#triple integrals

[semiaestheticmathematics]

7/13/17 - Saw another application of the Lagrange multiplier method. Reviewed double integrals with region specifications. Ran into triple integrals in the context of finding volume (will review this more tomorrow).

[1dxdydz]

The ring w/Spherical Coordinates

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.

[clarinette-lion]

me: *solves a triple integral*

me: "Hah! Awesome!"

me: *puts it into a comic*

[mathwork]

[homemaderaspberryjam]

I hate solving these awful triple integrals just to have the answer be 0. Very unsatisfying. 

[ilovepieandothersuchthings-blog]

so..tiiiiiiiiiiirrrrrreeed

you know what sucks? college...srsly..i have balloooooooooned like a whale. WHAM now my pants are like...uber tight and i get wedgies all the time. WHAT IS THIS?!??!!... le sigh..to the gym next quarter, i guess

ps. triple integrals suck hella butt

[abysso-pelagic]

DID SHE JUST SAY TRIPLE INTEGRALS...?! fuuuuuuuuuu