# Thread: Volume in polar coordinates

1. ## Volume in polar coordinates

These two questions are giving me some trouble. I can see that z = r^2, by using polar coordinates but after that i don't know what to do.

First Question
1.) Find the volume lying between the paraboloids:

z = x^2 + y^2 and 3z = 4-x^2-y^2

Second Question
2.) Find the volume lying inside both the sphere x^2 + y^2 + z^2 = 2a^2
and the cylinder x^2 + y^2 = a^2

Any help would be much appreciated, thanks guys/girls

2. Originally Posted by hammer
These two questions are giving me some trouble. I can see that z = r^2, by using polar coordinates but after that i don't know what to do.

First Question
1.) Find the volume lying between the paraboloids:

z = x^2 + y^2 and 3z = 4-x^2-y^2
The important thing to know is which one is one top.
Did you try to make some graph?

Note z=x^2+y^2 is a parabolid which opens up and the vertex is on the origin. And 3z=4-x^2-y^2 is a parabolid which opens down and vertex is at (0,0,4/3) if you divide by 3 to bring it to the form z=f(x,y). Thus, the second one is the top one.

How do they intersect? It looks like, if you graph them, that they intersect in a circle. We can actually show that,
z=x^2+y^2
3z=4-x^2-y^2
Then substitute equation (1) into equation (2):
3z = 4 - z
Thus,
z=1.
Thus, they intersect at z=1 (in that plane) and if you substitute that value in any equation say (1) then you get:
1=x^2+y^2
You get a circle with radius one.

In order to find the volume between these paraboloids we also need to know their projection onto the xy plane. Which is what we found above, it is the circle with radius 1.

Now we can find the volume.

This is Mine 54th Post!!!

3. THanks!

So for Question #2, the sphere is going to be the top one and the two surfaces will meet at a circle centered at the origin of radius a?

What would the integration of this look like for this question, specifically the limits of integration?

4. Originally Posted by hammer

Second Question
2.) Find the volume lying inside both the sphere x^2 + y^2 + z^2 = 2a^2
and the cylinder x^2 + y^2 = a^2
Certainly we can say a>0.

I recommend to draw a picture by hand to see what is going on. If you do that you shall see the cylinder is lying inside the sphere. The cylinder intersects the sphere, that is from where start measureing the volume, as the problem says. Let us see how they intersect.
x^2+y^2+z^2 = 2a^2
x^2+y^2 = a^2
Substitute equation (2) into equation (1):
a^2+z^2=2a^2
Thus,
z^2=a^2
Thus,
z=+/- a.
Hence they intersect in the plane z=a , above xy plane, and z=-a, below the xy plane.
Furthermore, the equation x^2+y^2=a^2, tells us they intersect in a circle with radius a.

To find the volume, we need to know the top surface, which is the top plane z=a. And the bottum surface which is z=-a. And the also need to know the projection of the intersections onto the xy plane, which we know is the circle at origin with radius a.

Hence.