Question

Use cylindrical or spherical coordinates, whichever seems more appropriate.

Find the volume of the smaller wedge cut from a sphere of radius 10 by two planes that intersect along a diameter at an angle of π/3.

Answer 1

What we've been doing with the "integrate with respect to z, convert the rest to polar" is ACTUALLY working with cylindrical coordinates, but we didn't really realize it yet! So, if you can't already tell, a change of coordinates is actually VERY useful to us because nobody wants to sit there and stare all day at radicals. Now, we've already seen spherical and cylindrical coordinates in a previous chapter (chapter 13), so let's see how to properly add them to the integral stuff.

The cylindrical coordinates of a point P are (r,θ,z), where r is the distance from the origin to the point, z is the height of that point or really the distance from it to the XY plane, and if you move P down to the XY plane so it's 2D, theta is the angle it makes between the radius and positive X-axis. We will use that convenience always.

Suppose a surface E is a type 1 region whose projection D on the XY plane is conveniently described in polar coordinates, which was basically what we had been saying, only instead of a point, we have a surface. Suppose f is continuous, and E = {(x,y,z) | (x,y) in D, u1(x,y) < z < u2(x,y)}, and D is given by the polar coordinates D = {(r,θ) | α < θ < β, h1(θ) < r < h2(θ)}.

We previously learned that

∫∫∫[E] (f(x,y,z)dV) = ∫∫[D] ((((∫[u1(x,y),u2(x,y)] (f(x,y,z)dz)))))dA, and then we also learned that after we did that, we can convert the remaining double into polar coordinates. If we combine it all, we have, then

∫∫∫[E] (f(x,y,z)dV) = ∫[α,β]∫[h1(θ),h2(θ)]∫[u1(rcos(θ),rsin(θ)),u2(rcos(θ),rsin(θ))] (f(rcos(θ), rsin(θ), z)*r*dzdrdθ)

That is the entire expression for a conversion from rectangular coordinates to cylindrical coordinates by saying x = rcos(θ) and y = rsin(θ) and z = z, just like we originally learned. This is EXTREMELY useful for dealing with surfaces like paraboloids, elliptical cones, elliptical cylinders, and other such things. Basically, when you see three variables, x,y,z, ONE is at degree 1 (so x, y, or z), and BOTH the OTHERS have degree 2 (so x2, y2, z2). But then, you COULD also solve for other such things, too, in cylindrical, if you really wanted to.

So, let's do an example!

We wanna evaluate ∫[-2,2]∫[-sqrt(4-x2),sqrt(4-x2)]∫[sqrt(x2+y2),2] ((x2 + y2)dzdydx)

What a nightmare, huh? This has E = {(x,y,z) | -2 < x < 2, -sqrt(4 - x2) < y < sqrt(4 - x2), sqrt(x2 + y2) < z < 2}.

Well, projecting this onto the XY plane basically means we ignore Z and look only at X and Y. We can EASILY see that it's a circle with radius 2, so we have 0 < r < 2 and 0 < θ < 2π.

Now, in IMMEDIATE conversion to cylindricals, we should remove that sqrt(x2 + y2) in the z mess. Well, that's not hard at all. We know x2 + y2 = r2 for circles, so we have sqrt(r2) = r, so bam. We NOW have

∫[0,2π]∫[0,2]∫[r,2] (r2rdzdrdθ)

The inside becomes r3, integrating that with respect to z becomes zr3 from z = r to 2, we have 2*r3 - r*r3 = 2r3 - r4. Integrating that with respect to r, we have r4/2 - r5/5, from 0 (trivial) to 2, we have 24/2 - 25/5 = 8 - 32/5 = 8/5. Integrating with respect to θ, we get 8θ/5 from 0 (trivial) to 2π, we have 8*2π/5 = 16π/5.

NOT HARD, HUH? Even got rid of the sqrt crap in the z, unlike the last section where we brute forced it!

Now for spherical stuff! We remember spherical coordinates (ρ,θ,φ) of a point, and we said x = ρ*sin(φ)cos(θ), y = ρ*sin(φ)cos(φ), and z = ρ*cos(φ), and thus we have a solid region

E = {(ρ,θ,φ) | a < ρ < b, α < θ < β, c < φ < d}, where a > 0, β - α < 2π, and d - c < π. Originally, we divided our surface into boxes for triple integrals. Now, we divide them into "spherical wedges," which look like a cube of Jell-O if you touch the top and slide it in a direction to make a sort of "bent" cube.

So, we divide E into smaller Eijk wedges by means of equally spaced spheres ρ = ρi, half-planes θ = θj, and half-cones φ = φk, so Eijk has approximate dimensions ∆ρ, ρi∆φ, and ρisin(φk)∆θ. The first point is the "length" of this wedge in the direction of the radius of a sphere centered at the origin. The second point is the "height" of the arc in the direction of the edge of the sphere (if you do the Jell-O reference, make it "bend" to the left. The left and right sides that went from straight to curve are this point). The third point is the "width" of this also in the direction of the edge of the sphere.

The VOLUME of this blasted wedge, then, is obnoxious, even if it's just the multiplication of them all--

∆Vijk ~ (∆ρ)(ρi∆φ)(ρisin(φk)∆θ) = ρi2sin(φk)∆ρ∆θ∆φ.

In fact, with the Mean Value Theorem, we can get a more exact value, which is the exact same thing as that, only put a little ~ over the ρi and φk, and then they represent just ANY point in Eijk. If we let (xijk*,yijk*,zijk*) be the rectangular coordinates of this point, then

∫∫∫[E] (f(x,y,z)dV) = lim ∑[i=1:l]∑[j=1:m]∑[k=1:n] (f(xijk*,yijk*,zijk*)∆Vijk) as l,m,n -> ∞
= lim ∑[i=1:l]∑[j=1:m]∑[k=1:n] (f(ρisin(φk)cos(θj), ρisin(φk)sin(θj), ρicos(φk))*ρi2sin(φk)∆ρ∆θ∆φ) as l,m,n -> ∞, where ALL those ρ,φ,θ have a ~ over them except those with a ∆.

This, of course, is a Riemann sum, so we have F(ρ,θ,φ) = ρ2sin(φ)*f(ρsin(φ)cos(θ), ρsin(φ)sin(θ), ρcos(φ)).

And so, we have our conversion of a triple integral into spherical coordinates!!!

∫∫∫[E] (f(x,y,z)dV) = ∫[c,d]∫[α,β]∫[a,b] (f(ρsin(φ)cos(θ), ρsin(φ)sin(θ), ρcos(φ))ρ2sin(φ)dρdθdφ),
where E is a spherical wedge given by E = {(ρ,θ,φ) | a < ρ < b, α < θ < β, c < φ < d}

Long, isn't it? It's actually fairly simple, unless we get into ugly things involving the integrals of cosn(θ)sinm(θ) or whatever.

Let's do an example!

We wanna evaluate ∫∫∫[B] (e(x^2 + y^2 + z^2)^(3/2)dV), where B is the unit ball B = {(x,y,z) | x2 + y2 + z2 < 1}.

Hah! Wouldn't you like to do that in rectangular? Well, we know x2 + y2 + z2 = ρ2, so that whole mess simplifies instantly to eρ^3, since the 3/2*2 in the exponents cancel. And since we're a unit ball, our boundaries are 0 < rho < 1, 0 < θ < 2π, 0 < φ < π. NOTE: typically, we always take θ around a full circle and phi only half. If you sweep a FULL circle only HALF of another circle, it generates a sphere. Always. We just prefer to give theta the 2π instead of φ 'cuz, well, we're used to polar and we just give that 2π for circles. It's just a standard reference, and you will probably see it in a LOT of your triple integrals in spherical coordinates.

Anyway, we now have ∫[0,π]∫[0,2π]∫[0,1] (eρ^3ρ2sin(φ)dρdθdφ), which we can break into

∫[0,π] (sin(φ)dφ) * ∫[0,2π] (dθ) * ∫[0,1] (eρ^3ρ2dρ).

The first is simple, it's just -cos(φ), from 0 to π, we have -cos(π) - (-cos(0)) = 1 + 1 = 2.
The second is also simple, it's just θ, from 0 (trivial) to 2π, it's just 2π. So, so far we have 2*2π*stuff = 4π*stuff.

The last is ALSO simple. We say u = ρ3, du = 3ρ2dρ, so dρ = du/(3ρ2), the ρs cancel, and we have (1/3)*∫ (eudu), which is just eu, or eρ^3 from 0 to 1, so we have e1 - e0 = e - 1. Putting it all together-- (4π)*(1/3)*(e - 1) = (4π/3)(e - 1). The end.

And that's it. A quick note, though. Don't get all gung-ho hung-up on the format. If you have a paraboloid (thus you'd go for cylindrical most often) going into, say, the Y direction, you are projecting a circle onto the XZ plane. So, instead of z = z and y = rsin(θ), switch 'em! y = y, z = rsin(θ). For spherical, that MIGHT be a little tough because of how crazy it all is, but just analyze your surface first.

With that, let's do some problems and wrap this thing up!

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1) Use cylindrical coordinates to evaluate ∫∫∫[E] (xdV), where E is the surface bounded by the planes z = 0, z = x + y + 5, and the cylinders x2 + y2 = 4 and x2 + y2 = 9.

In cylindrical coordinates, we see we are bounded by z = 0, z = rcos(θ) + rsin(θ) + 5, and the cylinders r = 2 and r = 3. So, our region E is, then, {(r,θ,z) | 0 < θ < 2π, 2 < r < 3, 0 < z < rcos(θ) + rsin(θ) + 5}

Okay, we have our boundaries, and the function xdV becomes rcos(θ)rdzdrdθ which is r2cos(θ)dzdrdθ.

Integrating with z first, we get zr2cos(θ) from z = 0 (trivial) to z = rcos(θ) + rsin(θ) + 5, and so we have (rcos(θ) + rsin(θ) + 5)r2cos(θ) = r3cos2(θ) + r3sin(θ)cos(θ) + 5r2cos(θ). Integrating with respect to r, we get (r4/4)cos2(θ) + (r4/4)sin(θ)cos(θ) + (5r3/3)cos(θ), from r = 2 to 3, so we have

[(81/4)cos2(θ) + (81/4)sin(θ)cos(θ) + 45cos(θ)] - [4cos2(θ) + 4sin(θ)cos(θ) + (40/3)cos(θ)] = (65/4)cos2(θ) + (65/4)sin(θ)cos(θ) + (95/3)cos(θ). (65/4)cos2(θ) = (65/8) + (65/8)cos(2θ). (65/4)sin(θ)cos(θ) = (65/8)sin(2θ). Justification: sin(2x) = 2sin(x)cos(x). cos2(x) = (1 + cos(2x))/2. USEFUL TRIG FORMULAS.

So, we need to integrate 65/8 + (65/8)cos(2θ) + (65/8)sin(2θ) + (95/3)cos(θ). That's not bad at all! That gives us 65θ/8 + (65/16)sin(2θ) - (65/16)cos(2θ) + (95/6)sin(θ), from θ = 0 to 2π, we get

[65π/4 + 0 - 65/16 + 0] - [0 + 0 - 65/16 + 0] = 65π/4.


2) Use spherical coordinates to evaluate ∫∫∫[E] (xyzdV), where E lies between the spheres ρ = 2 and ρ = 4 and above the cone φ = π/3.

Well, lucky you, you're already told everything from the get-go! You know the boundaries of ρ, you know that phi must lie between 0 and π/3, and then θ is stuck from 0 to 2π. Then you simply convert xyz = (ρsin(φ)cos(θ))(ρsin(φ)sin(θ))(ρcos(φ)) = ρ3sin2(φ)cos(φ)sin(θ)cos(θ), and then you multiply all that by ρ2sin(φ)dρdθdφ to get

∫[0,π/3]∫[0,2π]∫[2,4] (ρ5sin3(φ)cos(φ)sin(θ)cos(θ)dρdθdφ).

Integrating with ρ, we can ignore everything (for my damn typing convenience) and get a ρ6/6 from 2 to 4, so 46/6 - 26/6 = 4096/6 - 64/6 = 4032/6 = 672. We'll pull that outside of the integrals so we have 672*answer

The next stage is integrating with respect to θ. Now we can ignore the φ stuff for now and just look at sin(θ)cos(θ) = (1/2)sin(2θ). Integrating that, we get (-1/4)cos(2θ), from 0 to 2π, we have (-1/4)cos(4π) - (-1/4)cos(0) = 0. Well, great. We have 0*something. OUR FINAL ANSWER IS ZERO. Look at that? We didn't even HAVE to do the φ. MWAHAHAHAHAHA.

But for practice, maybe we should? It's just sin3(φ)cos(φ). You say u = sin(φ), du = cos(φ)dφ, so dφ = du/cos(φ), they cancel out, and you're left with u3du, which is just u4/4 from 0 to π/3, which is just (π/3)4/4 = π4/324.


3) Use cylindrical or spherical coordinates, whichever seems more appropriate, to find the volume of the smaller wedge cut from a sphere of radius a by two planes that intersect along a diameter at an angle of π/6.

Unfortunately for me, this is spherical, meaning I have to deal with more Greek letters (which is a real pain for me). Fortunately for you, this is not a difficult problem. We'll center our sphere at (0,0,0), so that'll be nice. We know the diameter is cut at π/6, and we will SAY it is on the Z-axis, and so that means we have a cut along the XZ plane at θ = 0 and another at θ = π/6. So, we have theta's boundaries. As for φ, well, phi is stuck from 0 to π since it has to sweep in that hemisphere.

So, we have V = ∫[0,π/6]∫[0,π]∫[0,a] (ρ2sin(φ)dρdφdθ), ahh, we chose to put θ on the outside! Not that it ACTUALLY matters, since we just wind up with

∫[0,π/6] (dθ) * ∫[0,π] (sin(φ)dφ) * ∫[0,a] (ρ2dρ)

Doing the first, we just get θ, which winds up just being π/6.
Doing the second, we just get -cos(φ), which winds up being -cos(π) - (-cos(0)) = 1 + 1 = 2.
Doing the last, we get ρ3/3 from 0 to a, which just winds up being a3/3. Multiplying everything together, we get

(π/6)*2*(a3/3) = πa3/9.

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Why do you care? Well, cylindrical and spherical coordinates are great, but you probably may never actually run into a need to use them UNLESS you're molding parts together, and your parts happen to be rather cylindrical or spherical in nature (a good example is engine part manufacturing, mostly small parts or things one would find in a stereotypical factory).