and x = 1 and whose top lies in the plane z = 2 - y.

SOLUTION:

I going to set the volume integrals up two different ways. The first way will be using cylindrical coordinates, and the second way will use rectangular coordinates.

First of all, I must convert x = 1 into polar coordinates.

r cos q = 1 ® r = sec q

Now to convert y = x.

y = x ® r sin q = r cos q ® tan q = 1 ® q = p /4

Finally, convert z = 2 - y.

z = 2 - y ® z = 2 - r sin q

 Rectangular coordinates

Which of the two methods was easier? The rectangular coordinates were. Only use cylindrical coordinates when working with cylinders, cones, or paraboloids. Any time that you are working with planes, use rectangular coordinates.

EXAMPLE 3:

Find the volume of the solid enclosed by the cone

between the planes z = 1 and z = 2.

SOLUTION:

First, I will convert the equation of the cone into polar coordinates.

z will go between 1 and 2, and r will go between 0 and z, so I will have to change the order of integration. I will integrate with respect to r first, then z, and finally q .

Here are the conversion formulas for spherical coordinates.

EXAMPLE 4:

 Find the volume of the solid bounded below by the sphere r = 2cos j and above by the cone

0 £ r £ 2cos j

EXAMPLE 5:

Find the volume of the smaller region cut from the solid sphere r £   2 by the plane z = 1.

EXAMPLE 6:

Find the volume of the solid enclosed by the cardiod of revolution r = 1 - cosj .

0 £ r £ 1 - cos j

0 £ j £ p

0 £ q £ 2p

EXAMPLE 7:

Find the centroid of the region that is bounded above by the surface

on the sides by the cylinder r = 4, and below by the xy-plane.

Since this is a centroid, then the density d = 1. If you look at the graph of this solid, you should notice that it is symmetric with respect to the x and y axis, so

All I need to find is the mass and M _{x y}. I will use cylindrical coordinates to do this.

EXAMPLE 8:

Find the centroid of a homogeneous solid hemisphere of radius 2.

The center of mass of a homogeneous sphere better be the origin. I could have used symmetry to determine that the center of mass was the origin, but I wanted to show you how to do a center of mass in spherical coordinates.