Study Guide

# Surface Area and Volume - Volume

## Volume

We know you're tired of surface area. You should be. We've been harping on it for the past four chapters, but we've only scraped the surface (area?) of the third dimension.

We took a look at 3D shapes, but found areas and circumferences and heights, all two-dimensional concepts that we already know. There's a whole other aspect to the third dimension that's been hidden from us. Like we've been living in the Matrix this whole time.

Don't worry. We're about to unplug from it. Wave goodbye to surface areas and distances, and say hello to volume.

Unlike Pantene commercials and Metallica concerts, volume isn't about big hair or torn eardrums. Volume is the amount of space that a figure encloses. It's about units cubed instead of squared. It's about thinking inside the box. Or boxes. Instead of calculating how much wrapping paper it takes to cover the outside of a gift box, we now have to think about what's inside it. A sweater vest? Three thousand jellybeans? A Toyota Prius? It all depends on volume.

• ### Volume of Prisms

For prisms, the fastest way to find the volume is to multiply the area of the base (whatever it is) times the height.

V = Bh

That's all volume is: base times height

### Sample Problem

What is the volume of this prism? Step 1: Find the area of the base. We're talking prisms here, so the bases are always the two parallel congruent shapes. This time around, our base shape is a triangle, so we know that the area equals half the triangle's base times the height. We only know the height of the triangle, though. Hmm, better back up a step.

Step 0: We know the height is 9 centimeters and the hypotenuse is 14 centimeters, but what about the base? Pythagorean Theorem, here we come.

a2 + b2 = c2
(9 cm)2 + b2 = (14 cm)2
b ≈ 10.72 cm

Step 1: Find the area of the base...take two.

B = ½bh
B = ½(10.72 cm)(9 cm)
B ≈ 48.24 cm2

Step 2: Find the volume. In this case, the height is the length of the prism, not the height of the triangle.

V = Bh
V = (48.24 cm2)(12 cm)

Prepare for centimeters cubed.

V ≈ 579 cm3

Awww, yeah.

If we're dealing with a cube, by the way, things get even simpler. All three dimensions of a cube are identical (length, width, and depth). So if we call our cube's side length s, our volume formula becomes:

V = Bh
V
= (s2)s
V
= s3

Yep, a cube's volume is just the side length cubed. Oh, so that's why raising something to the third power is called "cubing."

To mix up things a little, here's a video on the surface area and volume of a cube:

• ### Volume of Cylinders

The volume formula works not only for prisms, but for cylinders, too. The only difference is that the base is circular instead of triangular or rectangular or whatevular. If that's true (which it totes is), we can substitute the area of the base B for the area of a circle, πr2.

V = Bh = πr2h

### Sample Problem

What is the volume of this cylinder? V = πr2h
V = π(6 in)2(18 in)
V = 648π
V ≈ 2035.8 in2

Voluminously victorious.

Up until now, we've been dealing with right prisms and right cylinders. What about their oblique buddies? They probably feel a little left out. Don't worry, obliques. Just because you aren't right doesn't mean you're wrong. We can calculate your volumes just as easily as those goody-goody two-shoes right cylinders, thanks to an Italian mathematician named Signior Cavalieri.

The Life and Legend of Signior Bonaventura Cavalieri Bonaventura (but he goes by Bonnie) Cavalieri was born in Italy over 400 years ago as an oblique cylinder. All throughout school, his right cylinder classmates teased him for being different, so he tried whatever he could to straighten out his obliqueness and be like everyone else. Of course, that didn't work, so Cavalieri devoted his life to proving that he was made of the same stuff as any other right cylinder.

One day while making a sandwich, his knife slipped and he cut himself in half. (Don't worry. He's a cylinder. He just super-glued himself back together.) He realized that it didn't matter how the circles were arranged in a cylinder. All that mattered was that the circles were all the same.

Cavalieri named a principle (not a principal) after himself that said that as long as two cylinders have the same height and cross-sectional area at every level, they had the same volume. He proved it using two stacks of coins (after gluing himself back together, of course). You show 'em, Cavalieri. See? If both stacks have the same number of quarters, they've both got the same volume. Even if one of them is all wonky-looking.

So what did Signior Cavalieri's story teach us?

1. It doesn't matter what you look like on the outside. (Unless you're a Hollywood actor.)

2. If the height and radii for two cylinders are the same, their volumes are the same regardless of their rightness or obliqueness.

As tribute to Cavalieri and the difficult life he faced as an oblique cylinder, we'll calculate the volume of one.

### Sample Problem

What is the volume of this cylinder? Thanks to Cavalieri, we know the volume of this bad boy is just the area of the base times the height, no matter how oblique it looks.

Since we only have the slant height and we want the actual height, it's pretty clear what we need to do. Especially with that right angle staring us in the face.

a2 + b2 = c2
(9 m)2 + b2 = (15 m)2
b = 12 m

With the height, we can calculate the volume.

V = πr2h
V = π(4 m)2(12 m)
V = 192π
V ≈ 603.2 m3

Cavalieri would be proud.

• ### Volume of Pyramids

Although we might wonder how much tinfoil it takes to cover the Egyptian pyramids, finding out how many mummies fit inside it would be much more interesting. Provided they don't form an undead army and attack you, that is. If we have a pyramid with the same base (B) and height (h) as a prism, we can try and deduce its volume. The empty space in front and back of the pyramid can form another pyramid, and same with the empty space on either side of the pyramid. As it turns out, the prism can hold the volume of exactly three pyramids. Trippy, right? If you have trust issues with us, fill these prisms and pyramids with sand or water or something and see for yourself. We don't lie (that much). You should end up with this formula for the volume of a pyramid. Using this formula, we can calculate how much space we have inside the Egyptian pyramids.

### Sample Problem

If an Egyptian pyramid has a square base with an edge length of 700 feet and a height of 450 feet, what's the pyramid's volume? V = ⅓(700 ft)2(450)

V = 73,500,000 ft3

Or seven times as much in cubic dog-feet.

• ### Volume of Cones

The same formula we used for pyramids (V = ⅓Bh) applies to cones, only the base isn't a polygon anymore. It's a circle. Formula-wise, that means we replace B with πr2. That's it. ### Sample Problem

Along with a diet and strict workout regimen, your Aunt Bertha decides that any time she eats ice cream, she'll pack it into the cone, but won't have anything above it. The height of the cone is 10 centimeters and the radius is 3 centimeters. If 100 cubic centimeters of ice cream is 200 Calories and the cone itself is 50 Calories, how many Calories are in one of Aunt Bertha's ice cream cones? First, we can calculate the volume of the cone using our formula.  V ≈ 94.2 cm3

That's how much ice cream Aunt Bertha packs into the cone. To calculate how many Calories she eats per cone, we need to convert the volume to Calories. Don't forget to add the 50 Calories for the waffle cone itself. Calories = 50 Calories + 188.4 Calories
Calories = 238.4 Calories

Aunt Bertha eats 238.4 Calories per ice cream cone. She can treat herself to one every now and then. She'll work it off on the elliptical at the gym, anyway.

Bonnie Cavalieri worked to bring justice to the oblique cylinders and prisms of the world, but he succeeded in doing so much more. Oblique cones and pyramids also share in the glory of his principle, and today Cavalieri is known as the Knight in Slanting Armor, a hero for all oblique solids.

Except not really. His principle does apply to cones and pyramids just the same, though.

### Sample Problem

What is the volume of this oblique cone? Calculating the volume shouldn't be a problem. We just have to do a little dance, make a little love, and trig it up tonight.

First thing we have to do is find the height of the cone. We know the hypotenuse and the angle opposite the height. We don't need any other sines to tell us what trig function to use.  h = (19 in) × sin(56°)

People might think dogs are man's best friend, but they're wrong. Calculators are.

h ≈ 15.75 in

Now we can use the same formula to find the volume of the cone.  V ≈ 808.2 in3

Danger, Will Robinson! Make sure you remember the difference between surface area and volume. While we can apply Cavalieri's principle to oblique cylinders and pyramids for volume, we can't do the same with surface area.

That's all folks!

• ### Volume of Spheres

Everyone knows the moon isn't really made of cheese. But what if it were? Picture it. Sure, we'd have a bit of an issue with the ocean tides, but think of all that cheese. We could end world hunger (assuming no one was lactose-intolerant or vegan).

How much cheese would it even be? Well, we can find out using the volume of a sphere.

To find the volume of a sphere, we can chop it up into an infinite number of pyramids. It'll take a while (more like forever), so let's not and say we did. But if we're going to say we did, we better make it convincing. If we make the base of our pyramids the surface of the sphere and the vertex the very center, our radius ends up being the height of the pyramid. If we added up all these pyramids, we'd end up with an equation that looks like this: Since our heights all equal the radius, we can replace h with r and factor it out. Now we've got a hive of B's. All the bases added together equal the surface of the sphere, and we already have the formula for the surface area. Simplifying it slightly, we have our volume formula. We have our formula, so let's get down to business...to defeat the Huns.

### Sample Problem

What is the volume of this sphere if its radius is 4 inches long? Its radius is 4 inches and luckily, that's all we need to know.  V ≈ 268.1 in3

Piece of cake, right? Don't worry. We're hip and with-it. We can switch it up, yo.

### Sample Problem

What is the diameter of a sphere whose volume is 11,494 cubic feet? Since we're looking for the radius, it's probably best if we isolate for that first. Now we can plug in the volume and solve for that missing radius. Make sure we cube root that mess and not square root.

r ≈ 14 ft

We can find volume using the radius and vice versa. As bountiful (or bouncy-ful) as balls might be, there are other solids we should take into account.

Hemispheres, for instance. Since a hemisphere is exactly half of a sphere, its volume should be exactly half the volume of a sphere. That makes exactly 100% sense.

### Sample Problem

What's the volume of this hemisphere? We've got half a sphere on our hands, so it stands to reason that we wanna cut our volume formula in half.  Good deal. Now let's plug in that radius and get cracking. V ≈ 1526.8 cm3

Our triumph speaks volumes.

• ### Volume and Density

Volume is a gateway to a lot of other concepts like density, which is a fancy word for the ratio between mass and volume. If something has a high density, then it weighs a lot even if the volume is really small.

A tub of water and a tub of chocolate take up the same amount of space (volume), but the chocolate would weigh more than the water (mass) because of density. It would also be a lot tastier than water, but that has nothing to do with density.

Here's a handy formula to find an object's density, where d is density, M is mass, and V is our old buddy, volume:

d = MV

Pretty simple, right? Just divide the mass by the volume. We can also rearrange it to find mass or volume when we know the other two:

M = dV
V
= Md

### Sample Problem

We have a rectangular carton filled with strawberry Jell-O (after all, strawberry is the best flavor). If strawberry Jell-O weighs 5 pounds per cubic foot, how much does all the Jell-O weigh? Here's the game plan: We'll find the volume first, multiply by the density, and find the mass that way. Ready? Break!

V = Bh = l × w × h
V = (0.7 ft)(0.8 ft)(1 ft)
V = 0.56 ft3

We have 0.56 ft3 of strawberry Jell-O in the carton. To find the mass, we need to multiply the volume by the density. If we do that, we can see that the cubic feet will cancel each other out and we'll be left with pounds.

M = dV
M = (5 lbs/ft3)(0.56 ft3)
M = 2.8 lbs

Our carton of strawberry Jell-O weighs 2.8 lbs. Bon appétit.

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