Suppose we want to evaluate

where is the solid enclosed by the planes and and the cylinders We can use the cylindrical coordinates as

We see that the bottom of the given long washer is a region between to circles and so and We write

Suppose we want to evaluate

where is the solid enclosed by the plane and the cylinders We can use the cylindrical coordinates as

We see that the bottom of the given long washer is a region between to circles and so and We write

Give it a try and compute the rest.

Suppose we want use triple integral to find the volume of the solid enclosed by the intersection of the sphere and the cylinder Using cylindrical coordinates, we know

We see that

and

Note that and will provide the region of integration on -plane. Therefore, we write

Since the solid id symmetric with respect to -plane, we can compute half volume and multiply by Continue with the computation of this triple integral.

Suppose we are given

We want to use spherical coordinates to compute this triple integral. We start by writing

We see that the region of integration is a sphere of radius in the first octant. Therefore, and so

We would like to use spherical coordinates to integrate over the solid above the cone and inside the sphere We know the cone is heavy side down, pointing upward. The elements that we need are and First

since radius of the sphere is at max Next, to get the whole circle on -plane we need

To find the bounds for ask yourself how far the angle goes from the begining point on the positive side of the -axis to get to the cone. Consider the -plane only. If we travel one unit down that is if we take then and the obtained triangle provides that the angle that this cone makes with the -axis is However, the range of must be We now are ready to set up the triple integral and write