Cone w/Spherical and Cylindrical Coordinates
We are going to use spherical coordinates to find the volume of the cone in the picture.The blue angle is pi/3 and the height of the cone is 4. This is a very straight forward question to practice our technique of triple integrals in spherical coordinates. We are also going to solve the question with cylindrical coordinates to make sure that we get the same result for the volume of the object.
Let’s start with our inner most integral that’s respect to rho. In order to find the limits of the inner most integral, we can imagine that we have a light bulb in the origin, and we turn it on. We can also imagine that our solid,cone, is transparent so we can see what light rays hit first and last. If we turn on our light bulb in the origin, we actually just start in the solid. So, we don’t hit a surface. Our light rays start traveling from the inside of our cone. As we travel through our solid as light rays, we leave our solid by touching a plane which is the top face of our cone, z=4.
Now, we need to express our limits of integration in spherical coordinates. As we see, the light rays don’t hit a surface initially, and we are in the solid actually. So, we can say that our lower limit of integration is just 0 respect to rho. We turn on the bulb, and it just starts traveling through the cone. However, when we leave the cone, we leave it by touching the plane at z=4. We need to express it in spherical coordinates, and we know that planes can be expressed as asec(phi) where a is 4 for this instance because we have the plane z=4. We can also see it by multiply both sides with cos(phi). We have rho=4csc(phi) initially as we know. After multiplying both sides by cos(phi), we get rhocos(phi)=4. We know that rhocos(phi) is just z in cylindrical coordinates. So, we check that we expressed the plane z=4 right. You don’t have to do this process if you just want to memorize that we describe planes as acsc(phi) in spherical coordinates but very simple manipulations can show you where they come from.
So, we get 0 as our lower limit and 4csc(phi) as our upper limit for the inner most integral that’s respect to rho. The light rays from the bulb in the origin don’t hit a surface initially, and we just travel through our solid. We leave our solid by touching a plane. We expressed these two limits in terms of spherical coordinates. So, we are done with the first integral!
We need to find the limits of integration of the middle integral that’s respect to phi now. We know that phi is just the angle our rho makes with z axis. Or we can think about it as a radar that we cover our solid starting from z-axis. We see that the angle between our z-axis and the cone’s surface is just pi/3. So, we start from z which means we make 0 degree angle first. Then, we travel pi/3, and we completely cover our solid. We can write our limits of integration as 0 for our lower limit and pi/3 for our upper limit. We are actually done with the middle integral that’s respect to phi, too.
We are left with the outer most integral that’s respect to theta which is the easiest to find limits. We look at our cone and see it as a whole cone. It’s not a half cone. It’s not a quarter cone. When we look on top of the cone, we see a whole disk. Therefore, our limits range from 0 to 2pi. If we had a half disk instead from the top view, meaning we cut the cone in half, it would be 0 to pi. We write our limits of the last integral from 0 to 2 pi to get done with our triple integral. Now, we can solve it to find the volume of the cone.
In order to solve this question with cylindrical coordinates, we should have some sort of grasp on 3d graphs. We don’t have a surface function to define the cone. So, we will find it by using some basic trigonometry and algebra to determine our limits of integration.
We can write our triple integral in order of dzrdrdθ.
We are going to start with inner most integral that’s respect to z. We can imagine that we can send a beam of light that’s parallel to z-axis to get our limits. The first thing we hit is the cone. The light rays travel through the cone and leave it by touching the plane z=4. We need to remember that we can express a cone as z= sqrt(x^2+y^2). And it might be tempting to just stick this into the lower bond. However, if we take a closer look, we see that there’s an angle given. If we view the cone from the side, such as z-y plane, meaning that we plug in zero for x, we get z=sqrt(0+y^2) which is just z=y. However, z=y doesn’t seem identical to our cone. It’s quite similar though. The only difference is the slope of the line which results tighter or wider cones. We have a wider cone and we should have a slope that fits on our cone. We realize that the angle is pi/6 between the line and x-axis. I know that arctan(pi/6) is sqrt(3)/3. If I write my cone as sqrt(3)/3 sqrt(x^2+y^2), I get my cone! When we check it by viewing on z-y plane, we see z= sqrt(3)/3 that’s what we see from the side view of the cone.
We are luckily given the upper limit of the inner most integral which is z=4. We know that’s the upper limit because the light rays leave the solid by touching z=4 plane. I need to write my limits in terms of cylindrical coordinates so I leave my upper limit 4 as it is and I get sqrt(3)/3r for my lower limit. I’m done with the inner most integral!
We are in the middle integral that’s respect to r now. We can see that r ranges from zero to some value at most. When we view the cone from top, we see a disk with unknown radius but we can easily find it because we already have the equation for our cone. We know the disk is at z=4. If we set our equations equal, we can find the intersection of the plane z=4 and the cone which is the disk we are looking for.
We see that the radius of the disk is sqrt(48). We know that r ranges from 0 to sqrt(48) so we just plug 0 as our lower limit and sqrt(48) as our upper limit for the middle integral that’s respect to r.
We are finally on the outer most integral that’s respect to theta which is the easiest indeed. It’s a full solid. When we look at it from the top, we see a whole disk. So, the lower limit is 0 and the upper limit is 2pi.
As a result, we get the same results by using spherical and cylindrical coordinates. It’s expected if you have some troubles about graphs in 3d but who doesn’t? A very good way to practice 3d graphs is using some 3d graphers such as macbook’s grapher to see what they look like and how they change when you modify the equations. You can also download softwares like Google sketch-up and Adobe Fusion 360 which are free of charge for students to get some intuition about solids.
