Triple Integrals. (a) If is an solid (in space), what does the triple integral dv represent? Wh? (b) Suppose the shape of a solid object is described b the solid, and f(,, ) gives the densit of the object at the point (,, ) in kilograms per cubic meter. What does the triple integral f(,, ) dv represent? Wh?. Let be the solid tetrahedron bounded b the planes =, =, =, and + + 3 = 8. (The vertices of this tetrahedron are (,, ), (,, ), (6,, ), and (, 4, )). Write the triple integral f(,, ) dv as an iterated integral.,,,,, 4, 6,,
3. Let be the solid enclosed b the paraboloids = + and = 8 ( + ). (Note: The paraboloids intersect where = 4.) Write f(,, ) dv as an iterated integral in the order d d d. 4. In this problem, we ll look at the iterated integral (a) Rewrite the iterated integral in the order d d d. f(,, ) d d d. (b) Rewrite the iterated integral in the order d d d. 5. Let be the solid contained in + = 6 and ling between the planes = 3 and = 3. Sketch and write an iterated integral which epresses its volume. In which orders of integration can ou write just a single iterated integral (as opposed to a sum of iterated integrals)?
Triple Integrals in Clindrical or Spherical Coordinates. Let be the solid enclosed b the paraboloids = + and = 8 ( + ). (Note: The paraboloids intersect where = 4.) Write dv as an iterated integral in clindrical coordinates.. Find the volume of the solid ball + +. 3. Let be the solid inside both the cone = + and the sphere + + =. Write the triple integral dv as an iterated integral in spherical coordinates.
For the remaining problems, use the coordinate sstem (Cartesian, clindrical, or spherical) that seems easiest. 4. Let be the ice cream cone bounded below b = 3( + ) and above b + + = 4. Write an iterated integral which gives the volume of. (You need not evaluate.) 5. Write an iterated integral which gives the volume of the solid enclosed b = +, =, and =. (You need not evaluate.) 6. Let be the solid enclosed b = + and = 9. Rewrite the triple integral dv as an iterated integral. (You need not evaluate, but can ou guess what the answer is?) 7. The iterated integral in spherical coordinates π π/ π/ solid. Describe the solid (its shape and its densit at an point). ρ 3 sin 3 φ dρ dφ dθ computes the mass of a
Triple Integrals. (a) If is an solid (in space), what does the triple integral Solution. Remember that we are thinking of the triple integral Riemann sums, obtained from the following process:. Slice the solid into small pieces. dv represent? Wh? f(,, ) dv as a limit of. In each piece, the value of f will be approimatel constant, so multipl the value of f at an point b the volume V of the piece. (It s oka to approimate the volume V.) 3. Add up all of these products. (This is a Riemann sum.) 4. Take the limit of the Riemann sums as the volume of the pieces tends to. Now, if f is just the function f(,, ) =, then in Step, we end up simpl multipling b the volume of the piece, which gives us the volume of the piece. So, in Step 3, when we add all of these products up, we are just adding up the volume of all the small pieces, which gives the volume of the whole solid. So, dv represents the volume of the solid. (b) Suppose the shape of a solid object is described b the solid, and f(,, ) gives the densit of the object at the point (,, ) in kilograms per cubic meter. What does the triple integral f(,, ) dv represent? Wh? Solution. Following the process described in (a), in Step, we multipl the approimate densit of each piece b the volume of that piece, which gives the approimate mass of that piece. Adding those up gives the approimate mass of the entire solid object, and taking the limit gives us the eact mass of the solid object.. Let be the solid tetrahedron bounded b the planes =, =, =, and + + 3 = 8. (The vertices of this tetrahedron are (,, ), (,, ), (6,, ), and (, 4, )). Write the triple integral f(,, ) dv as an iterated integral. Solution. We ll do this in all 6 possible orders. Let s do it b writing the outer integral first, which means we think of slicing. There are three possible was to slice: parallel to the -plane, parallel to the -plane, and parallel to the -plane. (a) Slice parallel to the -plane. If we do this, we are slicing the interval [, 6] on the -ais, so the outer (single) integral will be 6 something d. To write the inner two integrals, we look at a tpical slice and describe it. When we do this, we think of as being constant (since, within a slice, is constant). Here is a tpical slice:
,,,,, 4, 6,, Each slice is a triangle, with one edge on the plane =, one edge on the plane =, and one edge on the plane + +3 = 8. (Since we are thinking of as being constant, we might rewrite this last equation as + 3 = 8.) Here s another picture of the slice, in D: 6, 3, 3 8 8, Now, we write a (double) iterated integral that describes this region. This will make up the inner two integrals of our final answer. There are two was to do this: i. If we slice verticall, we are slicing the interval [ ], 8 on the -ais, so the outer integral (of the two we are working on) will be (8 )/ something d. Each slice goes from = to the line + 3 = 8 (since we re tring to describe within a vertical slice, we ll rewrite this as = 8 3 ), so the inner integral will be iterated integral 6 (8 )/ (8 )/3 (8 )/3 f(,, ) d d d. f(,, ) d. This gives us the ii. If we slice horiontall, we are slicing the interval [ ], 6 on the -ais, so the outer integral (of the two we are working on) will be (6 )/3 3 something d. Each slice goes from = to the line + 3 = 8 (since we are tring to describe in a horiontal slice, we ll rewrite this as = 8 3 ), so the inner integral will be final answer 6 (6 )/3 (8 3)/ (b) Slice parallel to the -plane. (8 3)/ f(,, ) d d d. f(,, ) d. This gives the If we do this, we are slicing the interval [, 4] on the -ais. So, our outer (single) integral will be 4 something d. Each slice is a triangle with edges on the planes =, =, and ++3 = 8 (or + 3 = 8 ). Within a slice, is constant, so we can just look at and :
8, 3 3 8, 8, Our inner two integrals will describe this region. i. If we slice verticall, we are slicing the interval [, 8 ] on the -ais, so the outer integral (of the two we re working on) will be 8 = 8 3, which gives the iterated integral something d. Each slice goes from = to 4 8 (8 )/3 f(,, ) d d d. ii. If we slice horiontall, we are slicing the interval [, 8 ] on the -ais, so the outer integral (of the two we re working on) will be (8 )/3 = 8 3, which gives the iterated integral (c) Slice parallel to the -plane. 3 something d. Each slice goes from = to 4 (8 )/3 8 3 f(,, ) d d d. If we do this, we are slicing the interval [, ] on the -ais, so the outer (single) integral will be something d. Each slice is a triangle with edges on the planes =, =, and ++3 = 8 (or + = 8 3). Within a slice, is constant, so we can just look at and : 8 3, 8 3, 6 3, Our inner two integrals will describe this region. i. If we slice verticall, we are slicing the interval [, 6 3] on the -ais, so the outer integral (of the two we re working on) will be 6 3 something d. Each slice will go from = to the line + = 8 3 (which we write as = 8 3 since we re tring to describe ), which gives us the final integral 6 3 (8 3 )/ f(,, ) d d d. ii. If we slice horiontall, we are slicing the interval [ ], 8 3 on the -ais, so the outer integral 3
(of the two we re working on) will be (8 3)/ something d. Each slice will go from = to + = 8 3 (which we write as = 8 3 since we re tring to describe ), which gives us the answer (8 3)/ 8 3 f(,, ) d d d. 3. Let be the solid enclosed b the paraboloids = + and = 8 ( + ). (Note: The paraboloids intersect where = 4.) Write f(,, ) dv as an iterated integral in the order d d d. Solution. We can either do this b writing the inner integral first or b writing the outer integral first. In this case, it s probabl easier to write the inner integral first, but we ll show both methods. Writing the inner integral first: We are asked to have our inner integral be with respect to, so we want to describe how varies along a vertical line (where and are fied) to write the inner integral. We can see that, along an vertical line through the solid, goes from the bottom paraboloid ( = + ) to the top paraboloid ( = 8 ( + )), so the inner integral will be 8 ( + ) + f(,, ) d. To write the outer two integrals, we want to describe the projection of the region onto the plane. (A good wa to think about this is, if we moved our vertical line around to go through the whole solid, what and would we hit? Alternativel, if we could stand at the top of the -ais and look down at the solid, what region would we see?) In this case, the widest part of the solid is where the two paraboloids intersect, which is = 4 and + = 4. Therefore, the projection is the region + 4, a disk in the -plane: We want to write an iterated integral in the order d d to describe this region, which means we should slice verticall. We slice [, ] on the -ais, so the outer integral will be something d. 4
Along each slice, goes from the bottom of the circle ( = 4 ) to the top ( = 4 ), so we get the iterated integral 4 8 ( + ) Writing the outer integral first: 4 + f(,, ) d d d. We are asked to have our outer integral be with respect to, so we want to make slices parallel to the -plane. This amounts to slicing the interval [, ] on the -ais, so the outer integral will be something d. Each slice is a region bounded below b = + and above b = (8 ). (Remember that, within a slice, is constant.) Note that these curves intersect where + = (8 ), or = 8. This happens at = ± 4. At either of these -values, = 4. So, here is a picture of the region: 8 4 4 4 The two integrals describing this region are supposed to be in the order d d, which means we are slicing verticall. Slicing verticall amounts to slicing the interval [ 4, 4 ] on 4 the -ais, so the outer integral (of these two integrals) will be something d. Along 4 each vertical slice, goes from + to 8 ( + ), so we get the final iterated integral 4 8 ( + ) 4 + f(,, ) d d d. 4. In this problem, we ll look at the iterated integral (a) Rewrite the iterated integral in the order d d d. f(,, ) d d d. Solution. One strateg is to draw the solid region of integration and then write the integral as we did in #3. However, drawing the solid region of integration is rather challenging, so here s another approach. Remember that we can think of a triple integral as either a single integral of a double integral or a double integral of a single integral, and we know how to change the order of integration in a double integral. (See, for instance, #5 on the worksheet Double Integrals over General Regions.) This effectivel means that we can change the order of the inner two integrals b thinking of them together as a double integral, or we can change the order of the outer two integrals b thinking of them together as a double integral. 5
For this question, we just need to change the order of the outer two integrals, so we focus on those. The are fact that the outer integral is stuff d d. () Since this integral is d d, we should visualie the -plane. The the -ais. The fact that the inner integral is something d tells us that we are slicing the interval [, ] on stuff d tells us that each slice starts at = and goes to =. So, our region (with horiontal slices) looks like the picture on the left: To change the order of integration, we want to use vertical slices (the picture on the right). Now, we are slicing the interval [, ] on the -ais, so the outer integral will be slice has its bottom edge on = and its top edge on =, so we rewrite stuff d d. Remembering that stuff was the inner integral us the iterated integral f(,, ) d d d. (b) Rewrite the iterated integral in the order d d d. something d. Each stuff d d as f(,, ) d, this gives Solution. Let s continue from (a). As eplained there, we can change the order of the outer two integrals or of the inner two integrals. From (a), we have our iterated integral in the order d d d. If we change the order of the inner two integrals, then we ll have our iterated integral in the order d d d. If we then change the order of the outer two integrals of this, we ll get it into the order d d d. So, we reall have two steps. Step : Change the order of the inner double integral from (a). We had f(,, ) d d d, so we are going to focus on the inner double integral f(,, ) d d. Remember that, since this is the inner double integral and is the outer variable, we now think of as being a constant. () Then, the region of integration of the integral f(,, ) d d is just a rectangle (sliced horiontall): Z () Here, stuff is the inner integral f(,, ) d. () In fact, we should think of as being a constant between and, since the outer integral has going from to. 6
If we change to slicing horiontall, we rewrite this as outer integral back, we get the iterated integral Step : Change the order of the outer double integral. Now, we re working with the outer double integral, which is slices): f(,, ) d. The region of integration of f(,, ) d d. (3) Putting the f(,, ) d d d. f(,, ) d d d, and we want to change the order of stuff d d with stuff being the inner integral stuff d d looks like this (with horiontal If we change to slicing verticall, then the integral becomes that stuff was the inner integral, we get our final answer stuff d d. Remembering f(,, ) d d d. 5. Let be the solid contained in + = 6 and ling between the planes = 3 and = 3. Sketch and write an iterated integral which epresses its volume. In which orders of integration can ou write just a single iterated integral (as opposed to a sum of iterated integrals)? Solution. Here is a picture of : (4) (3) Another wa of thinking about it is that we re using Fubini s Theorem. (4) To remember how to sketch things like this, look back at #3 of the worksheet Quadric Surfaces. 7
B #(a), we know that a triple integral epressing the volume of is dv. We are asked to rewrite this as an iterated integral. Let s think about slicing the solid (which means we ll write the outer integral first). If we slice parallel to the -plane, then we are reall slicing [ 3, 3] on the -ais, and the outer integral is 3 3 something d. We use our inner two integrals to describe a tpical slice. Each slice is just the disk enclosed b the circle + = + 6, which is a circle of radius + 6: 6 6 6 6 We can slice this region verticall or horiontall; let s do it verticall. This amounts to slicing [ + 6, + 6 ] +6 on the -ais, so the outer integral is something d. Along each slice, +6 goes from the bottom of the circle ( = + 6 ) to the top of the circle ( = + 6 ). +6 So, the inner integral is f(,, ) d. +6 Putting this all together, we get the iterated integral 3 +6 +6 3 +6 +6 d d d. We are also asked in which orders we can write just a single iterated integral. Clearl, we ve done so with the order d d d. We also could have with d d d, since that would just be slicing the same disk horiontall. If we had d or d as our outer integral, then we would need to write multiple integrals. For instance, if we slice the hperboloid parallel to the -plane, some slices would look like this: 8
We would need to write a sum of integrals to describe such a slice. So, we can write a single iterated integral onl in the orders d d d and d d d. 9
Triple Integrals in Clindrical or Spherical Coordinates. Let be the solid enclosed b the paraboloids = + and = 8 ( + ). (Note: The paraboloids intersect where = 4.) Write dv as an iterated integral in clindrical coordinates. Solution. This is the same problem as #3 on the worksheet Triple Integrals, ecept that we are now given a specific integrand. It makes sense to do the problem in clindrical coordinates since the solid is smmetric about the -ais. In clindrical coordinates, the two paraboloids have equations = r and = 8 r. In addition, the integrand is equal to (r cos θ)(r sin θ). Let s write the inner integral first. If we imagine sticking vertical lines through the solid, we can see that, along an vertical line, goes from the bottom paraboloid = r to the top paraboloid = 8 r. So, our inner integral will be 8 r r (r cos θ)(r sin θ) d. To write the outer two integrals, we want to describe the projection of the solid onto the -plane. As we had figured out last time, the projection was the disk + 4. We can write an iterated integral in polar coordinates to describe this disk: the disk is r, θ < π, so our iterated integral will just be π 8 r π r (r cos θ)(r sin θ) r d dr dθ.. Find the volume of the solid ball + +. (inner integral) r dr dθ. Therefore, our final answer is Solution. Let be the ball. We know b #(a) of the worksheet Triple Integrals that the volume of is given b the triple integral dv. To compute this, we need to convert the triple integral to an iterated integral. The given ball can be described easil in spherical coordinates b the inequalities ρ, φ π, θ < π, so we can rewrite the triple integral dv as an iterated integral in spherical
coordinates π π ρ sin φ dρ dφ dθ = = = = π π π π π π ( ρ 3 3 3 sin φ φ= ρ= ρ= ) sin φ dφ dθ ( ) φ=π 3 cos φ dθ 3 dθ dφ dθ = 4 3 π 3. Let be the solid inside both the cone = + and the sphere + + =. Write the triple integral dv as an iterated integral in spherical coordinates. Solution. Here is a picture of the solid: We have to write both the integrand () and the solid of integration in spherical coordinates. We know that in Cartesian coordinates is the same as ρ cos φ in spherical coordinates, so the function we re integrating is ρ cos φ. The cone = + is the same as φ = π 4 in spherical coordinates.() The sphere + + = is ρ = in spherical coordinates. So, the solid can be described in spherical coordinates as ρ, φ π 4, θ π. This means that the iterated integral is π π/4 (ρ cos φ)ρ sin φ dρ dφ dθ. For the remaining problems, use the coordinate sstem (Cartesian, clindrical, or spherical) that seems easiest. 4. Let be the ice cream cone bounded below b = 3( + ) and above b + + = 4. Write an iterated integral which gives the volume of. (You need not evaluate.) () Wh? We could first rewrite = p + in clindrical coordinates: it s = r. In terms of spherical coordinates, this sas that ρ cos φ = ρ sin φ, so cos φ = sin φ. That s the same as saing that tan φ =, or φ = π 4.
Solution. We know b #(a) of the worksheet Triple Integrals that the volume of is given b the triple integral dv. The solid has a simple description in spherical coordinates, so we will use spherical coordinates to rewrite the triple integral as an iterated integral. The sphere + + = 4 is the same as ρ =. The cone = 3( + ) can be written as φ = π 6.() So, the volume is π π/6 ρ sin φ dρ dφ dθ. 5. Write an iterated integral which gives the volume of the solid enclosed b = +, =, and =. (You need not evaluate.) Solution. We know b #(a) of the worksheet Triple Integrals that the volume of is given b the triple integral dv. To compute this, we need to convert the triple integral to an iterated integral. Since the solid is smmetric about the -ais but doesn t seem to have a simple description in terms of spherical coordinates, we ll use clindrical coordinates. Let s think of slicing the solid, using slices parallel to the -plane. This means we ll write the outer integral first. We re slicing [, ] on the -ais, so our outer integral will be something d. To write the inner double integral, we want to describe each slice (and, within a slice, we can think of as being a constant). Each slice is just the disk enclosed b the circle + =, which is a circle of radius : () This is true because = p 3( + ) can be written in clindrical coordinates as = r 3. In terms of spherical coordinates, this sas that ρ cos φ = 3ρ sin φ. That s the same as saing tan φ = 3, or φ = π 6. 3
We ll use polar coordinates to write the iterated (double) integral describing this slice. The circle can be described as θ < π and r (and remember that we are still thinking of as a constant), so the appropriate integral is π r dr dθ. Putting this into our outer integral, we get the iterated integral π r dr dθ d. Note: For this problem, writing the inner integral first doesn t work as well, at least not if we want to write the integral with d as the inner integral. Wh? Well, if we tr to write the integral with d as the inner integral, we imagine sticking vertical lines through the solid. The problem is that there are different tpes of vertical lines. For instance, along the red line in the picture below, goes from the cone ( = + or = r) to = (in the solid). But, along the blue line, goes from = to =. So, we d have to write two separate integrals to deal with these two different situations. 6. Let be the solid enclosed b = + and = 9. Rewrite the triple integral iterated integral. (You need not evaluate, but can ou guess what the answer is?) dv as an Solution. = + describes a paraboloid, so the solid looks like this: Since the solid is smmetric about the -ais, a good guess is that clindrical coordinates will make things easier. In clindrical coordinates, the integrand is equal to r cos θ. 4
Let s think of slicing the solid, which means we ll write our outer integral first. If we slice parallel to the -plane, then we are slicing the interval [, 9] on the -ais, so our outer integral is 9 something d. We use the inner two integrals to describe a tpical slice; within a slice, is constant. Each slice is a disk enclosed b the circle + = (which has radius ). We know that we can describe this in polar coordinates as r, θ < π. So, the inner two integrals will be Therefore, the given triple integral is equal to the iterated integral 9 π 9 ( π r cos θ r dr dθ d = 3 r3 cos θ = = 9 π 9 = π r= r= ) 3 3/ cos θ dθ d ) θ=π 3 3/ sin θ d ( θ= dr dθ d (r cos θ) r dr dθ. That the answer is should not be surprising because the integrand f(,, ) = is anti-smmetric about the plane = (this is sort of like saing the function is odd: f(,, ) = f(,, )), but the solid is smmetric about the plane =. Note: If ou decided to do the inner integral first, ou probabl ended up with d as our inner integral. In this case, a valid iterated integral is 7. The iterated integral in spherical coordinates π 3 9 π π/ r r cos θ r d dr dθ. π/ solid. Describe the solid (its shape and its densit at an point). ρ 3 sin 3 φ dρ dφ dθ computes the mass of a Solution. The shape of the solid is described b the region of integration. We can read this off from the bounds of integration: it is π θ π, φ π, ρ. We can visualie ρ b imagining a solid ball of radius with a solid ball of radius taken out of the middle. φ π tells us we ll onl have the top half of that, and π θ π tells us that we ll onl be looking at one octant: the one with negative and positive: To figure out the densit, remember that we get mass b integrating the densit. If we call this solid, then the iterated integral in the problem is the same as the triple integral ρ sin φ dv since dv is ρ sin φ dρ dφ dθ. So, the densit of the solid at a point (ρ, φ, θ) is ρ sin φ. 5