dxdydz

We drilled out a cylindrical piece out of our sphere with radius 2 for the ring example with spherical coordinates. Here, we have a very similar sphere but we don’t drill the cylindrical piece from the centre. Still, the cylinder has a radius of 1 in the sphere but our drill is offset. We need to calculate the volume of this drilled sphere.

We might consider spherical coordinates again to calculate the volume but it’s actually easier to calculate it with cylindrical coordinates. It might seem that it’s an excuse out of thin air to use cylindrical coordinates but, look, when we view our solid from top, it’s screaming that it’s very suitable for polar coordinates. Then, we need to consider the spherical surfaces on top and bottom but we also know that we can express these surfaces in cylindrical coordinates easily.

We got the idea what the solid looks like in general but let’s spend few more seconds to think if we can ease our burden of calculation. The half of the sphere is literally untouched. There’s no drilled out in the first half. So, we actually know the volume of a hemisphere. The drilled half is the one we need to deal with. If we calculate the drilled half and add the hemisphere’s volume, we get the volume of our solid.

Let’s start setting up our triple integrals with respect to z first. Probably, it’s the most straight forward one because when we send a beam of light that’s parallel to z-axis, we hit the lower hemisphere first. We travel through our solid and leave it by touching the upper hemisphere. Whether I send my beam of light close to the centre or edges, I always hit these two spherical surfaces. With one exception, you might argue. We don’t hit anything in the drilled cylindrical section. It’s true and it’s what we are going to subtract in the next integral.

So, we got the lower hemisphere for our lower limit for our inner most integral with respect to z. We know that we can express a sphere as r2=x2+y2+z2.In this case, the radius is 2, so we have 4=x2+y2+z2 We solve for z,our integral is respect to z remember, and we get z=±sqrt(4-x2-y2). It’s an expression from heaven to convert into polar coordinates because x2+y2=r2 as we know. So, we just get the ±sqrt(4-r2) as our limits of integration with respect to z. We are done with our first integral. Now, it’s time to deal with area that we are integration over.

We need to express these two circles in terms of polar coordinates. The big one from the sphere is easy because it’s just a circle with radius 2 and centered at the origin. So, it’s just r=2. It’s now time to take out the old but good polar coordinates from our arsenal. We have a circle with radius 1 and its centre is at x=1. If we wanted to express this circle in cartesian coordinates, we would just indicate it 12=(x-1)2+y2. It’s r= 2cosθ. If you don’t really recall where it comes from or choose not to believe in me, we can multiply both sides by r and substitute x2+y2 in r2 on the left hand side. We have 2rcosθ on the right and we know rcosθ is just x in polar coordinates. If I rearrange my equation as y2+x2-2x=0 and complete the square, I get the 12=(x-1)2+y2. As a result, we express our circle with radius 2 as r=2 and 12=(x-1)2+y2 one as 2cosθ.

When we send a beam of light from the origin, we hit these two circles. It’s also very important to see that this occurs over [pi/2,-pi/2]! When our light ray hits 3pi/2 for example, we never touch our little circle. We are also interested in the half of the circle because we already know the volume of the undrilled half.

We are actually done with our middle integral with respect to r. We know that our upper limit is r=2 and lower limit is r=2cosθ because our light rays first hit the little circle and leave the area by touching the circle w/r=2. It’s also saying that we cover the half of the big circle and subtract the little one because this is what we are trying to do. We drilled out that piece and we need to subtract the solid over 2cosθ.

Our last step is to find the limits of the outer most integral with respect to θ. We are calculating half of the object. When we look from top, we see that our θ is ranging between pi/2 to -pi/2. We are done with our setting up our triple integral!

We are going to solve the ring problem. Imagine we have a sphere with radius 2. Inside of our sphere, we have a cylinder with radius 1. We take out the cylindrical piece, and we are left with an object like a ring. We can imagine it like we have a drill on top of our sphere, and we drill out through the center. We need to find the volume of this object.

Let’s use the spherical coordinates for this problem because we have a sphere in the end! Before writing our triple integral in spherical coordinates, let’s examine the object a bit. We know that we have two distinct surfaces for this object. We have a cylindrical surface as a result of our drilling out operation. We have also a surface from our sphere with radius 2. We have a full object, meaning we don’t have a half or quarter of the object.

We have three integrals, so we need to find limits for each one. Let’s start with our inner most integral that’s respect to rho. When we deal with rho, imagining a light bulb in the origin can help us. Let’s imagine that our object, ring, is transparent. If we turn the light bulb on in the origin, what surface does it hit first? It would hit the cylindrical surface first because we have nothing before the cylindrical surface but empty space that’s coming from our drill. Alright, now our light rays penetrate through our solid. It hit the cylindrical surface first as we know it. Now, light rays from the light bulb in the origin traveled through the solid, and they are about to leave. What surface does it touch right before leaving the solid? It’s the spherical surface. My light rays hit the cylindrical surface first and left the solid by touching the spherical surface.

We actually identified the limits of the inner most integral that’s respect to rho. We need to write it in spherical coordinates though. Anything we have in spherical coordinates, limits and integrand, must be in terms of spherical coordinates! First, we hit cylindrical surface with radius 1. How can I describe it in terms of spherical coordinates? It’s 1 csc(phi) because our radius is 1 for the cylinder. So, what about the csc(phi) part? If we say rho=csc(phi), I can multiply both sides by sin(phi) to cancel out csc(phi). I got 1 on the right side, and sin(phi)rho on the left side which equal r in polar coordinates. r=1 represent a cylinder with radius 1 in cylindrical coordinates,the 3d version of polar coordinates. Or you can just accept it as acsc(phi) represents a cylindrical surface with radius a in spherical coordinates. My upper limit is quite easy. It is our spherical surface from our sphere with radius 2. So, the sphere with radius 2 is just rho=2. So my lower limit is csc(phi) and my upper limit is just 2 for the inner most integral that’s respect to rho. We are done for our first integral!

For my middle integral that’s respect to phi, I need to know the angle interval that my solid exists. We can think it as radar. It can start from 0 to pi, meaning that I can make 0 degree to z-axis. For this instance, I am actually touching my z axis. Then, I can let my radar cover starting from the top point on z-axis,0, to the bottom point,phi. I’d make a total coverage of phi for a full sphere. However, we drilled out the cylindrical part in our sphere. We don’t have a full sphere. Our coverage starts from an angle,phi, we cover our solid. We stop our coverage where the solid ends. We need to find this angle. It’s not hard.We’ll use trigonometry for finding the angles. Let’s imagine out solid in z-y or z-x plane, meaning we look from the sides. I see a circle with radius 2(my sphere) and I see two straight lines(from my cylinder with radius 1).I think one of the most challenging/confusing concepts in spherical coordinates is the presence of phi. I heard that spherical coordinates are confusing or hard from my friends but it’s because we are so used to work in cartesian coordinates. We spend a couple weeks to adapt into a total different coordinates system so, it’s very normal to be confused at first. 

We can plug in out x value, 1, in the equation. We ended up with square root of 3. I know the vertical and horizontal components of the triangle. Thus, we know that tangent of square root 3 is the angle in my triangle. It’s not phi though!!! Phi is the angle between my rho and z-axis. I got pi/3 for my angle in the triangle. So, phi is pi/6. This is the angle where we begin to cover the solid. It’s our lower limit. This is a symmetric solid. So, the angle is 5pi/6 which is my upper limit. So, in the interval of pi/6 and 5pi/6, I cover my solid. These are my limit for my middle integral that’s respect to phi. So, my lower limit is pi/6 because I first begin covering the solid at this angle and my upper limit is 5pi/6 where I completely cover the solid. We are done with our middle integral, too.

Finally, the last and sweetest one, the most outer integral that’s respect to theta. It’s a full solid, not a half or quarter one. Our limits will be 0 and 2pi for the last integral.

We can finally start solving our triple integral.

It’s always good to check our answers by using other methods. What method can we use for this problem rather than another triple integral? The old and good washer method can be a good fit. We have circles with different radii stacked up on top of each other along the y-axis. I know my range for my y-axis from the previous method which is from negative square root of 3 to positive square root of 3. I’ll subtract the circle inside the bigger circle. I end up with the same answer. It’s always good to check your answers because a small mistake in triple integrals can give you a different result because it’s a long process and can be tricky.

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.

Cylindrical coordinates:

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.