Geometry · G.GMD.B.4

Solids of revolution

A vase, a bowl, a table leg and a ball are all round in the same way. Is there one flat shape that describes each of them completely?

Where you are 8 in Solids

Geometry11

Solids8

Three-dimensional shapes7 Volume of solids7 Cross sections of solids Nets of solids ▾ Show all 8 (3 more)

Why this exists

A potter’s wheel and a woodworker’s lathe make round things in the same way. The material spins about a fixed line, and the tool, held at one place, cuts the same shape all the way round. The finished vase or table leg is described completely by one flat outline: the shape of a single slice taken through its middle.

That is a large saving. A solid is hard to draw, hard to measure and hard to describe in words. A flat region and a line are easy. If every round solid comes from spinning a flat region about a line, then questions about the solid (what shape is it, what does a slice look like, how much does it hold) can be answered by looking at the flat region instead. The cylinder, the cone and the sphere all turn out to be made this way, and so do many solids that have no name of their own.

Definition

Start with a flat region R and a straight line \ell lying in the same plane, with R entirely on one side of \ell. The region may touch \ell, but it may not cross it.

Turn R one full turn about \ell, the way a door turns on its hinges. Every point P of R travels round a circle. The circle lies in the plane through P perpendicular to \ell, its center is the point of \ell nearest to P, and its radius is the distance d from P to \ell.

Figure 1As the region turns about the axis, the point P travels round a circle. Its center is on the axis and its radius is d, the distance from P to the axis.

The solid of revolution swept out by R about \ell is the set of every point on every one of these circles. The line \ell is the axis of revolution. A point of R that lies on the axis has distance 0 from it, so it does not move at all: its circle is a single point.

In examples the region is usually given by its corners or by an inequality in the coordinate plane, and the axis is usually the x-axis, the y-axis, or a line such as x = 1 parallel to one of them.

Intuition

Picture the region as a flag on a flagpole and spin the pole. Each edge of the flag paints a surface as it goes round, and the cloth fills the space between those surfaces.

An edge on the axis stays put and becomes the solid’s center line.
An edge perpendicular to the axis sweeps a flat disk, or a flat ring if it does not reach the axis. These are the solid’s flat ends.
An edge parallel to the axis, at distance r from it, sweeps the curved side of a cylinder of radius r.
A slanted edge sweeps the curved side of a cone, or of a cone with its top cut off.
A semicircle whose center is on the axis sweeps a sphere, because every point of it is the same distance from that center and turning never changes that distance.

Figure 2A rectangle sweeps a cylinder, a right triangle sweeps a cone, and a half-disk sweeps a sphere. Each solid is outlined by its region and the region’s mirror image across the axis.

Cut the solid by a plane through its axis and the cut is always the region together with its mirror image across the axis. That gives a quick way to sketch any solid of revolution: draw the region, reflect it in the axis, and join matching corners with ellipses for the circles they sweep.

Two things decide the solid: the region and the axis. The same rectangle spun about two different sides gives two different cylinders, and the same rectangle spun about a line that does not touch it gives a tube with a hole down the middle.

Results

Every point travels round a circle. A point at distance d from the axis sweeps a circle of radius d, centered on the axis, in the plane through the point perpendicular to the axis.

Slices across the axis are circles. Cut a solid of revolution with a plane perpendicular to its axis. The cut is a disk when the region reaches the axis at that level, and a ring (the flat band between two circles with the same center, also called an annulus) when it does not. If the region meets that level in several separate pieces, the cut is several rings with the same center. A disk of radius r has area \pi r^2, and a ring with outer radius R and inner radius r has area \pi R^2 - \pi r^2.

Slices through the axis are the region twice. Cut a solid of revolution with a plane that contains its axis. The cut is a copy of the region together with its mirror image across the axis.

The standard solids. For a region against the y-axis, or any axis:

Region	Placed against the axis	Solid
rectangle	one side on the axis	cylinder
right triangle	one leg on the axis	cone
half-disk	its diameter on the axis	sphere
quarter disk	one straight edge on the axis	hemisphere
right trapezoid	the leg perpendicular to the parallel sides on the axis	frustum
rectangle	one side parallel to the axis, not touching it	tube
disk	not touching the axis	torus (a ring doughnut)

For the cylinder, the side along the axis is the height and the side across is the radius. For the cone, the leg on the axis is the height and the other leg is the radius. The volumes then come from the formulas already known: V = \pi r^2 h for a cylinder, V = \tfrac{1}{3}\pi r^2 h for a cone and V = \tfrac{4}{3}\pi r^3 for a sphere. A region built from these pieces gives a solid whose volume is the sum, or the difference, of theirs.

Symmetry. A solid of revolution is carried onto itself by a turn of any angle about its axis. This is rotational symmetry of the strongest kind: not for a few special angles but for all of them.

Why it is true

The circle. Turning about \ell is a rigid motion that leaves every point of \ell where it is, so it keeps every distance to \ell the same. It also keeps each plane perpendicular to \ell where it is, moving points around inside that plane. Inside such a plane the motion is just a rotation about a point, the point where \ell crosses the plane. A point P at distance d from that center is carried round the circle of radius d about it, and a full turn carries it all the way round.

Slices across the axis. Fix a plane perpendicular to \ell. Since points never leave their own perpendicular plane, the only points of the solid in this plane are swept by the points of R that were in it to begin with. Those points form a piece of a line in the plane of R, running straight out from the axis, so their distances from the axis fill an interval from some smallest value a to some largest value b. Each of them sweeps a circle about the same center, with every radius from a to b turning up, so together they fill the ring between radius a and radius b. When a = 0 the region touches the axis at that level and the ring has no hole: it is a disk. When R meets the plane in several pieces, each piece gives its own ring.

Slices through the axis. A plane containing \ell meets each circle of radius d > 0 about \ell in exactly two points, on opposite sides of \ell and both at distance d from it. Turn the solid so that R lies in that plane. One of the two points on each circle is then a point of R, and the other is its reflection across \ell. So the cut is R together with its mirror image.

The standard solids. Take a rectangle with one side, of length h, on the axis, and the other side of length r. Every slice across the axis between its ends is a disk of radius r, which is exactly a cylinder of radius r and height h. For a right triangle with one leg on the axis, the slanted side runs from a point on the axis to the far corner, and each position it turns to is a segment from that point to the circle swept by the far corner. A solid made of all the segments from one point to a circular disk is a cone. For a half-disk of radius r with its diameter on the axis, every point of the semicircle is at distance r from the center, which is on the axis and never moves, so the semicircle sweeps the sphere of radius r and the inside of the half-disk sweeps the inside of the sphere.

Worked examples

Example 1

One rectangle, two axes

The rectangle with vertices (0, 0), (3, 0), (3, 5) and (0, 5) is spun one full turn, first about the y-axis and then about the x-axis. Name each solid and find its volume.

Solution

About the y-axis, the side from (0, 0) to (0, 5) lies on the axis, so the solid is a cylinder. That side, of length 5, is the height. The side across, from (0, 0) to (3, 0), is swept into the base, so the radius is 3:

V = \pi r^2 h = \pi (3)^2 (5) = 45\pi \approx 141.4

About the x-axis, the side from (0, 0) to (3, 0) lies on the axis. Now the height is 3 and the radius is 5:

V = \pi (5)^2 (3) = 75\pi \approx 235.6

The same region gives two different cylinders. The volume is larger about the x-axis because the longer side has become the radius, and the radius is squared while the height is not.

Example 2

A triangle about each leg

The triangle with vertices (0, 0), (6, 0) and (0, 4) is spun about the y-axis. Name the solid, find its volume, and describe its slice at height y = 2 and its slice by a plane containing the axis. Then find the volume when the same triangle is spun about the x-axis.

Solution

The leg from (0, 0) to (0, 4) lies on the y-axis, so the solid is a cone. That leg is the height, h = 4, and the leg along the x-axis is swept into the base, so r = 6. The apex is (0, 4).

V = \tfrac{1}{3}\pi r^2 h = \tfrac{1}{3}\pi (6)^2 (4) = 48\pi

The slanted side joins (6, 0) to (0, 4). It drops 6 in x while y rises 4, so at height y it is at x = 6 - \tfrac{3}{2}y. At y = 2 the region runs from the axis out to x = 6 - 3 = 3. It touches the axis, so the slice is a disk of radius 3, with area 9\pi.

A plane containing the axis cuts out the triangle and its mirror image across the y-axis: the triangle with vertices (-6, 0), (6, 0) and (0, 4). It has base 12 and height 4, so its area is \tfrac{1}{2}(12)(4) = 24.

About the x-axis the roles of the legs swap: the height is 6 and the radius is 4.

V = \tfrac{1}{3}\pi (4)^2 (6) = 32\pi

Example 3

A half-disk and its slices

The half-disk x \ge 0, x^2 + y^2 \le 25 is spun about the y-axis. Name the solid and find its volume. Then describe the slice perpendicular to the axis at height y = 3, and the slice by a plane containing the axis.

Solution

The straight edge of the half-disk is the diameter from (0, -5) to (0, 5), which lies on the axis. Its curved edge is a semicircle of radius 5 centered at the origin, which is on the axis. So the solid is a sphere of radius 5:

V = \tfrac{4}{3}\pi r^3 = \tfrac{4}{3}\pi (5)^3 = \tfrac{500\pi}{3}

At height y = 3 the half-disk runs from the axis out to the point where x^2 + 3^2 = 25, which is x = 4. The region touches the axis there, so the slice is a disk of radius 4, with area 16\pi. The radius comes from the Pythagorean theorem: the point (4, 3) is 5 from the center.

The slice through the axis is the half-disk together with its mirror image, which is the whole disk of radius 5, with area 25\pi. Every slice of a solid sphere through its center is a disk of the sphere’s own radius.

Example 4

A gap between the region and the axis

The triangle with vertices (0, 0), (4, 0) and (4, 6) is spun about the y-axis. Describe the solid, find its volume, and find the area of its slice at height y = 3.

Figure 3The triangle never reaches the axis except at its lowest corner, so the solid is a cylinder with a cone-shaped hollow.

Solution

This triangle touches the y-axis only at the origin. Its vertical side, from (4, 0) to (4, 6), is parallel to the axis at distance 4, so it sweeps the side of a cylinder of radius 4 and height 6. Its slanted side runs from (0, 0) up to (4, 6), so it sweeps a cone with its apex at the origin, opening upward, with radius 4 at the top. The region lies between the two, so the solid is the cylinder with the cone carved out of it:

\begin{aligned} V &= \pi (4)^2 (6) - \tfrac{1}{3}\pi (4)^2 (6) \\ &= 96\pi - 32\pi = 64\pi \end{aligned}

The slanted side rises 6 while x goes out 4, so at height y it is at x = \tfrac{2}{3}y. At y = 3 the region runs from x = 2 to x = 4. It does not reach the axis, so the slice is a ring with inner radius 2 and outer radius 4:

\pi (4)^2 - \pi (2)^2 = 16\pi - 4\pi = 12\pi

The ring tells the same story as the volume: the hole at each height is the slice of the cone, and the outer circle is the slice of the cylinder.

Example 5

A trapezoid and the frustum

The trapezoid with vertices (0, 0), (6, 0), (3, 4) and (0, 4) is spun about the y-axis. Find the volume of the solid.

Figure 4A frustum is a cone with its top cut off. Continuing the slanted side to the axis finds the whole cone.

Solution

The side from (0, 0) to (0, 4) is on the axis, the bottom edge sweeps a disk of radius 6, the top edge sweeps a disk of radius 3, and the slanted side sweeps a surface that narrows from one to the other. The solid is a frustum: a cone with its top cut off by a plane parallel to its base.

No new formula is needed. Continue the slanted side until it meets the axis. It goes in 3 as it rises 4, so at height y it is at x = 6 - \tfrac{3}{4}y, and it reaches x = 0 when y = 8. The whole cone has radius 6 and height 8. The part cut off is a smaller cone above y = 4, with radius 3 and height 8 - 4 = 4. So

\begin{aligned} V &= \tfrac{1}{3}\pi (6)^2 (8) - \tfrac{1}{3}\pi (3)^2 (4) \\ &= 96\pi - 12\pi = 84\pi \end{aligned}

Check. A frustum with height h and end radii R and r has volume \tfrac{1}{3}\pi h (R^2 + Rr + r^2), and here that is \tfrac{1}{3}\pi (4)(36 + 18 + 9), which is \tfrac{4}{3}\pi(63) = 84\pi.

Example 6

An axis that is not a coordinate axis

The rectangle with vertices (3, 1), (5, 1), (5, 4) and (3, 4) is spun about the line x = 1. Describe the solid and find its volume.

Figure 5A rectangle held away from the axis sweeps a tube. Every slice across the axis is the same ring.

Solution

What matters is distance from the axis, not the coordinates themselves. The axis is the vertical line x = 1. The rectangle’s left side, at x = 3, is 3 - 1 = 2 from it, and its right side, at x = 5, is 5 - 1 = 4 from it. The height runs from y = 1 to y = 4, which is 3.

Both sides are parallel to the axis and neither touches it, so the solid is a tube: a cylinder of radius 4 with a cylinder of radius 2 removed from its middle, both of height 3.

\begin{aligned} V &= \pi (4)^2 (3) - \pi (2)^2 (3) \\ &= 48\pi - 12\pi = 36\pi \end{aligned}

There is a second way to see this. Every slice across the axis, at any height from 1 to 4, is the same ring, with area \pi(4^2 - 2^2) = 12\pi. A solid whose slices all have area 12\pi over a height of 3 has volume 12\pi \times 3 = 36\pi, by the same stacking argument behind the cylinder formula and Cavalieri’s principle.

Example 7

A right triangle about its hypotenuse

A right triangle has legs 3 and 4 and hypotenuse 5. Find the volume of the solid swept when it is spun about each leg in turn, and when it is spun about its hypotenuse.

Figure 6Spun about its hypotenuse, a right triangle sweeps two cones that share one base.

Solution

About the leg of length 4, the solid is a cone with height 4 and radius 3: \tfrac{1}{3}\pi(3)^2(4) = 12\pi. About the leg of length 3, the height is 3 and the radius is 4: \tfrac{1}{3}\pi(4)^2(3) = 16\pi.

About the hypotenuse the triangle touches the axis only along that side, and its right-angle corner is the point farthest out. Drop a perpendicular from that corner to the hypotenuse. It splits the triangle into two right triangles, each with a leg on the axis, so the solid is two cones glued base to base, and the shared base is the circle swept by the corner.

The common radius is the length of that perpendicular, the height of the triangle measured from the hypotenuse. The area of the triangle is \tfrac{1}{2}(3)(4) = 6, and it is also \tfrac{1}{2}(5)h, so

\tfrac{1}{2}(5)h = 6 \qquad\Longrightarrow\qquad h = \tfrac{12}{5}

The two cones have this radius and heights that add to the whole hypotenuse, 5. Adding their volumes, the radius part is shared:

\begin{aligned} V &= \tfrac{1}{3}\pi \left(\tfrac{12}{5}\right)^2 (5) \\ &= \tfrac{1}{3}\pi \cdot \tfrac{144}{25} \cdot 5 = \tfrac{48\pi}{5} \end{aligned}

That is 9.6\pi, smaller than either of the others. The hypotenuse is the longest side, so it makes the tallest solid, but the radius, which is squared, is the shortest of the three choices. The heights of the two cones themselves are \tfrac{9}{5} and \tfrac{16}{5}, though the volume never needed them separately.

Boundaries

The axis lies in the plane of the region. Spin a disk about the line through its center perpendicular to it, and the disk only turns within its own plane: it sweeps nothing new. The definition needs the axis and the region in one plane.

The region stays on one side of the axis. If it crosses, the two parts sweep solids that overlap. The rectangle from x = -1 to x = 2 and y = 0 to y = 3, spun about the y-axis, sweeps a cylinder of radius 2, and the part on the left adds nothing, because it lies inside what the part on the right already sweeps. Adding the two volumes counts the overlap twice.

A full turn. Turning a region only part of the way round sweeps a wedge of the solid, not the whole of it, and none of the shapes in the table is formed.

A flat region, not a curve. Spinning a curve with no area inside it sweeps a surface rather than a solid: a segment parallel to the axis sweeps the side of a cylinder and nothing inside it. The area of such a surface is its own subject, the surface area of a solid of revolution.

The formulas need the standard pieces. The volume can be read off with the cylinder, cone and sphere formulas only when the region is built from rectangles, right triangles, right trapezoids and parts of disks set against the axis. A region bounded by a curve such as y = x^2 still sweeps a perfectly good solid, but finding its volume needs the disk method. The torus in the table is another solid whose volume no formula in this lesson gives.

Where this goes next

The slices across the axis are what make solids of revolution so workable. Every slice is a disk or a ring, and a disk’s area is \pi r^2. Slice a solid into many thin disks, add up their volumes, and let the slices get thinner: that is the disk method of calculus, and with a hole down the middle it becomes the washer method. They find the volume of any solid of revolution, whatever curve bounds the region, and they agree with the cylinder, cone and sphere formulas used here.

Comparing slices also proves things. The solid of Example 4, built with its height equal to its radius, has at each height a ring of exactly the same area as the slice of a hemisphere of that radius, resting on its flat face, at the same height. By Cavalieri’s principle the two have the same volume, and that is one of the classic ways to find the volume of a sphere, set out in why the volume formulas hold.

Spinning a curve rather than a region gives a surface, and its area is the surface area of a solid of revolution. Spinning a whole line that crosses the axis sweeps a double cone, and slicing that at different angles gives the circle, ellipse, parabola and hyperbola: conic sections from a double cone.

And the lathe from the start of this lesson is a real design tool. A bottle, a bowl or a chair leg can be described by its outline and its axis, and its volume and weight follow from that outline: modeling with solids.

Exercises

Naming the solid

Each region is spun one full turn about the line given. Name the solid, give its dimensions, and find its exact volume.

The rectangle with vertices (0, 0), (2, 0), (2, 7) and (0, 7), about the y-axis.
The triangle with vertices (0, 0), (3, 0) and (0, 5), about the y-axis.
The half-disk x \ge 0, x^2 + y^2 \le 16, about the y-axis.
The quarter disk x \ge 0, y \ge 0, x^2 + y^2 \le 25, about the y-axis.
The rectangle with vertices (1, 0), (3, 0), (3, 2) and (1, 2), about the y-axis.
The disk of radius 1 centered at (3, 0), about the y-axis. Name and describe the solid; no volume is needed.

Finding the region

Describe a region and an axis that sweep out a cylinder of radius 6 and height 2.
Describe a region and an axis that sweep out a cone of radius 2 and height 9.
A capsule is a cylinder of radius 2 and length 6 with a hemisphere of radius 2 on each end. Describe a region and an axis that sweep it out, and find its volume.
A pipe is 10 long, with outer radius 5 and inner radius 4. Describe a region and an axis that sweep it out, and find the volume of the material in the pipe.

Volumes

Find the exact volume of each solid.

The rectangle with vertices (0, 0), (3, 0), (3, 8) and (0, 8), spun about the x-axis.
The triangle with vertices (0, 0), (5, 0) and (0, 12), spun about the x-axis, and then about the y-axis.
The rectangle with vertices (0, 0), (3, 0), (3, 10) and (0, 10), with the quarter disk x \ge 0, y \ge 10, x^2 + (y - 10)^2 \le 9 sitting on top of it, spun about the y-axis.
The triangle with vertices (0, 0), (3, 0) and (3, 9), spun about the y-axis.
The triangle of Exercise 14, spun about the x-axis.
The trapezoid with vertices (0, 0), (4, 0), (2, 6) and (0, 6), spun about the y-axis.
The L-shaped region made of the rectangle with vertices (0, 0), (4, 0), (4, 1), (0, 1) and the rectangle with vertices (0, 1), (1, 1), (1, 5), (0, 5), spun about the y-axis.
The square with vertices (1, 0), (3, 0), (3, 2) and (1, 2), spun about the line x = 1, and then about the line x = -1.

Slices

The half-disk x \ge 0, x^2 + y^2 \le 36 is spun about the y-axis. Describe the slice by the plane perpendicular to the axis at height y = 4, and find its area.
The solid of Exercise 14 is cut by the plane perpendicular to the axis at height y = 6. Describe the slice and find its area.
The cone of Exercise 2 is cut by a plane that contains its axis. Describe the slice and find its area.

Reasoning

Explain why every slice of a solid of revolution by a plane perpendicular to its axis is made of circles: a disk, a ring, or several rings with one center. Why can such a slice never be a square?
A right triangle has legs a and b. Show that the solid swept by spinning it about leg a has \tfrac{b}{a} times the volume of the solid swept by spinning it about leg b. Which leg gives the larger solid?
The rectangle with vertices (-1, 0), (2, 0), (2, 3) and (-1, 3) crosses the y-axis. Explain why spinning it about the y-axis gives a cylinder of radius 2, find its volume, and explain why adding the volumes swept by the parts on either side of the axis gives the wrong answer.
Is a cube a solid of revolution? Explain.

Answers

A cylinder with radius 2 and height 7, and V = \pi (2)^2 (7) = 28\pi. The side on the axis is the height and the side across is the radius.
A cone with radius 3, height 5 and apex (0, 5), and V = \tfrac{1}{3}\pi (3)^2 (5) = 15\pi.
A sphere of radius 4, and V = \tfrac{4}{3}\pi (4)^3 = \tfrac{256\pi}{3}. The diameter lies on the axis and the semicircle is centered at the origin.
A hemisphere of radius 5, flat face down on the plane y = 0, and V = \tfrac{1}{2} \cdot \tfrac{4}{3}\pi (5)^3 = \tfrac{250\pi}{3}.
A tube of height 2: a cylinder of radius 3 with a cylinder of radius 1 removed from its middle. V = \pi (3)^2 (2) - \pi (1)^2 (2), which is 18\pi - 2\pi = 16\pi.
A torus, shaped like a ring doughnut or an inflated inner tube. Its center circle has radius 3, the circle the disk’s center travels, and its hole has radius 3 - 1 = 2, its outer edge radius 3 + 1 = 4. Every slice across the axis between y = -1 and y = 1 is a ring.
A rectangle 6 by 2 with a side of length 2 on the axis, for example the rectangle with vertices (0, 0), (6, 0), (6, 2) and (0, 2) spun about the y-axis.
A right triangle with legs 9 and 2 and the leg of length 9 on the axis, for example the triangle with vertices (0, 0), (2, 0) and (0, 9) spun about the y-axis. The apex is (0, 9).
The rectangle with vertices (0, 0), (6, 0), (6, 2) and (0, 2), with a quarter disk of radius 2 on each end (centered at (0, 0) on the left and (6, 0) on the right), spun about the x-axis. The two hemispheres together make a sphere of radius 2, so V = \pi (2)^2 (6) + \tfrac{4}{3}\pi (2)^3, which is 24\pi + \tfrac{32\pi}{3} = \tfrac{104\pi}{3}.
A rectangle 10 long and 1 wide, parallel to the axis and 4 from it, for example the one with vertices (0, 4), (10, 4), (10, 5) and (0, 5) spun about the x-axis. V = \pi (5)^2 (10) - \pi (4)^2 (10), which is 250\pi - 160\pi = 90\pi.
192\pi. The side on the x-axis has length 3, so it is the height, and the radius is 8: \pi (8)^2 (3) = 192\pi.
About the x-axis: a cone with radius 12 and height 5, \tfrac{1}{3}\pi (12)^2 (5) = 240\pi. About the y-axis: radius 5 and height 12, \tfrac{1}{3}\pi (5)^2 (12) = 100\pi.
108\pi. The rectangle sweeps a cylinder, \pi (3)^2 (10) = 90\pi, and the quarter disk sweeps a hemisphere of radius 3 on top, \tfrac{2}{3}\pi (3)^3 = 18\pi. The solid is shaped like a grain silo.
54\pi. As in Example 4, the solid is a cylinder of radius 3 and height 9 with a cone of the same radius and height carved out: 81\pi - 27\pi = 54\pi.
81\pi. The base lies on the x-axis, the vertical side sweeps a disk of radius 9, and the slanted side sweeps a cone with apex at the origin. So it is a cone with radius 9 and height 3: \tfrac{1}{3}\pi (9)^2 (3) = 81\pi.
56\pi. A frustum. The slanted side is x = 4 - \tfrac{1}{3}y, which meets the axis at y = 12. The whole cone is \tfrac{1}{3}\pi (4)^2 (12) = 64\pi, and the cone cut off above y = 6 has radius 2 and height 6, \tfrac{1}{3}\pi (2)^2 (6) = 8\pi. So 64\pi - 8\pi.
20\pi. A flat round base, \pi (4)^2 (1) = 16\pi, with a post on top, \pi (1)^2 (4) = 4\pi.
About x = 1: the left side is on the axis, so the solid is a cylinder with radius 2 and height 2, \pi (2)^2 (2) = 8\pi. About x = -1: the sides are 2 and 4 from the axis, so the solid is a tube, \pi (4)^2 (2) - \pi (2)^2 (2), which is 32\pi - 8\pi = 24\pi.
A disk of radius \displaystyle \sqrt{36 - 16} = \sqrt{20} = 2\sqrt{5}, with area 20\pi. The region touches the axis at that height, so the slice has no hole.
A ring with inner radius 2 and outer radius 3, with area \pi (3)^2 - \pi (2)^2 = 5\pi. The slanted side is at x = \tfrac{1}{3}y, which is 2 when y = 6.
The triangle together with its mirror image across the axis: an isosceles triangle with base 6 and height 5, with area \tfrac{1}{2}(6)(5) = 15.
Turning about the axis moves each point round a circle inside the plane through it perpendicular to the axis. So the slice in that plane is swept only by the points of the region at that level, and each sweeps a full circle centered on the axis. The circles from one piece of the region fill a ring, or a disk if the piece reaches the axis. A square has corners, which are farther from its center than the middles of its sides, and no union of whole circles about one center has that shape.
About leg a, the height is a and the radius is b, so V_a = \tfrac{1}{3}\pi b^2 a. About leg b, V_b = \tfrac{1}{3}\pi a^2 b. Then \tfrac{b}{a} V_b = \tfrac{1}{3}\pi a b^2 = V_a. Spinning about the shorter leg gives the larger solid, because the longer leg becomes the radius, and the radius is squared.
The part from x = 0 to x = 2 sweeps a cylinder of radius 2 and height 3. The part from x = -1 to x = 0 sweeps a cylinder of radius 1 and the same height, which lies inside the first. So the solid is the cylinder of radius 2, with V = \pi (2)^2 (3) = 12\pi. Adding 12\pi + 3\pi = 15\pi counts the inner cylinder twice.
No. Every slice of a solid of revolution across its axis is a disk or rings, but every slice of a cube by a plane through its inside is a polygon, so no line can serve as an axis. Another way to see it: a solid of revolution is carried onto itself by a turn of any angle about its axis, while a cube is carried onto itself only by turns of a quarter, a third or a half of a full turn, or whole multiples of those, about suitable lines.