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